## Gravitation

### 7.1 INTRODUCTION

Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything thrown up falls down towards the earth, going uphill is lot more tiring than going downhill, raindrops from the clouds above fall towards the earth and there are many other such phenomena. Historically it was the Italian Physicist Galileo (1564-1642) who recognised the fact that all bodies, irrespective of their masses, are accelerated towards the earth with a constant acceleration. It is said that he made a public demonstration of this fact. To find the truth, he certainly did experiments with bodies rolling down inclined planes and arrived at a value of the acceleration due to gravity which is close to the more accurate value obtained later.

A seemingly unrelated phenomenon, observation of stars, planets and their motion has been the subject of attention in many countries since the earliest of times. Observations since early times recognised stars which appeared in the sky with positions unchanged year after year. The more interesting objects are the planets which seem to have regular motions against the background of stars. The earliest recorded model for planetary motions proposed by Ptolemy about 2000 years ago was a ‘geocentric’ model in which all celestial objects, stars, the sun and the planets, all revolved around the earth. The only motion that was thought to be possible for celestial objects was motion in a circle. Complicated schemes of motion were put forward by Ptolemy in order to describe the observed motion of the planets. The planets were described as moving in circles with the centre of the circles themselves moving in larger circles. Similar theories were also advanced by Indian astronomers some 400 years later. However a more elegant model in which the Sun was the centre around which the planets revolved - the ‘heliocentric’ model - was already mentioned by Aryabhatta ( $5^{\text {th }}$ century A.D.) in his treatise. A thousand years later, a Polish monk named Nicolas Copernicus (1473-1543) proposed a definitive model in which the planets moved in circles around a fixed central sun. His theory was discredited by the church, but notable amongst its supporters was Galileo who had to face prosecution from the state for his beliefs.

It was around the same time as Galileo, a nobleman called Tycho Brahe (1546-1601) hailing from Denmark, spent his entire lifetime recording observations of the planets with the naked eye. His compiled data were analysed later by his assistant Johannes Kepler (15711640). He could extract from the data three elegant laws that now go by the name of Kepler’s laws. These laws were known to Newton and enabled him to make a great scientific leap in proposing his universal law of gravitation.

### 7.2 KEPLER’S LAWS

The three laws of Kepler can be stated as follows:

**Law of orbits :**All planets move in elliptical orbits with the Sun situated at one of the foci (Fig 7.1a) of the ellipse (Fig. 7.1a). This law was a deviation from the Copernican model which allowed only circular orbits. The ellipse, of which the circle is a special case, is a closed curve which can be drawn very simply as follows.

Fig. 7.1(a) An ellipse traced out by a planet around the sun. The closest point is P and the farthest point is A, P is called the perihelion and A the aphelion. The semimajor axis is half the distance AP

Fig. 7.1(b) Drawing an ellipse. A string has its ends fixed at F1 and F2 . The tip of a pencil holds the string taut and is moved around

Select two points $\mathrm{F}_1$ and $\mathrm{F}_2$. Take a length of a string and fix its ends at $F_1$ and $F_2$ by pins. With the tip of a pencil stretch the string taut and then draw a curve by moving the pencil keeping the string taut throughout.(Fig. 7.1(b)) The closed curve you get is called an ellipse. Clearly for any point $\mathrm{T}$ on the ellipse, the sum of the distances from $\mathrm{F}_1$ and $\mathrm{F}_2$ is a constant. $\mathrm{F}_1, \mathrm{~F}_2$ are called the focii. Join the points $\mathrm{F}_1$ and $\mathrm{F}_2$ and extend the line to intersect the ellipse at points $\mathrm{P}$ and $\mathrm{A}$ as shown in Fig. 7.1(b). The midpoint of the line PA is the centre of the ellipse $\mathrm{O}$ and the length $\mathrm{PO}=$ AO is called the semi-major axis of the ellipse. For a circle, the two focii merge into one and the semi-major axis becomes the radius of the circle.

**2. Law of areas :** The line that joins any planet to the sun sweeps equal areas in equal intervals of time (Fig. 7.2). This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.

Fig. 7.2 The planet P moves around the sun in an elliptical orbit. The shaded area is the area ∆A swept out in a small interval of time ∆t.

**3.Law of periods :** The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

Table 7.1 gives the approximate time periods of revolution of eight* planets around the sun along with values of their semi-major axes.

**Table 7.1**

Data from measurement of planetary motions given below confirm Kepler’s Law of Periods

$$ \begin{aligned} & (a \equiv \text{Semi-major axis in units of } 10^{10} \mathrm{~m}. \\ & T \equiv \text{Time period of revolution of the planet in years }(y). \\ & Q \equiv \text{The quotient } ( T^{2} / a^{3})\\ & \text{in units of } 10^{-34} \mathrm{y}^{2} \mathrm{~m}^{-3}.) \end{aligned} $$

Planet | $\mathbf{a}$ | $\mathbf{T}$ | $\mathbf{Q}$ |
---|---|---|---|

Mercury | 5.79 | 0.24 | 2.95 |

Venus | 10.8 | 0.615 | 3.00 |

Earth | 15.0 | 1 | 2.96 |

Mars | 22.8 | 1.88 | 2.98 |

Jupiter | 77.8 | 11.9 | 3.01 |

Saturn | 143 | 29.5 | 2.98 |

Uranus | 287 | 84 | 2.98 |

Neptune | 450 | 165 | 2.99 |

The law of areas can be understood as a consequence of conservation of angular momentum whch is valid for any central force. A central force is such that the force on the planet is along the vector joining the Sun and the planet. Let the Sun be at the origin and let the position and momentum of the planet be denoted by $\mathbf{r}$ and $\mathbf{p}$ respectively. Then the area swept out by the planet of mass $\mathrm{m}$ in time interval $\Delta t$ is (Fig. 7.2) $\Delta \mathbf{A}$ given by

$$ \begin{equation*} \Delta \mathbf{A}=1 / 2(\mathbf{r} \times \mathbf{v} \Delta t) \tag{7.1} \end{equation*} $$

Hence

$$ \Delta \mathbf{A} / \Delta \mathrm{t}=1 / 2(\mathbf{r} \times \mathbf{p}) / \mathrm{m},(\text { since } \mathbf{v}=\mathbf{p} / \mathrm{m}) $$ $$ \begin{equation*} =\mathrm{L} /(2 \mathrm{~m}) \tag{7.2} \end{equation*} $$

where $\mathbf{v}$ is the velocity, $\mathbf{L}$ is the angular momentum equal to $(\mathbf{r} \times \mathbf{p})$. For a central force, which is directed along $\mathbf{r}, \mathbf{L}$ is a constant as the planet goes around. Hence, $\Delta \mathbf{A} / \Delta t$ is a constant according to the last equation. This is the law of areas. Gravitation is a central force and hence the law of areas follows.

**Example 7.1** Let the speed of the planet at the perihelion $P$ in Fig. 7.1(a) be $V_P$ and the Sun-planet distance SP be $r_P$. Relate $\{r_P, V_P\}$ to the corresponding quantities at the aphelion $\{r_A, V_A\}$. Will the planet take equal times to traverse $B A C$ and $C P B$ ?

**Answer** The magnitude of the angular momentum at $P$ is $L_p=m_p r_p V_p$, since inspection tells us that $\mathbf{r}_p$ and $\mathbf{v}_p$ are mutually perpendicular. Similarly, $L_A=m_p r_A V_A$. From angular momentum conservation

$$ m_{p} r_{p} v_{p}=m_{p} r_{A} v_{A} $$

or $\frac{v_{p}}{v_{A}}=\frac{r_{A}}{r_{p}}$

Since $r_{A}>r_{p}, V_{p}>v_{A}$.

The area $S B A C$ bounded by the ellipse and the radius vectors $S B$ and $S C$ is larger than $\mathrm{SBPC}$ in Fig. 7.1. From Kepler’s second law, equal areas are swept in equal times. Hence the planet will take a longer time to traverse $B A C$ than $C P B$.

### 7.3 UNIVERSAL LAW OF GRAVITATION

Legend has it that observing an apple falling from a tree, Newton was inspired to arrive at an universal law of gravitation that led to an explanation of terrestrial gravitation as well as of Kepler’s laws. Newton’s reasoning was that the moon revolving in an orbit of radius $R_{m}$ was subject to a centripetal acceleration due to earth’s gravity of magnitude

$$ \begin{equation*} a_{m}=\frac{V^{2}}{R_{m}}=\frac{4 \pi^{2} R_{m}}{T^{2}} \tag{7.3} \end{equation*} $$

where $V$ is the speed of the moon related to the time period $T$ by the relation $V=2 \pi R_{m} / T$. The time period $T$ is about 27.3 days and $R_{m}$ was already known then to be about $3.84 \quad 10^{8} \mathrm{~m}$. If we substitute these numbers in Eq. (7.3), we get a value of $a_{m}$ much smaller than the value of acceleration due to gravity $g$ on the surface of the earth, arising also due to earth’s gravitational attraction. This clearly shows that the force due to earth’s gravity decreases with distance. If one assumes that the gravitational force due to the earth decreases in proportion to the inverse square of the distance from the centre of the earth, we will have $a_{m} \alpha R_{m}^{-2} ; g \alpha R_{E}^{-2}$ and we get

$$ \begin{equation*} \frac{g}{a_{m}}=\frac{R_{m}^{2}}{R_{E}^{2}} \simeq 3600 \tag{7.4} \end{equation*} $$

in agreement with a value of $g \simeq 9.8 \mathrm{~m} \mathrm{~s}^{-2}$ and the value of $a_{\mathrm{m}}$ from Eq. (7.3). These observations led Newton to propose the following Universal Law of Gravitation :

Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The quotation is essentially from Newton’s famous treatise called ‘Mathematical Principles of Natural Philosophy’ (Principia for short).

Stated Mathematically, Newton’s gravitation law reads : The force $\mathbf{F}$ on a point mass $m_{2}$ due to another point mass $m_{1}$ has the magnitude

$$ \begin{equation*} |\mathbf{F}|=G \frac{m_{1} m_{2}}{r^{2}} \tag{7.5} \end{equation*} $$

Equation (7.5) can be expressed in vector form as

$$ \begin{aligned} \mathbf{F} & =G \frac{m_{1} m_{2}}{r^{2}}(-\hat{\mathbf{r}})=-G \frac{m_{1} m_{2}}{r^{2}} \hat{\mathbf{r}} \\ \\ & =-G \frac{m_{1} m_{2}}{|\mathbf{r}|^{3}} \hat{\mathbf{r}} \end{aligned} $$

where $\mathrm{G}$ is the universal gravitational constant, $\hat{\mathbf{r}}$ is the unit vector from $m_1$ to $m_2$ and $\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1$ as shown in Fig. 7.3.

Fig. 7.3 Gravitational force on m1 due to m2 is along r where the vector r is (r2 – r1 ).

Gravitational force on $m_1$ due to $m_2$ is along $\mathbf{r}$ where the vector $\mathbf{r}$ is ($\mathbf{r}_2-\mathbf{r}_1$).
The gravitational force is attractive, i.e., the force $\mathbf{F}$ is along $-\mathbf{r}$. The force on point mass $m_1$ due to $m_2$ is of course $-\mathbf{F}$ by Newton’s third law. Thus, the gravitational force **F**_{12} on the body 1 due to 2 and **F**_{21} on the body 2 due to 1 are related as

**F**_{12}=-**F**_{21}.

Before we can apply Eq. (7.5) to objects under consideration, we have to be careful since the law refers to point masses whereas we deal with extended objects which have finite size. If we have a collection of point masses, the force on any one of them is the vector sum of the gravitational forces exerted by the other point masses as shown in Fig 7.4.

Fig. 7.4 Gravitational force on point mass m1 is the vector sum of the gravitational forces exerted by m2 , m3 and m4 .

The total force on $m_1$ is

$$ F_1=\frac{G m_2 m_1}{r_{21}^2} \hat{r_{21}}+\frac{G m_3 m_1}{r_{31}^2} \hat{r_{31}}+\frac{G m_4 m_1}{r_{41}^2} \hat{r_{41}} $$

**Example 7.2** Three equal masses of $m \mathrm{~kg}$ each are fixed at the vertices of an equilateral triangle $\mathrm{ABC}$.

(a) What is the force acting on a mass $2 m$ placed at the centroid $\mathrm{G}$ of the triangle?

(b) What is the force if the mass at the vertex $\mathrm{A}$ is doubled?

Take $\mathrm{AG}=\mathrm{BG}=\mathrm{CG}=1 \mathrm{~m}$ (see Fig. 7.5)

**Answer** (a) The angle between GC and the positive $x$-axis is $30^{\circ}$ and so is the angle between GB and the negative $x$-axis. The individual forces in vector notation are

Fig. 7.5 Three equal masses are placed at the three vertices of the ∆ ABC. A mass 2m is placed at the centroid G.

$$ \begin{aligned} & \mathbf{F_\mathrm{GA}}=\frac{G m(2 m)}{1} \hat{\mathbf{j}} \\ & \mathbf{F_\mathrm{GB}}=\frac{G m(2 m)}{1}\left(\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right) \\ & \mathbf{F_\mathrm{GC}}=\frac{G m(2 m)}{1}\left(+\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right) \end{aligned} $$

From the principle of superposition and the law of vector addition, the resultant gravitational force $\mathbf{F}_{\mathrm{R}}$ on $(2 m)$ is

$$ \begin{aligned} & \mathbf{F_\mathrm{R}}= \mathbf{F_\mathrm{GA}}+\mathbf{F_\mathrm{GB}}+\mathbf{F_\mathrm{GC}} \\ & \mathbf{F_\mathrm{R}}=2 G m^{2} \hat{\mathbf{j}}+2 G m^{2}\left(-\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right) \\ &+2 G m^{2}\left(\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right)=0 \end{aligned} $$

Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero.

(b) Now if the mass at vertex A is doubled then

$$ \begin{aligned} & \mathrm{F_{G A}^{\prime}}=\frac{\mathrm{G} 2 m \cdot 2 m}{1} \hat{\mathrm{j}}=4 \mathrm{Gm}^{2} \hat{\mathrm{j}} \\ & \mathrm{F_{G B}^{\prime}}=\mathrm{F_G B} \text { and } \mathrm{F_G C}^{\prime}=\mathrm{F_G C} \\ & \mathrm{~F_{R}^{\prime}}=\mathrm{F_G A}^{\prime}+\mathrm{F_G B}^{\prime}+\mathrm{F_G C}^{\prime} \\ & \mathrm{F_{\mathrm{R}}^{\prime}}=2 G m^{2} \hat{\mathrm{j}} \end{aligned} $$

For the gravitational force between an extended object (like the earth) and a point mass, Eq. (7.5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in the same direction. We have to add up these forces vectorially for all the point masses in the extended object to get the total force. This is easily done using calculus. For two special cases, a simple law results when you do that :

(1) The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell.

Qualitatively this can be understood as follows: Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components prependicular to this line cancel out when summing over all regions of the shell leaving only a resultant force along the line joining the point to the centre. The magnitude of this force works out to be as stated above.

(2) The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero.

Gualitatively, we can again understand this result. Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.

### 7.4 THE GRAVITATIONAL CONSTANT

The value of the gravitational constant $G$ entering the Universal law of gravitation can be determined experimentally and this was first done by English scientist Henry Cavendish in 1798. The apparatus used by him is schematically shown in Fig.7.6

Fig. 7.6 Schematic drawing of Cavendish’s experiment. S1 and S2 are large spheres which are kept on either side (shown shades) of the masses at A and B. When the big spheres are taken to the other side of the masses (shown by dotted circles), the bar AB rotates a little since the torque reverses direction. The angle of rotation can be measured experimentally.

The bar $\mathrm{AB}$ has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire. Two large lead spheres are brought close to the small ones but on opposite sides as shown. The big spheres attract the nearby small ones by equal and opposite force as shown. There is no net force on the bar but only a torque which is clearly equal to $\mathrm{F}$ times the length of the bar,where $\mathrm{F}$ is the force of attraction between a big sphere and its neighbouring small sphere. Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque. If $\theta$ is the angle of twist of the suspended wire, the restoring torque is proportional to $\theta$, equal to $\tau \theta$. Where $\tau$ is the restoring couple per unit angle of twist. $\tau$ can be measured independently e.g. by applying a known torque and measuring the angle of twist. The gravitational force between the spherical balls is the same as if their masses are concentrated at their centres. Thus if $d$ is the separation between the centres of the big and its neighbouring small ball, $\mathrm{M}$ and $\mathrm{m}$ their masses, the gravitational force between the big sphere and its neighouring small ball is.

$$ \begin{equation*} F=G \frac{M m}{d^{2}} \tag{7.6} \end{equation*} $$

If $L$ is the length of the bar $A B$, then the torque arising out of $F$ is $F$ multiplied by L. At equilibrium, this is equal to the restoring torque and hence

$$ \begin{equation*} G \frac{M m}{d^{2}} L=\tau \theta \tag{7.7} \end{equation*} $$

Observation of $\theta$ thus enables one to calculate $G$ from this equation.

Since Cavendish’s experiment, the measurement of $G$ has been refined and the currently accepted value is

$$ \begin{equation*} G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2} \tag{7.8} \end{equation*} $$

### 7.5 ACCELERATION DUE TO GRAVITY OF THE EARTH

The earth can be imagined to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the largest one at its surface. A point outside the earth is obviously outside all the shells. Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre according to the result stated in section 7.3. The total mass of all the shells combined is just the mass of the earth. Hence, at a point outside the earth, the gravitational force is just as if its entire mass of the earth is concentrated at its centre.

For a point inside the earth, the situation is different. This is illustrated in Fig. 7.7.

Fig. 7.7 The mass m is in a mine located at a depth d below the surface of the Earth of mass ME and radius RE. We treat the Earth to be spherically symmetric

Again consider the earth to be made up of concentric shells as before and a point mass $m$ situated at a distance $r$ from the centre. The point $\mathrm{P}$ lies outside the sphere of radius $r$. For the shells of radius greater than $r$, the point $\mathrm{P}$ lies inside. Hence according to result stated in the last section, they exert no gravitational force on mass $m$ kept at $\mathrm{P}$. The shells with radius $\leq r$ make up a sphere of radius $r$ for which the point $P$ lies on the surface. This smaller sphere therefore exerts a force on a mass $m$ at $\mathrm{P}$ as if its mass $M_{r}$ is concentrated at the centre. Thus the force on the mass $m$ at $\mathrm{P}$ has a magnitude

$$ \begin{equation*} F=\frac{G m\left(M_{\mathrm{r}}\right)}{r^{2}} \tag{7.9} \end{equation*} $$

We assume that the entire earth is of uniform density and hence its mass is $M_{\mathrm{E}}=\frac{4 \pi}{3} R_{E}^{3} \rho$ where $M_{\mathrm{E}}$ is the mass of the earth $R_{\mathrm{E}}$ is its radius and $\rho$ is the density. On the other hand the mass of the sphere $M_{r}$ of radius $r$ is $\frac{4 \pi}{3} \rho r^{3}$ and hence

$$ \begin{align*} F & =G m\left(\frac{4 p}{3} r\right) \frac{r^{3}}{r^{2}}=G m\left(\frac{M_{E}}{R_{E}^{3}}\right) \frac{r^{3}}{r^{2}} \\ & =\frac{G m M_{\mathrm{E}}}{R_{E}{ }^{3}} r \tag{7.10} \end{align*} $$

If the mass $m$ is situated on the surface of earth, then $r=R_{E}$ and the gravitational force on it is, from Eq. (7.10)

$$ \begin{equation*} F=G \frac{M_{E} m}{R_{E}^{2}} \tag{7.11} \end{equation*} $$

The acceleration experienced by the mass $\mathrm{m}$, which is usually denoted by the symbol $g$ is related to $\mathrm{F}$ by Newton’s $2^{\text {nd }}$ law by relation $F=m g$. Thus

$$ \begin{equation*} g=\frac{F}{m}=\frac{G M_{E}}{R_{E}^{2}} \tag{7.12} \end{equation*} $$

Acceleration $g$ is readily measurable. $R_{E}$ is a known quantity. The measurement of $G$ by Cavendish’s experiment (or otherwise), combined with knowledge of $g$ and $R_{E}$ enables one to estimate $M_{E}$ from Eq. (7.12). This is the reason why there is a popular statement regarding Cavendish : “Cavendish weighed the earth”.

### 7.6 ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH

Consider a point mass $m$ at a height $h$ above the surface of the earth as shown in Fig. 7.8(a). The radius of the earth is denoted by $R_{E}$. Since this point is outside the earth, Its distance from the centre of the earth is $\left(R_{E}+h\right)$. If $F(h)$ denoted the magnitude of the force on the point mass $m$, we get from Eq. (7.5) :

Fig. 7.8 (a) g at a height h above the surface of the earth

$$ \begin{equation*} F(h)=\frac{G M_{E} m}{\left(R_{E}+h\right)^{2}} \tag{7.13} \end{equation*} $$

The acceleration experienced by the point mass is $F(h) / m \equiv g(h)$ and we get

$$ \begin{equation*} g(h)=\frac{F(h)}{m}=\frac{G M_{E}}{\left(R_{E}+h\right)^{2}} . \tag{7.14} \end{equation*} $$

This is clearly less than the value of $g$ on the surface of earth : $g=\frac{G M_{E}}{R_{E}^{2}}$. For $h \ll R_{E}$, we can expand the RHS of Eq. (7.14) :

$$ g(h)=\frac{G M_{E}}{R_{E}^{2}\left(1+h / R_{E}\right)^{2}}=g\left(1+h / R_{E}\right)^{-2} $$

$\text{For}\frac{h}{R _{E}} \ll 1$,Using binomial expression

$$ \begin{equation*} g(h) \cong g \quad 1-\frac{2 h}{R _{E}} \tag{7.15} \end{equation*} $$

Equation (7.15) thus tells us that for small heights $h$ above the value of $g$ decreases by $a$ factor $\left(1-2 h / R_{E}\right)$.

Now, consider a point mass $m$ at a depth $d$ below the surface of the earth (Fig. 7.8(b)), so that its distance from the centre of the earth is $\left(R_{E}-d\right)$ as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius $\left(R_{E}-d\right)$ and a spherical shell of thickness $d$. The force on $m$ due to the outer shell of thickness $d$ is zero because the result quoted in the previous section. As far as the smaller sphere of radius ( $ R_{E}-d$ )is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If $M_{s}$ is the mass of the smaller sphere, then,

$$ \begin{equation*} M_{s} / M_{E}=\left(R_{E}-\mathrm{d}\right)^{3} / R_{E}^{3} \tag{7.16} \end{equation*} $$

Since mass of a sphere is proportional to be cube of its radius.

(b)

Fig. 7.8 (b) g at a depth d. In this case only the smaller sphere of radius (RE –d) contributes to g.

Thus the force on the point mass is

$$ \begin{equation*} F(d)=G M_{s} m /\left(R_{E}-d\right)^{2} \tag{7.17} \end{equation*} $$

Substituting for $\mathrm{M}_{\mathrm{s}}$ from above, we get

$$ \begin{equation*} F(d)=G M_{E} m\left(R_{E}-d\right) / R_{E}^{3} \tag{7.18} \end{equation*} $$

and hence the acceleration due to gravity at a depth $d$,

$$ \begin{align*} & g(d)=\frac{F(d)}{m} \text { is } \\ & g(d)=\frac{F(d)}{m}=\frac{G M_{E}}{R_{E}^{3}}\left(R_{E}-d\right) \\ & =g \frac{R_{E}-d}{R_{E}}=g\left(1-d / R_{E}\right) \tag{7.19} \end{align*} $$

Thus, as we go down below earth’s surface, the acceleration due gravity decreases by a factor $\left(1-d / R_{E}\right)$. The remarkable thing about acceleration due to earth’s gravity is that it is maximum on its surface decreasing whether you go up or down.

### 7.7 GRAVITATIONAL POTENTIAL ENERGY

We had discussed earlier the notion of potential energy as being the energy stored in the body at its given position. If the position of the particle changes on account of forces acting on it, then the change in its potential energy is just the amount of work done on the body by the force. As we had discussed earlier, forces for which the work done is independent of the path are the conservative forces.

The force of gravity is a conservative force and we can calculate the potential energy of a body arising out of this force, called the gravitational potential energy. Consider points close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to $\mathrm{mg}$, directed towards the centre of the earth. If we consider a point at a height $h_{1}$ from the surface of the earth and another point vertically above it at a height $h_{2}$ from the surface, the work done in lifting the particle of mass $m$ from the first to the second position is denoted by $W_{12}$

$$ \begin{align*} W_{12} & =\text { Force } \times \text { displacement } \\ & =m g\left(h_{2}-h_{1}\right) \tag{7.20} \end{align*} $$

If we associate a potential energy $W(h)$ at a point at a height $h$ above the surface such that

$$ \begin{equation*} W(h)=m g h+W_{o} \tag{7.21} \end{equation*} $$

(where $W_{\mathrm{o}}=$ constant) ;

then it is clear that

$$ \begin{equation*} W_{12}=W\left(h_{2}\right)-W\left(h_{1}\right) \tag{7.22} \end{equation*} $$

The work done in moving the particle is just the difference of potential energy between its final and initial positions.Observe that the constant $W_{\text {o cancels out in Eq. (7.22). Setting } h}$ $=0$ in the last equation, we get $W(h=0)=W_{\text {o }}$ . $h=0$ means points on the surface of the earth. Thus, $W_{\mathrm{o}}$ is the potential energy on the surface of the earth. If we consider points at arbitrary distance from the surface of the earth, the result just derived is not valid since the assumption that the gravitational force $m g$ is a constant is no longer valid. However, from our discussion we know that a point outside the earth, the force of gravitation on a particle directed towards the centre of the earth is

$$ \begin{equation*} F=\frac{G M_{E} m}{r^{2}} \tag{7.23} \end{equation*} $$

where $M_{E}=$ mass of earth, $m=$ mass of the particle and $r$ its distance from the centre of the earth. If we now calculate the work done in lifting a particle from $r=r_{1}$ to $r=r_{2}\left(r_{2}>r_{1}\right)$ along a vertical path, we get instead of Eq. (7.20)

$$ \begin{align*} & W_{12}=\int_{r_{1}}^{r_{2}} \frac{G M m}{r^{2}} \mathrm{~d} r \ & =-G M_{\mathrm{E}} m \frac{1}{r_{2}}-\frac{1}{r_{1}} \tag{7.24} \end{align*} $$

In place of Eq. (7.21), we can thus associate a potential energy $W(r)$ at a distance $r$, such that

$$ \begin{equation*} W(r)=-\frac{G M_{\mathrm{E}} m}{r}+W_{1} \tag{7.25} \end{equation*} $$

valid for $r>R$,

so that once again $W_{12}=W\left(r_{2}\right)-W\left(r_{1}\right)$. Setting $r$ =infinity in the last equation, we get $\mathrm{W}(r=$ infinity $)=W_{1}$. Thus, $W_{1}$ is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (7.22) and (7.24). One conventionally sets $W_{1}$ equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point.

We have calculated the potential energy at a point of a particle due to gravitational forces on it due to the earth and it is proportional to the mass of the particle. The gravitational potential due to the gravitational force of the earth is defined as the potential energy of a particle of unit mass at that point. From the earlier discussion, we learn that the gravitational potential energy associated with two particles of masses $m_{1}$ and $m_{2}$ separated by distance by a distance $r$ is given by

$$ V=-\frac{G m_{1} m_{2}}{r} \text { (if we choose } V=0 \text { as } r \rightarrow \infty \text { ) } $$

It should be noted that an isolated system of particles will have the total potential energy that equals the sum of energies (given by the above equation) for all possible pairs of its constituent particles. This is an example of the application of the superposition principle.

**Example 7.3** Find the potential energy of a system of four particles placed at the vertices of a square of side $l$. Also obtain the potential at the centre of the square.

**Answer** Consider four masses each of mass $m$ at the corners of a square of side $l$; See Fig. 7.9. We have four mass pairs at distance $l$ and two diagonal pairs at distance $\sqrt{2} l$ Hence,

$$ \begin{aligned} & W(r)=-4 \frac{G m^{2}}{l}-2 \frac{G m^{2}}{\sqrt{2} l} \\ & =-\frac{2 G m^{2}}{l} 2+\frac{1}{\sqrt{2}}=-5.41 \frac{G m^{2}}{l} \end{aligned} $$

Fig. 7.9

The gravitational potential at the centre of the square $(r=\sqrt{2} l / 2)$ is

$$ U(r)=-4 \sqrt{2} \frac{\mathrm{Gm}}{l} . $$

### 7.8 ESCAPE SPEED

If a stone is thrown by hand, we see it falls back to the earth. Of course using machines we can shoot an object with much greater speeds and with greater and greater initial speed, the object scales higher and higher heights. A natural query that arises in our mind is the following: “can we throw an object with such high initial speeds that it does not fall back to the earth?’

The principle of conservation of energy helps us to answer this question. Suppose the object did reach infinity and that its speed there was $\mathrm{V}_{f}$. The energy of an object is the sum of potential and kinetic energy. As before $W_1$ denotes that gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is

$$ \begin{equation*} E(\infty)=W_{1}+\frac{m V_{f}^{2}}{2} \tag{7.26} \end{equation*} $$

If the object was thrown initially with a speed $V_{i}$ from a point at a distance $\left(h+R_{E}\right)$ from the centre of the earth $\left(R_{E}=\right.$ radius of the earth $)$, its energy initially was

$$ \begin{equation*} E\left(h+R_{E}\right)=\frac{1}{2} m V_{i}^{2}-\frac{G m M_{E}}{\left(h+R_{E}\right)}+W_{1} \tag{7.27} \end{equation*} $$

By the principle of energy conservation Eqs. (7.26) and (7.27) must be equal. Hence

$$ \begin{equation*} \frac{m V_{i}^{2}}{2}-\frac{G m M_{E}}{\left(h+R_{E}\right)}=\frac{m V_{f}^{2}}{2} \tag{7.28} \end{equation*} $$

The R.H.S. is a positive quantity with a minimum value zero hence so must be the L.H.S. Thus, an object can reach infinity as long as $\mathrm{V}_i$ is such that

$$ \begin{equation*} \frac{m V_{i}^{2}}{2}-\frac{G m M_{E}}{\left(h+R_{E}\right)} \geq 0 \tag{7.29} \end{equation*} $$

The minimum value of $\mathrm{V}_i$ corresponds to the case when the L.H.S. of Eq. (7.29) equals zero. Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) corresponds to

$$ \begin{equation*} \frac{1}{2} m(V_i^2)_{\min }=\frac{G m M_E}{h+R_E} \tag{7.30} \end{equation*} $$

If the object is thrown from the surface of the earth, $h=0$, and we get

$$ \begin{equation*} (V_i)_{\min }=\sqrt{\frac{2 G M_E}{R_E}} \tag{7.31} \end{equation*} $$

Using the relation $g=G M_{E} / R_{E}^{2}$, we get

$$ \begin{equation*} (V_i)_{\min }=\sqrt{2 g R_E} \tag{7.32} \end{equation*} $$

Using the value of $g$ and $R_{E}$, numerically $\left(V_{i}\right)_{\min } \approx 1.2 \mathrm{~km} / \mathrm{s}$. This is called the escape speed, sometimes loosely called the escape velocity.

Equation (7.32) applies equally well to an object thrown from the surface of the moon with $g$ replaced by the acceleration due to Moon’s gravity on its surface and $r_{E}$ replaced by the radius of the moon. Both are smaller than their values on earth and the escape speed for the moon turns out to be $2.3 \mathrm{~km} / \mathrm{s}$, about five times smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the surface of the moon having velocities larger than this will escape the gravitational pull of the moon.

**Example 7.4** Two uniform solid spheres of equal radii $R$, but mass $M$ and $4 M$ have a centre to centre separation $6 R$, as shown in Fig. 7.10. The two spheres are held fixed. A projectile of mass $m$ is projected from the surface of the sphere of mass $M$ directly towards the centre of the second sphere. Obtain an expression for the minimum speed $v$ of the projectile so that it reaches the surface of the second sphere.

**Answer** The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 7.10) is defined as the position where the two forces cancel each other exactly. If $\mathrm{ON}=r$, we have

$$ \begin{aligned} & \frac{G M m}{r^{2}}=\frac{4 G M m}{(6 R-r)^{2}} \\ & (6 R-r)^{2}=4 r^{2} \\ & 6 R-r= \pm 2 r \\ & r=2 R \text { or }-6 R . \end{aligned} $$

The neutral point $r=-6 R$ does not concern us in this example. Thus $\mathrm{ON}=r=2 R$. It is sufficient to project the particle with a speed which would enable it to reach $\mathrm{N}$. Thereafter, the greater gravitational pull of $4 M$ would suffice. The mechanical energy at the surface of $M$ is

$$ E_{i}=\frac{1}{2} m v^{2}-\frac{G M m}{R}-\frac{4 G M m}{5 R} . $$

At the neutral point $N$, the speed approaches zero. The mechanical energy at $\mathrm{N}$ is purely potential.

$$ E_{N}=-\frac{G M m}{2 R}-\frac{4 G M m}{4 R} $$

From the principle of conservation of mechanical energy

$$ \frac{1}{2} v^{2}-\frac{G M}{R}-\frac{4 G M}{5 R}=-\frac{G M}{2 R}-\frac{G M}{R} $$

or

$$ \begin{aligned} & v^{2}=\frac{2 G M}{R} \frac{4}{5}-\frac{1}{2} \\ & v=\frac{3 G M}{5 R}^{1 / 2} \end{aligned} $$

A point to note is that the speed of the projectile is zero at $\mathrm{N}$, but is nonzero when it strikes the heavier sphere $4 M$. The calculation of this speed is left as an exercise to the students.

### 7.9 EARTH SATELLITES

Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Since, 1957, advances in technology have enabled many countries including India to launch artificial earth satellites for practical use in fields like telecommunication, geophysics and meteorology.

We will consider a satellite in a circular orbit of a distance $\left(R_{E}+h\right)$ from the centre of the earth, where $R_{E}=$ radius of the earth. If $m$ is the mass of the satellite and $V$ its speed, the centripetal force required for this orbit is

$$ \begin{equation*} \mathrm{F}(\text { centripetal })=\frac{m V^{2}}{\left(R_{E}+h\right)} \tag{7.33} \end{equation*} $$

directed towards the centre. This centripetal force is provided by the gravitational force, which is

$$ \begin{equation*} \mathrm{F}(\text { gravitation })=\frac{G m M_{E}}{\left(R_{E}+h\right)^{2}} \tag{7.34} \end{equation*} $$

where $M_{E}$ is the mass of the earth.

Equating R.H.S of Eqs. (7.33) and (7.34) and cancelling out $m$, we get

$$ \begin{equation*} V^{2}=\frac{G M_{E}}{\left(R_{E}+h\right)} \tag{7.35} \end{equation*} $$

Thus $V$ decreases as $h$ increases. From equation (7.35), the speed $V$ for $h=0$ is

$$ \begin{equation*} V^{2}(h=0)=G M / R_{E}=g R_{E} \tag{7.36} \end{equation*} $$

where we have used the relation $g=G M / R_{E}{ }^{2}$. In every orbit, the satellite traverses a distance $2 \pi\left(R_{E}+h\right)$ with speed $V$. Its time period $T$ therefore is

$$ \begin{equation*} T=\frac{2 \pi\left(R_{E}+h\right)}{V}=\frac{2 \pi\left(R_{E}+h\right)^{3 / 2}}{\sqrt{G M_{E}}} \tag{7.37} \end{equation*} $$

on substitution of value of $V$ from Eq. (7.35).

Squaring both sides of Eq. (7.37), we get

$T^{2}=k\left(R_{E}+h\right)^{3} \quad\left(\right.$ where $\left.k=4 \pi^{2} / G M_{E}\right)$

which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a satellite very close to the surface of earth $h$ can be neglected in comparison to $R_{E}$ in Eq. (7.38). Hence, for such satellites, $T$ is $T_{0}$, where

$$ \begin{equation*} T_{0}=2 \pi \sqrt{R_{E} / g} \tag{7.39} \end{equation*} $$

If we substitute the numerical values $\mathrm{g} \simeq 9.8 \mathrm{~m} \mathrm{~s}^{-2}$ and $R_{E}=6400 \mathrm{~km}$. , we get

$$ T_{0}=2 \pi \sqrt{\frac{6.4 \times 10^{6}}{9.8}} \mathrm{~s} $$

Which is approximately 85 minutes.

Example 7.5 The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of $9.4 \times 10^{3} \mathrm{~km}$. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days?

Answer (i) We employ Eq. (7.38) with the sun’s mass replaced by the martian mass $M_{m}$

$$ \begin{aligned} & T^{2}=\frac{4 \pi^{2}}{G M_{m}} R^{3} \\ \\ & \mathrm{M}_m=\frac{4 \pi^{2}}{G} \frac{R^3}{T^2} \\ \\ & =\frac{4 \times(3.14)^2 \times(9.4)^3 \times 10^{18}}{6.67 \times 10^{-11} \times(459 \times 60)^2} \\ \\ & \mathrm{M}_m=\frac{4 \times(3.14)^2 \times(9.4)^3 \times 10^{18}}{6.67 \times(4.59 \times 6)^2 \times 10^{-5}} \\ \\ & =6.48 \times 10^{23} \mathrm{~kg} . \end{aligned} $$

(ii) Once again Kepler’s third law comes to our aid,

$$ \frac{T_{M}^{2}}{T_{E}^{2}}=\frac{R_{M S}^{3}}{R_{E S}^{3}} $$

where $R_{M S}$ is the mars -sun distance and $R_{E S}$ is the earth-sun distance.

$$ \therefore \quad T _{M}=(1.52)^{3 / 2} \times 365 $$

$$ =684 \text { day } $$

We note that the orbits of all planets except Mercury and Mars are very close to being circular. For example, the ratio of the semiminor to semi-major axis for our Earth is, $b / a=0.99986$.

**Example 7.6** Weighing the Earth: You are given the following data: $g=9.81 \mathrm{~ms}^{-2}$, $R_{E}=6.37 \times 10^{6} \mathrm{~m}$, the distance to the moon $R$ $=3.84 \times 10^{8} \mathrm{~m}$ and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth $M_{E}$ in two different ways.

**Answer** From Eq. (7.12) we have

$$ \begin{aligned} & M_{E}=\frac{g R_{E}^{2}}{G} \\ \\ & =\frac{9.81 \times\left(6.37 \times 10^{6}\right)^{2}}{6.67 \times 10^{-11}} \\ \\ & =5.97 \times 10^{24} \mathrm{~kg} . \end{aligned} $$

The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq. (7.38)]

$$ \begin{aligned} & T^{2}=\frac{4 \pi^{2} R^{3}}{G M_{E}} \\ \\ & M_{E}=\frac{4 \pi^{2} R^{3}}{G T^{2}} \\ \\ & =\frac{4 \times 3.14 \times 3.14 \times(3.84)^{3} \times 10^{24}}{6.67 \times 10^{-11} \times(27.3 \times 24 \times 60 \times 60)^{2}} \\ \\ & =6.02 \times 10^{24} \mathrm{~kg} \end{aligned} $$

Both methods yield almost the same answer. the difference between them being less than $1 %$.

**Example 7.7** Express the constant k of Eq. (7.38) in days and kilometres. Given $\mathrm{k}=10^{-13} \mathrm{~s}^{2} \mathrm{~m}^{-3}$. The moon is at a distance of $3.84 \times 10^{5} \mathrm{~km}$ from the earth. Obtain its time-period of revolution in days.

**Answer** Given

$$ \begin{aligned} & k=10^{-13} \mathrm{~s}^{2} \mathrm{~m}^{-3} \\ = & 10^{-13} \frac{1}{(24 \times 60 \times 60)^{2}} \mathrm{~d}^{2} \frac{1}{(1 / 1000)^{3} \mathrm{~km}^{3}} \\ = & 1.33 \times 10^{-14} \mathrm{~d}^{2} \mathrm{~km}^{-3} \end{aligned} $$

Using Eq. (7.38) and the given value of $k$, the time period of the moon is

$$ \begin{aligned} & T^{2}=\left(1.33 \times 10^{-14}\right)\left(3.84 \times 10^{5}\right)^{3} \\ & T=27.3 \mathrm{~d} \end{aligned} $$

Note that Eq. (7.38) also holds for elliptical orbits if we replace $\left(R_{E}+h\right)$ by the semi-major axis of the ellipse. The earth will then be at one of the foci of this ellipse.

### 7.10 ENERGY OF AN ORBITING SATELLITE

Using Eq. (7.35), the kinetic energy of the satellite in a circular orbit with speed $v$ is

$$ K . E=\frac{1}{2} m v^{2} $$

$$ \begin{equation*} =\frac{G m M _{E}}{2\left(R _{E}+h\right)} \tag{7.40} \end{equation*} $$

Considering gravitational potential energy at infinity to be zero, the potential energy at distance $(\mathrm{R}+\mathrm{h})$ from the centre of the earth is

$$ \begin{equation*} P . E=-\frac{G m M_{E}}{\left(R_{E}+h\right)} \tag{7.41} \end{equation*} $$

The K.E is positive whereas the P.E is negative. However, in magnitude the K.E. is half the P.E, so that the total E is

$$ \begin{equation*} E=K . E+P . E=-\frac{G m M_{E}}{2\left(R_{E}+h\right)} \tag{7.42} \end{equation*} $$

The total energy of an circularly orbiting satellite is thus negative, with the potential energy being negative but twice is magnitude of the positive kinetic energy.

When the orbit of a satellite becomes elliptic, both the K.E. and P.E. vary from point to point. The total energy which remains constant is negative as in the circular orbit case. This is what we expect, since as we have discussed before if the total energy is positive or zero, the object escapes to infinity. Satellites are always at finite distance from the earth and hence their energies cannot be positive or zero.

**Example 7.8** A $400 \mathrm{~kg}$ satellite is in a circular orbit of radius $2 R_{E}$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4 R_{E}$ ? What are the changes in the kinetic and potential energies?

**Answer** Initially,

$$ E_{i}=-\frac{G M_{E} m}{4 R_{E}} $$

While finally

$$ E_{f}=-\frac{G M_{E} m}{8 R_{E}} $$

The change in the total energy is

$$\Delta E=E_{f}-E_{i}$$

$$ =\frac{G M_{E} m}{8 R_{E}}=\left(\frac{G M_{E}}{R_{E}^{2}}\right) \frac{m R_{E}}{8} $$

$$ \Delta E=\frac{g m R_{E}}{8}=\frac{9.81 \times 400 \times 6.37 \times 10^{6}}{8}=3.13 \times 10^{9} \mathrm{~J} $$

The kinetic energy is reduced and it mimics $\Delta E$, namely, $\Delta K=K_{f}-K_{i}=-3.13 \times 10^{9} \mathrm{~J}$.

The change in potential energy is twice the change in the total energy, namely

$$ \Delta V=V_{f}-V_{i}=-6.25 \times 10^{9} \mathrm{~J} $$

### Summary

**1.** Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses $m_{1}$ and $m_{2}$ separated by a distance $r$ has the magnitude

$$ F=G \frac{m_{1} m_{2}}{r^{2}} $$

where $G$ is the universal gravitational constant, which has the value $6.672 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.

**2.** If we have to find the resultant gravitational force acting on the particle $\mathrm{m}$ due to a number of masses $M_{1}, M_{2}, \ldots . M_{n}$ etc. we use the principle of superposition. Let $F_{1}, F_{2}, \ldots . F_{n}$ be the individual forces due to $M_{1}, M_{2}, \ldots . M_{n}$, each given by the law of gravitation. From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force $F_{R}$ is then found by vector addition

$$ F_{R}=F_{1}+F_{2}+\ldots \ldots+F_{n}=\sum_{i=1}^{n} F_{i} $$

where the symbol ’ $\Sigma$ ’ stands for summation.

**3.** Kepler’s laws of planetary motion state that

(a) All planets move in elliptical orbits with the Sun at one of the focal points

(b) The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. This follows from the fact that the force of gravitation on the planet is central and hence angular momentum is conserved.

(c) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet. The period $T$ and radius $R$ of the circular orbit of a planet about the Sun are related by

$$ T^{2}=\left(\frac{4 \pi^{2}}{G M_{s}}\right) R^{3} $$

where $M_{s}$ is the mass of the Sun. Most planets have nearly circular orbits about the Sun. For elliptical orbits, the above equation is valid if $R$ is replaced by the semi-major axis, $a$.

**4.** The acceleration due to gravity.

(a) at a height $h$ above the earth’s surface

$$ \begin{aligned} & g(h)=\frac{G M_{E}}{\left(R_{E}+h\right)^{2}} \\ \\ & = \frac{G M_{E}}{R_{E}^{2}} \left(1-\frac{2 h}{R_{E}}\right) \quad \text { for } h < < R_{E} \end{aligned} $$

$$ g(h)=g(0) 1-\frac{2 h}{R_{E}} \quad \text { where } g(0)=\frac{G M_{E}}{R_{E}^{2}} $$

(b) at depth $d$ below the earth’s surface is

$ g(d)=\frac{G M_{E}}{R_{E}^{2}} 1-\frac{d}{R_{E}}=g(0) \quad 1-\frac{d}{R_{E}} $

**5.** The gravitational force is a conservative force, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance $r$ is given by

$$ V=-\frac{G m_{1} m_{2}}{r} $$

where $V$ is taken to be zero at $r \rightarrow \infty$. The total potential energy for a system of particles is the sum of energies for all pairs of particles, with each pair represented by a term of the form given by above equation. This prescription follows from the principle of superposition.

**6.** If an isolated system consists of a particle of mass $m$ moving with a speed $v$ in the vicinity of a massive body of mass $M$, the total mechanical energy of the particle is given by

$$ E=\frac{1}{2} m v^{2}-\frac{G M m}{r} $$

That is, the total mechanical energy is the sum of the kinetic and potential energies. The total energy is a constant of motion.

**7.** If $m$ moves in a circular orbit of radius a about $M$, where $M \gg m$, the total energy of the system is

$$ E=-\frac{G M m}{2 a} $$

with the choice of the arbitrary constant in the potential energy given in the point 5., above. The total energy is negative for any bound system, that is, one in which the orbit is closed, such as an elliptical orbit. The kinetic and potential energies are

$$ \begin{aligned} K & =\frac{G M m}{2 a} \\ \\ V & =-\frac{G M m}{a} \end{aligned} $$

**8.** The escape speed from the surface of the earth is

$$ v_{e}=\sqrt{\frac{2 G M_{E}}{R_{E}}}=\sqrt{2 g R_{E}} $$

and has a value of $11.2 \mathrm{~km} \mathrm{~s}^{-1}$.

**9.** If a particle is outside a uniform spherical shell or solid sphere with a spherically symmetric internal mass distribution, the sphere attracts the particle as though the mass of the sphere or shell were concentrated at the centre of the sphere.

**10.** If a particle is inside a uniform spherical shell, the gravitational force on the particle is zero. If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mass interior to the particle.

Physical Quantity | Symbol | Dimensions | Unit | Remarks |
---|---|---|---|---|

Gravitational Constant | $G$ | $\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]$ | $\mathrm{N} \mathrm{m}^{2} \mathrm{~kg}^{-2}$ | $6.67 \times 10^{-11}$ |

Gravitational Potential Energy |
$V(\mathrm{r})$ | $\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]$ | $\mathrm{J}$ | $-\frac{G M m}{r}$ (scalar) |

Gravitational Potential |
$U(\mathrm{r})$ | $\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]$ | $\mathrm{J} \mathrm{kg}^{-1}$ | $-\frac{G M}{r}$ |

Gravitational Intensity |
$\mathbf{E}$ or $\mathbf{g}$ |
$\left[\mathrm{LT}^{-2}\right]$ | $\mathrm{m} \mathrm{s}^{-2}$ | $\frac{G M}{r^{2}} \hat{\mathrm{r}}$ (vector) |

### POINTS TO PONDER

**1.** In considering motion of an object under the gravitational influence of another object the following quantities are conserved:

(a) Angular momentum

(b) Total mechanical energy

Linear momentum is not conserved

**2.** Angular momentum conservation leads to Kepler’s second law. However, it is not special to the inverse square law of gravitation. It holds for any central force.

**3.** In Kepler’s third law (see Eq. (7.1) and $T^{2}=K_{S} R^{3}$. The constant $K_{S}$ is the same for all planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (7.38)].

**4.** An astronaut experiences weightlessness in a space satellite. This is not because the gravitational force is small at that location in space. It is because both the astronaut and the satellite are in “free fall” towards the Earth.

**5.** The gravitational potential energy associated with two particles separated by a distance $r$ is given by

$$ V=-\frac{G m_{1} m_{2}}{r}+\text { constant } $$

The constant can be given any value. The simplest choice is to take it to be zero. With this choice

$$ V=-\frac{G m_{1} m_{2}}{r} $$

This choice implies that $V \rightarrow 0$ as $r \rightarrow \infty$. Choosing location of zero of the gravitational energy is the same as choosing the arbitrary constant in the potential energy. Note that the gravitational force is not altered by the choice of this constant.

**6.** The total mechanical energy of an object is the sum of its kinetic energy (which is always positive) and the potential energy. Relative to infinity (i.e. if we presume that the potential energy of the object at infinity is zero), the gravitational potential energy of an object is negative. The total energy of a satellite is negative.

**7.** The commonly encountered expression $m g h$ for the potential energy is actually an approximation to the difference in the gravitational potential energy discussed in the point 6, above.

**8.** Although the gravitational force between two particles is central, the force between two finite rigid bodies is not necessarily along the line joining their centre of mass. For a spherically symmetric body however the force on a particle external to the body is as if the mass is concentrated at the centre and this force is therefore central.

**9.** The gravitational force on a particle inside a spherical shell is zero. However, (unlike a metallic shell which shields electrical forces) the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible.

### Exercises

**7.1** Answer the following :

(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?

(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?

## Show Answer

**Answer**

**(a)** No

**(b)** Yes

**(c)** Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects.

If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).

Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

**7.2** Choose the correct alternative :

(a) Acceleration due to gravity increases/decreases with increasing altitude.

(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.

(d) The formula $-\operatorname{Mm}\left(1 / r_{2}-1 / r_{1}\right)$ is more/less accurate than the formula $m g\left(r_{2}-r_{1}\right)$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the earth.

## Show Answer

**Answer**

**(a)** Decreases

**(b)** Decreases

**(c)** Mass of the body

**(s)** More

**Explanation:**

(a) Acceleration due to gravity at depth $h$ is given by the relation:

$ g_h=(1-\frac{2 h}{R_e}) g $

Where,

$ R_e=\text{ Radius of the Earth } $

$g=$ Acceleration due to gravity on the surface of the Earth

It is clear from the given relation that acceleration due to gravity decreases with an increase in height.

(b) Acceleration due to gravity at depth $d$ is given by the relation:

$ g_d=(1-\frac{d}{R_e}) g $

It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

(c) Acceleration due to gravity of body of mass $m$ is given by the relation: $g=\frac{G M}{R^{2}}$

Where,

$G=$ Universal gravitational constant

$M=$ Mass of the Earth

$R=$ Radius of the Earth

Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) Gravitational potential energy of two points $r_2$ and $r_1$ distance away from the centre of the Earth is respectively given by:

$V(r_1)=-\frac{G m M}{r_1}$

$V(r_2)=-\frac{G m M}{r_2}$

$\therefore$ Difference in potential energy, $V=V(r_2)-V(r_1)=-G m M(\frac{1}{r_2}-\frac{1}{r_1})$

Hence, this formula is more accurate than the formula $m g(r_2-r_1)$.

**7.3** Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?

## Show Answer

**Answer**

Lesser by a factor of 0.63

Time taken by the Earth to complete one revolution around the Sun, $T_e=1$ year

Orbital radius of the Earth in its orbit, $R_e=1 AU$

Time taken by the planet to complete one revolution around the Sun, $T_p=\frac{1}{2} T_e=\frac{1}{2}$ year Orbital radius of the planet $=R_p$

From Kepler’s third law of planetary motion, we can write:

$ \begin{aligned} & (\frac{R_p}{R_e})^{3}=(\frac{T_P}{T_e})^{2} \\ & \frac{R_p}{R_e}=(\frac{T_P}{T_e})^{\frac{2}{3}} \\ & \quad=(\frac{1}{1})^{\frac{2}{3}}=(0.5)^{\frac{2}{3}}=0.63 \end{aligned} $

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

**7.4** Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is $4.22 \times 10^{8} \mathrm{~m}$. Show that the mass of Jupiter is about one-thousandth that of the sun

## Show Answer

**Answer**

Orbital period of $I_0, T _{l o}=1.769$ days $=1.769 \times 24 \times 60 \times 60 s$

Orbital radius of $I_0, R _{l o}=4.22 \times 10^{8} m$

Satellite $I_0$ is revolving around the Jupiter

Mass of the latter is given by the relation:

$$ \begin{equation*} M_J=\frac{4 \pi^{2} R_{ \pm}^{3}}{G T_{l o}^{2}} \tag{i} \end{equation*} $$

Where,

$M_J=$ Mass of Jupiter

$G=$ Universal gravitational constant

Orbital period of the Earth,

$T_e=365.25$ days $=365.25 \times 24 \times 60 \times 60 s$

Orbital radius of the Earth,

$ R_e=1 AU=1.496 \times 10^{11} m $

Mass of Sun is given as:

$$ \begin{align*} & M_s=\frac{4 \pi^{2} R_e^{3}}{G T_e^{2}} \tag{ii}\\ \\ & \begin{align*} & \therefore \frac{M_s}{M_J}=\frac{4 \pi^{2} R_e^{3}}{G T_e^{2}} \times \frac{G T _{l o}^{2}}{4 \pi^{2} R _{l o}^{3}}=\frac{R_e^{3}}{R _{l o}^{3}} \times \frac{T _{l o}^{2}}{T_e^{2}} \\ \\ & \quad=(\frac{1.769 \times 24 \times 60 \times 60}{365.25 \times 24 \times 60 \times 60})^{2} \times(\frac{1.496 \times 10^{11}}{4.22 \times 10^{8}})^{3} \\ \\ & \quad=1045.04 \end{align*} \\ \\ & \therefore \frac{M_S}{M_J} \sim 1000 \\ \\ & M_S \sim 1000 \times M_J \end{align*} $$

Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.

**7.5** Let us assume that our galaxy consists of $2.5 \times 10^{11}$ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be $10^{5} \mathrm{ly}$.

## Show Answer

**Answer**

Mass of our galaxy Milky Way, $M=2.5 \times 10^{11}$ solar mass

Solar mass $=$ Mass of Sun $=2.0 \times 10^{36} kg$

Mass of our galaxy, $M=2.5 \times 10^{11} \times 2 \times 10^{36}=5 \times 10^{41} kg$

Diameter of Milky Way, $d=10^{5}$ ly

Radius of Milky Way, $r=5 \times 10^{4} ly$

$1 ly=9.46 \times 10^{15} m$

$\therefore r=5 \times 10^{4} \times 9.46 \times 10^{15}$

$=4.73 \times 10^{20} m$

Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:

$ \begin{aligned} & T=(\frac{4 \pi^{2} r^{3}}{G M})^{\frac{1}{2}} \\ \\ & =(\frac{4 \times(3.14)^{2} \times(4.73)^{3} \times 10^{60}}{6.67 \times 10^{-11} \times 5 \times 10^{41}})^{\frac{1}{2}}=(\frac{39.48 \times 105.82 \times 10^{30}}{33.35})^{\frac{1}{2}} \\ \\ & =(125.27 \times 10^{30})^{\frac{1}{2}}=1.12 \times 10^{16} s \\ \\ & 1 \text{ year }=365 \times 324 \times 60 \times 60 s \\ \\ & 1 s=\frac{1}{365 \times 24 \times 60 \times 60} \text{ years } \\ \\ & \therefore 1.12 \times 10^{16} s=\frac{1.12 \times 10^{16}}{365 \times 24 \times 60 \times 60} \\ \\ & =3.55 \times 10^{8} \text{ years } \end{aligned} $

**7.6** Choose the correct alternative:

(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.

(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

## Show Answer

**Answer**

**(a)** Kinetic energy

**(b)** Less

**(a)** Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (may be negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative.

Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.

**(b)** An orbiting satellite acquires a certain amount of energy that enables it to revolve around the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out of the influence of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

**7.7** Does the escape speed of a body from the earth depend on

(a) the mass of the body,

(b) the location from where it is projected,

(c) the direction of projection,

(d) the height of the location from where the body is launched?

## Show Answer

**Answer**

**(a)** No

**(b)** No

**(c)** No

**(d)** Yes

Escape velocity of a body from the Earth is given by the relation:

$v_{\text{esc }}=\sqrt{2 g R}$

$g=$ Acceleration due to gravity

$R=$ Radius of the Earth

It is clear from equation (i) that escape velocity $v_{\text{esc }}$ is independent of the mass of the body and the direction of its projection. However, it depends on gravitational potential at the point from where the body is launched. Since this potential marginally depends on the height of the point, escape velocity also marginally depends on these factors.

**7.8** A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

## Show Answer

**Answer**

**(a)** No

**(b)** No

**(c)** Yes

**(d)** No

**(e)** No

Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Its linear speed, angular speed, kinetic, and potential energy varies from point to point in the orbit.

**7.9** Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.

## Show Answer

**Answer**

(b), (c), and (d)

Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.

A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.

Headaches are caused because of mental strain. It can affect the working of an astronaut in space.

Space has different orientations. Therefore, orientational problem can affect an astronaut in space.

**7.10** In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0.

Fig 7.11

## Show Answer

**Answer**

(iii)

Gravitational potential $(V)$ is constant at all points in a spherical shell. Hence, the gravitational potential gradient $(\frac{d V}{d r})$ is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.

If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at centre $O$ will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at centre $O$ of the given hemispherical shell has the direction as indicated by $\to \mathbf{c}$.

**7.11** For the above problem, the direction of the gravitational intensity at an arbitrary point $\mathrm{P}$ is indicated by the arrow (i) $\mathrm{d}$, (ii) e, (iii) $\mathrm{f}$, (iv) $\mathrm{g}$.

## Show Answer

**Answer**

(ii)

Gravitational potential $(V)$ is constant at all points in a spherical shell. Hence, the gravitational potential gradient $(\frac{d V}{d r})$ is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.

If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point $P$ will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point $P$ of the hemispherical shell has the direction as indicated by \to $\mathbf{e}$.

**7.12** A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun $=2 \times 10^{30} \mathrm{~kg}$, mass of the earth $=6 \times 10^{24} \mathrm{~kg}$. Neglect the effect of other planets etc. (orbital radius $=1.5 \times 10^{11} \mathrm{~m}$ ).

## Show Answer

**Answer**

Mass of the Sun, $M_s=2 \times 10^{30} kg$

Mass of the Earth, $M_e=6 \times 10^{24} kg$

Orbital radius, $r=1.5 \times 10^{11} m$

Mass of the rocket $=m$

Let $x$ be the distance from the centre of the Earth where the gravitational force acting on satellite $P$ becomes zero.

From Newton’s law of gravitation, we can equate gravitational forces acting on satellite $P$ under the influence of the Sun and the Earth as:

$\frac{G m M_s}{(r-x)^{2}}=G m \frac{M_e}{x^{2}}$

$(\frac{r-x}{x})^{2}=\frac{M_s}{M_e}$

$\frac{r-x}{x}=(\frac{2 \times 10^{30}}{60 \times 10^{24}})^{\frac{1}{2}}=577.35$

$1.5 \times 10^{11}-x=577.35 x$

$578.35 x=1.5 \times 10^{11}$

$x=\frac{1.5 \times 10^{11}}{578.35}=2.59 \times 10^{8} m$

**7.13** How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5 \times 10^{8} \mathrm{~km}$.

## Show Answer

**Answer**

Orbital radius of the Earth around the Sun, $r=1.5 \times 10^{11} m$

Time taken by the Earth to complete one revolution around the Sun,

$T=1$ year $=365.25$ days

$=365.25 \times 24 \times 60 \times 60 s$

Universal gravitational constant, $G=6.67 \times 10^{-11} Nm^{2} kg^{-2}$

Thus, mass of the Sun can be calculated using the relation,

$ \begin{aligned} M & =\frac{4 \pi^{2} r^{3}}{G T^{2}} \\ & =\frac{4 \times(3.14)^{2} \times(1.5 \times 10^{11})^{3}}{6.67 \times 10^{-11} \times(365.25 \times 24 \times 60 \times 60)^{2}} \\ & =\frac{133.24 \times 10}{6.64 \times 10^{4}}=2.0 \times 10^{30} kg \end{aligned} $

Hence, the mass of the Sun is $2 \times 10^{30} kg$.

**7.14** A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is $1.50 \times 10^{8} \mathrm{~km}$ away from the sun?

## Show Answer

**Answer**

Distance of the Earth from the Sun, $r_e=1.5 \times 10^{8} km=1.5 \times 10^{11} m$

Time period of the Earth $=T_e$

Time period of Saturn, $T_s=29.5 T_e$

Distance of Saturn from the Sun $=r_s$

From Kepler’s third law of planetary motion, we have

$ T=(\frac{4 \pi^{2} r^{3}}{G M})^{\frac{1}{2}} $

For Saturn and Sun, we can write

$ \begin{aligned} \frac{r_s^{3}}{r_e^{3}} & =\frac{T_s^{2}}{T_e^{2}} \\ r_s & =r_e(\frac{T_s}{T_e})^{\frac{2}{3}} \\ & =1.5 \times 10^{11}(\frac{29.5 T_e}{T_e})^{\frac{2}{3}} \\ & =1.5 \times 10^{11}(29.5)^{\frac{2}{3}} \\ & =1.5 \times 10^{11} \times 9.55 \\ = & 14.32 \times 10^{11} m \end{aligned} $

Hence, the distance between Saturn and the Sun is $1.43 \times 10^{12} m$.

**7.15** A body weighs $63 \mathrm{~N}$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?

## Show Answer

**Answer**

Weight of the body, $W=63 N$

Acceleration due to gravity at height $h$ from the Earth’s surface is given by the relation:

$ g^{\prime}=\frac{g}{(\frac{1+h}{R_e})^{2}} $

Where,

$g=$ Acceleration due to gravity on the Earth’s surface

$R e=$ Radius of the Earth

For $h=\frac{R_e}{2}$

$g^{\prime}=\frac{g}{(1+\frac{R_e}{2 \times R_e})^{2}}=\frac{g}{(1+\frac{1}{2})^{2}}=\frac{4}{9} g$

Weight of a body of mass $m$ at height $h$ is given as:

$ \begin{aligned} W^{\prime} & =m g \\ & =m \times \frac{4}{9} g=\frac{4}{9} \times m g \\ & =\frac{4}{9} W \\ & =\frac{4}{9} \times 63=28 N \end{aligned} $

**7.16** Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed $250 \mathrm{~N}$ on the surface?

## Show Answer

**Answer**

Weight of a body of mass $m$ at the Earth’s surface, $W=m g=250 N$

Body of mass $m$ is located at depth, $\quad d=\frac{1}{2} R_e$

Where,

$R_e=$ Radius of the Earth

Acceleration due to gravity at depth $g(d)$ is given by the relation:

$ \begin{aligned} g^{\prime} & =(1-\frac{d}{R_e}) g \\ & =(1-\frac{R_e}{2 \times R_e}) g=\frac{1}{2} g \end{aligned} $

Weight of the body at depth $d$,

$ \begin{aligned} W^{\prime} & =m g^{\prime} \\ & =m \times \frac{1}{2} g=\frac{1}{2} m g=\frac{1}{2} W \\ & =\frac{1}{2} \times 250=125 N \end{aligned} $

**7.17** A rocket is fired vertically with a speed of $5 \mathrm{~km} \mathrm{~s}^{-1}$ from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $=6.0 \times 10^{24} \mathrm{~kg}$; mean radius of the earth $=6.4 \times 10^{6} \mathrm{~m} ; G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.

## Show Answer

**Answer**

$8 \times 10^{6} m$ from the centre of the Earth

Velocity of the rocket, $v=5 km / s=5 \times 10^{3} m / s$

Mass of the Earth, $M_e=6.0 \times 10^{24} kg$

Radius of the Earth, $R_e=6.4 \times 10^{6} m$

Height reached by rocket mass, $m=h$

At the surface of the Earth,

Total energy of the rocket $=$ Kinetic energy + Potential energy $=\frac{1}{2} m v^{2}+(\frac{-G M_e m}{R_e})$

At highest point $h$,

$v=0$

And, Potential energy $=-\frac{G M_e m}{R_e+h}$

Total energy of the rocket $=0+(-\frac{G M_e m}{R_e+h})=-\frac{G M_e m}{R_e+h}$

From the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface $=$ Total energy at height $h$

$ \begin{aligned} & \frac{1}{2} m v^{2}+(-\frac{G M_e m}{R_e})=-\frac{G M_e m}{R_e+h} \\ \\ & \frac{1}{2} v^{2}=G M_e(\frac{1}{R_e}-\frac{1}{R_e+h}) \\ \\ & =G M_e(\frac{R_e+h-R_e}{R_e(R_e+h)}) \\ \\ & \frac{1}{2} v^{2}=\frac{G M_e h}{R_e(R_e+h)} \times \frac{R_e}{R_e} \\ \\ & \frac{1}{2} \times v^{2}=\frac{g R_e h}{R_e+h} \end{aligned} $

Where $g=\frac{G M}{R_e^{2}}=9.8 m / s^{2}$ (Acceleration due to gravity on the Earth’s surface)

$\therefore v^{2}(R_e+h)=2 g R_e h$

$v^{2} R_e=h(2 g R_e-v^{2})$

$h=\frac{R_e v^{2}}{2 g R_e-v^{2}}$

$=\frac{6.4 \times 10^{6} \times(5 \times 10^{3})^{2}}{2 \times 9.8 \times 6.4 \times 10^{6}-(5 \times 10^{3})^{2}}$

$h=\frac{6.4 \times 25 \times 10^{12}}{100.44 \times 10^{6}}=1.6 \times 10^{6} m$

Height achieved by the rocket with respect to the centre of the Earth

$ \begin{aligned} & =R_e+h \\ & =6.4 \times 10^{6}+1.6 \times 10^{6} \\ & =8.0 \times 10^{6} m \end{aligned} $

**7.18** The escape speed of a projectile on the earth’s surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

## Show Answer

**Answer**

Escape velocity of a projectile from the Earth, $v_{esc}=11.2 km / s$

Projection velocity of the projectile, $v_p=3 v_{esc}$

Mass of the projectile $=m$

Velocity of the projectile far away from the Earth $=v_f$

Total energy of the projectile on the Earth $=\frac{1}{2} m v_p^{2}-\frac{1}{2} m v_esc^{2}$

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth $=\frac{1}{2} m v_f^{2}$

From the law of conservation of energy, we have

$ \begin{aligned} \frac{1}{2} & m v_p^{2}-\frac{1}{2} m v_esc^{2}=\frac{1}{2} m v_f^{2} \\ v_f & =\sqrt{v_p^{2}-v_esc^{2}} \\ & =\sqrt{(3 v_{esc})^{2}-(v_{esc})^{2}} \\ & =\sqrt{8} v_{esc} \\ & =\sqrt{8} \times 11.2=31.68 km / s \end{aligned} $

**7.19** A satellite orbits the earth at a height of $400 \mathrm{~km}$ above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite $=200 \mathrm{~kg}$; mass of the earth $=6.0 \times 10^{24} \mathrm{~kg}$; radius of the earth $=6.4 \times 10^{6} \mathrm{~m} ; G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.

## Show Answer

**Answer**

Mass of the Earth, $M=6.0 \times 10^{24} kg$

Mass of the satellite, $m=200 kg$

Radius of the Earth, $R_e=6.4 \times 10^{6} m$

Universal gravitational constant, $G=6.67 \times 10^{-11} Nm^{2} kg^{-2}$

Height of the satellite, $h=400 km=4 \times 10^{5} m=0.4 \times 10^{6} m$

Total energy of the satellite at height $h=\frac{1}{2} m v^{2}+(\frac{-G M_e m}{R_e+h})$

Orbital velocity of the satellite, $v=\sqrt{\frac{G M_e}{R_e+h}}$

Total energy of height, $h=\frac{1}{2} m(\frac{G M_e}{R_e+h})-\frac{G M_e m}{R_e+h}=-\frac{1}{2}(\frac{G M_e m}{R_e+h})$

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.

Energy required to send the satellite out of its orbit $=-$ (Bound energy)

$ \begin{aligned} & =\frac{1}{2} \frac{G M_e m}{(R_e+h)} \\ \\ & =\frac{1}{2} \times \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{(6.4 \times 10^{6}+0.4 \times 10^{6})} \\ \\ & =\frac{1}{2} \times \frac{6.67 \times 6 \times 2 \times 10}{6.8 \times 10^{6}}=5.9 \times 10^{9} J \end{aligned} $

**7.20** Two stars each of one solar mass $\left(=2 \times 10^{30} \mathrm{~kg}\right)$ are approaching each other for a head on collision. When they are a distance $10^{9} \mathrm{~km}$, their speeds are negligible. What is the speed with which they collide ? The radius of each star is $10^{4} \mathrm{~km}$. Assume the stars to remain undistorted until they collide. (Use the known value of $G$ ).

## Show Answer

**Answer**

Mass of each star, $M=2 \times 10^{30} kg$

Radius of each star, $R=10^{4} km=10^{7} m$

Distance between the stars, $r=10^{9} km=10^{12} m$

For negligible speeds, $v=0$ total energy of two stars separated at distance $r$

$$ \begin{align*} & =\frac{-G M M}{r}+\frac{1}{2} m v^{2} \\ & =\frac{-G M M}{r}+0 \tag{i} \end{align*} $$

Now, consider the case when the stars are about to collide:

Velocity of the stars $=v$

Distance between the centers of the stars $=2 R$

Total kinetic energy of both stars $=\frac{1}{2} M v^{2}+\frac{1}{2} M v^{2}=M v^{2}$

Total potential energy of both stars $=\frac{-G M M}{2 R}$

Total energy of the two stars $=M v^{2}-\frac{G M M}{2 R} \quad \quad \quad \text{(ii)}$

Using the law of conservation of energy, we can write:

$ \begin{aligned} & M v^{2}-\frac{G M M}{2 R}=\frac{-G M M}{r} \\ \\ & v^{2}=\frac{-G M}{r}+\frac{G M}{2 R}=G M(-\frac{1}{r}+\frac{1}{2 R}) \\ \\ & \quad=6.67 \times 10^{-11} \times 2 \times 10^{30}[-\frac{1}{10^{12}}+\frac{1}{2 \times 10^{7}}] \\ \\ & \quad=13.34 \times 10^{19}[-10^{-12}+5 \times 10^{-8}] \\ \\ & \sim 13.34 \times 10^{19} \times 5 \times 10^{-8} \\ \\ & \sim 6.67 \times 10^{12} \\ \\ & v=\sqrt{6.67 \times 10^{12}}=2.58 \times 10^{6} m / s \end{aligned} $

**7.21** Two heavy spheres each of mass $100 \mathrm{~kg}$ and radius $0.10 \mathrm{~m}$ are placed $1.0 \mathrm{~m}$ apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

## Show Answer

**Answer**

0 ;

$-2.7 \times 10^{-8} J / kg$;

Yes;

Unstable

**Explanation:**

The situation is represented in the given figure:

Mass of each sphere, $M=100 kg$

Separation between the spheres, $r=1 m$

$X$ is the mid point between the spheres. Gravitational force at point $X$ will be zero. This is because gravitational force exerted by each sphere will act in opposite directions.

Gravitational potential at point $X$ :

$ \begin{aligned} & =\frac{-G M}{(\frac{r}{2})}-\frac{G M}{(\frac{r}{2})}=-4 \frac{G M}{r} \\ & =\frac{4 \times 6.67 \times 10^{-11} \times 100}{1} \\ & =-2.67 \times 10^{-8} J / kg \end{aligned} $

Any object placed at point $X$ will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.