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.

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 $m g(r_2-r_1)$.

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.

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.

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} $

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.

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.

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.

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.

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 $\to \mathbf{c}$.

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 $\mathbf{e}$.

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:

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

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

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

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

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

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

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 $2 \times 10^{30} kg$.

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 $1.43 \times 10^{12} m$.

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} $

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} $

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} $

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} $

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} $

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} $

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.