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 (
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.
8.2 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
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.
Select two points
- 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.
- 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
(a
Planet | |||
---|---|---|---|
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
Hence
where
Example 7.1 Let the speed of the planet at the perihelion
Answer The magnitude of the angular momentum at
or
Since
The area
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
where
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
in agreement with a value of
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
Equation (7.5) can be expressed in vector form as
where
Gravitational force on
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.
The total force on
Example 7.2 Three equal masses of
(a) What is the force acting on a mass
(b) What is the force if the mass at the vertex
Take
Answer (a) The angle between GC and the positive
From the principle of superposition and the law of vector addition, the resultant gravitational force
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
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
The bar
If
Observation of
Since Cavendish’s experiment, the measurement of
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.
Again consider the earth to be made up of concentric shells as before and a point mass
We assume that the entire earth is of uniform density and hence its mass is
If the mass
The acceleration experienced by the mass
Acceleration
7.6 ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH
Consider a point mass
Its distance from the centre of the earth is
The acceleration experienced by the point mass is
This is clearly less than the value of
Equation (7.15) thus tells us that for small heights
Now, consider a point mass
Since mass of a sphere is proportional to be cube of its radius.
Thus the force on the point mass is
Substituting for
and hence the acceleration due to gravity at a depth
Thus, as we go down below earth’s surface, the acceleration due gravity decreases by a factor
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
If we associate a potential energy
(where
then it is clear that
The work done in moving the particle is just the difference of potential energy between its final and initial positions.Observe that the constant
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
where
In place of Eq. (7.21), we can thus associate a potential energy
valid for
so that once again
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
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
Answer Consider four masses each of mass
Hence,
The gravitational potential at the centre of the square
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
If the object was thrown initially with a speed
By the principle of energy conservation Eqs. (7.26) and (7.27) must be equal. Hence
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
The minimum value of
If the object is thrown from the surface of the earth,
Using the relation
Using the value of
Equation (7.32) applies equally well to an object thrown from the surface of the moon with
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
The neutral point
At the neutral point
From the principle of conservation of mechanical energy
or
A point to note is that the speed of the projectile is zero at
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
directed towards the centre. This centripetal force is provided by the gravitational force, which is
where
Equating R.H.S of Eqs. (7.33) and (7.34) and cancelling out
Thus
where we have used the relation
on substitution of value of
Squaring both sides of Eq. (7.37), we get
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
If we substitute the numerical values
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
Answer (i) We employ Eq. (7.38) with the sun’s mass replaced by the martian mass
(ii) Once again Kepler’s third law comes to our aid,
where
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,
Example 7.6 Weighing the Earth: You are given the following data:
Answer From Eq. (7.12) we have
The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq. (7.38)]
Both methods yield almost the same answer. the difference between them being less than
Example 7.7 Express the constant k of Eq. (7.38) in days and kilometres. Given
Answer Given
Using Eq. (7.38) and the given value of
Note that Eq. (7.38) also holds for elliptical orbits if we replace
7.10 ENERGY OF AN ORBITING SATELLITE
Using Eq. (7.35), the kinetic energy of the satellite in a circular orbit with speed
Considering gravitational potential energy at infinity to be zero, the potential energy at distance
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
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. Answer Given
Example 7.8 A
Answer Initially,
While finally
The change in the total energy is
The kinetic energy is reduced and it mimics
The change in potential energy is twice the change in the total energy, namely
SUMMARY
1. Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses
where
2. If we have to find the resultant gravitational force acting on the particle
where the symbol ’
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
where
4. The acceleration due to gravity.
(a) at a height
(b) at depth
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
where
6. If an isolated system consists of a particle of mass
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
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
8. The escape speed from the surface of the earth is
and has a value of
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 | ||||
Gravitational Potential Energy |
(scalar) |
|||
Gravitational Potential |
||||
Gravitational Intensity |
or |
(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
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
The constant can be given any value. The simplest choice is to take it to be zero. With this choice
This choice implies that
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
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?
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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.
(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
Show Answer
Answer
(a) Decreases
(b) Decreases
(c) Mass of the body
(s) More
Explanation:
(a) Acceleration due to gravity at depth
Where,
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
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
Where,
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
Hence, this formula is more accurate than the formula
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Answer
Lesser by a factor of 0.63
Time taken by the Earth to complete one revolution around the Sun,
Orbital radius of the Earth in its orbit,
Time taken by the planet to complete one revolution around the Sun,
From Kepler’s third law of planetary motion, we can write:
Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
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Answer
Orbital period of
Orbital radius of
Satellite
Mass of the latter is given by the relation:
Where,
Orbital period of the Earth,
Orbital radius of the Earth,
Mass of Sun is given as:
Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.
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Answer
Mass of our galaxy Milky Way,
Solar mass
Mass of our galaxy,
Diameter of Milky Way,
Radius of Milky Way,
Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
(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.
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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.
(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?
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Answer
(a) No
(b) No
(c) No
(d) Yes
Escape velocity of a body from the Earth is given by the relation:
It is clear from equation (i) that escape velocity
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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.
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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.
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Answer
(iii)
Gravitational potential
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
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
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Answer
(ii)
Gravitational potential
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
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
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Answer
Mass of the Sun,
Mass of the Earth,
Orbital radius,
Mass of the rocket
Let
From Newton’s law of gravitation, we can equate gravitational forces acting on satellite
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Answer
Orbital radius of the Earth around the Sun,
Time taken by the Earth to complete one revolution around the Sun,
Universal gravitational constant,
Thus, mass of the Sun can be calculated using the relation,
Hence, the mass of the Sun is
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Answer
Distance of the Earth from the Sun,
Time period of the Earth
Time period of Saturn,
Distance of Saturn from the Sun
From Kepler’s third law of planetary motion, we have
For Saturn and Sun, we can write
Hence, the distance between Saturn and the Sun is
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Answer
Weight of the body,
Acceleration due to gravity at height
Where,
For
Weight of a body of mass
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Answer
Weight of a body of mass
Body of mass
Where,
Acceleration due to gravity at depth
Weight of the body at depth
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Answer
Velocity of the rocket,
Mass of the Earth,
Radius of the Earth,
Height reached by rocket mass,
At the surface of the Earth,
Total energy of the rocket
At highest point
And, Potential energy
Total energy of the rocket
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface
Where
Height achieved by the rocket with respect to the centre of the Earth
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Answer
Escape velocity of a projectile from the Earth,
Projection velocity of the projectile,
Mass of the projectile
Velocity of the projectile far away from the Earth
Total energy of the projectile on the Earth
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth
From the law of conservation of energy, we have
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Answer
Mass of the Earth,
Mass of the satellite,
Radius of the Earth,
Universal gravitational constant,
Height of the satellite,
Total energy of the satellite at height
Orbital velocity of the satellite,
Total energy of height,
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
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Answer
Mass of each star,
Radius of each star,
Distance between the stars,
For negligible speeds,
Now, consider the case when the stars are about to collide:
Velocity of the stars
Distance between the centers of the stars
Total kinetic energy of both stars
Total potential energy of both stars
Total energy of the two stars
Using the law of conservation of energy, we can write:
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Answer
0 ;
Yes;
Unstable
Explanation:
The situation is represented in the given figure:
Mass of each sphere,
Separation between the spheres,
Gravitational potential at point
Any object placed at point