# Question-1

**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?**

**Solution:**

A body cannot be shielded from the influence of gravitational forces of nearby matter by putting it inside a hollow sphere or by some other means, because the gravitational force always attracts all the particles or bodies towards it.

# Question-2

**If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sunâ€™s pull is greater than the moonâ€™s pull. However, the tidal effect of the moonâ€™s pull is greater than the tidal effect of sun. Why?**

**Solution:**

Generally the gravitational force of attraction between two bodies is directly proportional to their masses and inversely proportional to the square of the distance between them. The gravitational force between the sun and the earth or between the moon and the earth varies inversely to the square of the distance. But due to very large mass of the sun, the gravitational force on the earth due to the sun is about 7x10

^{4}times the force due to moon on it. The tidal effect varies inversely as the cube of the distance. Therefore, though the mass of the moon is quite small as compared to the mass of the sun, yet the tidal effect of moonâ€™s pull is very large (as compared to that due to the sun) due to small distance between moon and earth.

# Question-3

**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?**

**Solution:**

Due to the small space ship, the force of attraction between the earth and the ship is too less, the astronaut cannot detect gravity. But if the space station orbiting around the earth has a large size, then he can detect the gravity because of the increase in size, which in turn enhances the force of attraction between them. If the size of the space station is very large, the magnitude of the gravity will also become appreciable and hence hope to detect it.

# Question-4

**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?**

**Solution:**

Given that the velocity of the planet is twice the velocity of earth. So the orbital size of the planet is nearly one fourth of the orbital size of the earth.

The planet revolves around the sun due to the centripetal force provided by the gravitational pull of the sun. If '2v' is the velocity of the existed planet and 'v' is the velocity of the earth.

---- (1)

When the velocity of the planet is twice the earth, we have

---- (2)

equating (1) and (2), we get

4 =

Therefore

Where r

_{p }and r

_{e }are the radii of the orbits. Hence, the orbital size of the planet is one-fourth of the orbital size of the earth.

# Question-5

**Io. one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10**

^{8}m. Show that the mass of Jupiter is about one-thousandth that of the sun.**Solution:**

Given data: radius of the orbit = 4.22 x 10

^{8}m

Orbital period = 1.769 days

Universal constant of Gravitation = 6.67 x 10

^{-11}Nm

^{2}kg

^{-2}

Orbital period of Jupiter satellite (T

_{1}) is 1.769 days. Orbital radius of Jupiterâ€™s satellite is (r

_{1}) is 4.22 x 10

^{8}m. Orbital period of sunâ€™s satellite is 1 year = 365 days, and the orbital radius of sunâ€™s satellite (earth) r

_{2}is 1.496 x 10

^{11}. Therefore, the mass of Jupiter is

Substituting the values, we get = 1046. Thus the mass of the Jupiter is about 1/1000 that of the sun.

# Question-6

**If we assume that our galaxy consists of 2.5x10**

^{11}_{ }stars each of one solar mass, how long will a star at a distance of 5x10^{4}light years from the galactic centre take to complete one revolution? Given one solar mass = 1.99 x10^{30}kg, 1 light year = 9.46x10^{15}m and gravitation constant G = 6.67 x10^{-11}Nm^{2}kg^{-2}**Solution:**

Mass of our galaxy (M) = mass of a star Ã— number of stars

(1.99 x 10

^{30}) (2.5 x 10

^{11}) = 4.975 Ã— 10

^{41}kg.

Let 'm' be the mass of the star and 'v' its velocity round its orbit. The centripetal force is provided to the star for its circular orbit by the gravitational pull of the galaxy. Hence

^{}

^{}

We know

Substituting the values of r, G and M and solving, we get

^{ }=

^{ }3.55Ã— 10

^{8}years.

# Question-7

**Choose the correct alternative:**

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

(ii) Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to a sphere of uniform density)

(iii) The effect of rotation on the effective value of acceleration due to gravity is greatest at the equator/poles.

(iv) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.

(v) The formula -G M m (1/r

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

(ii) Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to a sphere of uniform density)

(iii) The effect of rotation on the effective value of acceleration due to gravity is greatest at the equator/poles.

(iv) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.

(v) The formula -G M m (1/r

_{2}- 1/r_{1}) is more/less accurate than the formula mg (r_{2}- r_{1}), for the difference of potential energy between two points r_{2}and r_{1}distances away from the centres of Earth.**Solution:**

(i) decreases.

(ii) decreases.

(iii) equator

(iv) mass of the body

(v) more accurate.

# Question-8

**Choose the correct alternative:**

(i) The gravitational potential energy of two mass points infinite distance away is taken to be zero, the gravitational potential energy of a galaxy is (positive/negative/zero)

(ii) The universe on the large scale is shaped by (gravitational/electromagnetic) forces on the atomic scale by (gravitation/electromagnetic) forces, on the nuclear scale by (gravitational/ electromagnetic/strong nuclear ) forces.

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

(iv) The energy required to rocket 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.

(i) The gravitational potential energy of two mass points infinite distance away is taken to be zero, the gravitational potential energy of a galaxy is (positive/negative/zero)

(ii) The universe on the large scale is shaped by (gravitational/electromagnetic) forces on the atomic scale by (gravitation/electromagnetic) forces, on the nuclear scale by (gravitational/ electromagnetic/strong nuclear ) forces.

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

(iv) The energy required to rocket 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.

**Solution:**

(i) negative.

(ii) Gravitation, electromagnetic, strong nuclear.

(iii) Kinetic energy.

(iv) Less.

# Question-9

**Does the escape velocity 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?

Explain your answer?

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?

Explain your answer?

**Solution:**

Escape velocity V = âˆš (2 G M / R)

a. Escape velocity is independent of the mass of the body.

b. The escape velocity depends upon the gravitational potential at the point of projection. As the potential depends, to a minor extent, on the altitude and height of the point, the location affects it to a minor extent.

c. The escape velocity is independent of the direction of projection.

d. The escape velocity depends on the height of the location. For a projection point at a height h from the ground level,

V = âˆš (2 G M / R+h). Hence it depends on the height of projection point.

# Question-10

**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.

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.

**Solution:**

a. No. The linear speed of the comet is = R w. In a highly elliptical orbit, R is not constant.

b. No. Angular speed is not constant.

c. Angular momentum is constant since it is under a radial force only.

d. No. When the comet comes near the sun, the potential energy decreases because of the reduced radius and the kinetic energy increases due to increased speed.

e. No, for the same reason as in d. above.

f. Total energy remains constant.

# Question-11

**Which of the following symptoms is likely to afflict an astronaut in space**

(a) swollen feet, (b) swollen face (c) headache (d) orientational problem.

(a) swollen feet, (b) swollen face (c) headache (d) orientational problem.

**Solution:**

(b) swollen face,(c) headache and (d) orientational problem.

# Question-12

**The gravitation intensity at the centre of the drumhead defined by a hemispherical shell has the direction indicated by the arrow (in the figure) (i) a, (ii) b, (iii) c, (iv) zero.**

**Solution:**

When the hemisphere is made to a sphere, potential is constant at both P and C. Hence (iii) c is correct.

# Question-13

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

**Solution:**

(ii) e

# Question-14

**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=2Ã—10**

Neglect the effect of other planets etc. (Orbital radius = 1.5 Ã—10

^{30}kg, mass of the Earth=6.0Ã—10^{24}kg.Neglect the effect of other planets etc. (Orbital radius = 1.5 Ã—10

^{11}m)**Solution:**

Mass of the sun , M

_{s}= 2

**Ã—**10

^{30}kg.

Mass of the earth , M

_{e}= 6.0

**Ã—**10

^{ 24}kg.

Distance from earth to sun R = 1.5

**Ã—**10

^{11}m

Let the gravitational forces of earth and sun be equal at a point N at a distance r from earth

At the point N,

Gravitational force of sun = gravitational force of earth

G M

_{s}m / (R - r)

^{2}= G M

_{e}m / r

^{2}

M

_{s}/ (R -r)

^{2}= M

_{e}/ r

^{2}

(R -r)

^{2}/ r

^{2}= M

_{s}/ M

_{e}

= 2

**Ã—**10

^{30}/ 6

**Ã—**10

^{24}= 10

^{6}/ 3

(R -r) / r = 10

^{3}/ âˆš 3

R/r = (10

^{3}/ âˆš 3) + 1 â‰ˆ 10

^{3}/ âˆš 3

r = R / (103 / âˆš 3)

= 1.732

**Ã—**1.5

**Ã—**10

^{11}/ 10

^{3}

= 2.60

**Ã—**10

^{8}m

Gravitational force on the rocket is zero at a point 2.60

**Ã—**10

^{8}m from ear.

# Question-15

**How will you "weigh the Sun", i.e. estimate its mass? You will need to know the period of one of its planets and the radius of the planetary orbit. The mean orbital radius of the Earth around the Sun is 1.5 Ã—10**

^{8}km. Estimate the mass of the Sun.**Solution:**

The universal gravitational constant is defined in such a way that it is applicable to all objects and also in such a way to make gravitational mass equal to inertial mass. Because of this we will be able to make equality between the formulae derived for the two masses and solve them for the unknown factor in it.

Centripetal force F = M

_{e}R Ï‰

^{2 }[ M

_{e}= mass of the earth; R = radius of earth's orbit around sun]

= M

_{e}R (2Ï€ /T)

^{2}[ since Ï‰ = 2Ï€ /T, where T = Time period of one revolution]

Also, Gravitational force F = G M

_{s}M

_{e}/ R

^{2}

M

_{e}R (2Ï€ /T)

^{2}= G M

_{s}M

_{e}/ R

^{2}

.

^{.}. M

_{s}= 4 Ï€

^{2}R

^{3}/ G T

^{2}

G = 6.67 Ã— 10

^{ -11}N m

^{2}kg

^{ -2}

T = 365.25 days = 365.25 Ã— 24 Ã— 60 Ã— 60 s

R = 1.5 Ã— 10

^{8}km = 1.5 Ã— 10

^{11}m

Substituting in the above equation,

Mass of sun = M

_{s}

= {4 Ã— 3.14

^{2}Ã— (1.5 Ã— 10

^{11})

^{3}} / {(365.25 Ã— 24 Ã— 3600)

^{2}Ã— 6.67 Ã— 10

^{-11})} kg

= 2.0038 Ã— 10

^{30}kg.

# Question-16

**A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 Ã— 10**

^{8}km away from the Sun?**Solution:**

For any heavenly body in orbit,

Centripetal force F = M

_{1}R Ï‰

^{2}[ M

_{1}= mass of heavenly body; R = radius of orbit of that body]

= M

_{1}R (2Ï€ /T)

^{2}[ since Ï‰ = 2Ï€ /T where T = Time period of one revolution]

= M

_{1}R ( 4Ï€

^{2}/ T

^{2}) â€¦.................(1)

Also the gravitational force between two heavenly bodies = G M

_{1}M

_{2}/ R

^{2}............(2)

Equating (1) and (2)

M

_{1}R ( 4Ï€

^{2}/ T

^{2}) = G M

_{1}M

_{2}/ R

^{2}

R

^{3}= G M

_{2}T

_{2}/ 4Ï€

^{2}â€¦....................................(3)

Applying the above equation(3) between Saturn and Sun, we get,

R

_{saturn}

^{3}= G M

_{sun}T

_{saturn}

^{2}/ 4Ï€

^{2}â€¦..........................(4)

Applying equation(3) between Earth and Sun, we get,

R

_{earth}

^{3}= G M

_{sun}T

_{earth}

^{2}/ 4Ï€

^{2}â€¦.......................(5)

(4)/ (5) â‡’ R

_{saturn}

^{3}/ R

_{earth}

^{3}= T

_{saturn}

^{2}/ T

_{earth}

^{2 }

R

_{saturn}

^{3}= R

_{earth}

^{3}(T

_{saturn}/ T

_{earth})

^{2}

R

_{saturn}= R

_{earth}(T

_{saturn}/ T

_{earth})

^{2/3}.

= (1.5 Ã— 10

^{11}) Ã— ( 29.5)

^{2/3}m

= 1.432 Ã— 10

^{12}m

= Distance of Saturn from the Sun.

# Question-17

**A body weighs 63 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?**

**Solution:**

According to Newton's law of gravitation,

W

_{1}= 63 N

m g = G M m / R

^{2}[ m = mass of the body; M = mass of earth]

g = G M / R

^{2}â€¦.(1)

If g

_{1}and g

_{2}are the accelerations due to gravity at distances R

_{1}and R

_{2}from the centre of earth,

g

_{1}/ g

_{2}= ( R

_{2}/ R

_{1})

^{2}

Comparing g on earth's surface (R

_{1}= R ie. earth's radius) and at required height ( R

_{2}= 2/3 R) ,

g

_{1}/ g

_{2}= ( 2/3 )

^{2}= 4/9

If W

_{1}and W

_{2}are respective weights,

W

_{2}/ W

_{1}= (m g

_{1}) / (m g

_{2}) = 4/9

Weight at the height Â½ R from earth's surface = W

_{2}= W

_{1}Ã— 4/9

For the given body W

_{2}= 63 Ã— 4/9 = 28 N.

# Question-18

**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 N on the surface?**

**Solution:**

[When one goes down the surface of the earth, the influencing mass of the earth is only the inner sphere of reduced radius at that depth]

On the surface of earth, m g = G M m / R

^{2}

[ M is the total mass of earth, R the full radius and g the acceleration due to gravity at the surface.]

g = G M / R

^{2}.........................(1)

Halfway to the centre of the earth radius = R

_{1}= R/2 and mass of the earth = (Â½)

^{3}M = M/8

g

_{1}= G (M/8) / (R/2)

^{2}

= Â½ G M / R

^{2}.....................(2)

From (1) and (2) we have,

g

_{1}/ g = Â½

Mass halfway down the earth = m g

_{1}

Â½ mg = Â½ 250 N

= 125 N

# Question-19

**A rocket is fired vertically with a speed of 5 km s**

^{-1}from the Earth surface. How far from the Earth does the rocket go before returning to the Earth? Mass of the earth = 6.0 Ã— 10^{24}kg, mean radius of the Earth R = 6.4 Ã— 10^{6}m, G = 6.67 Ã— 10^{-11}N m^{2}kg^{-2}.**Solution:**

[Work done in bringing a mass

**m**against a mass

**M**at a distance '

**x**' through a distance dx

= gravitational force Ã—distance = (G M m / x

^{2}) dx

Work done in bringing it from âˆž to r = G M m / x

^{2}dx

= G M m [ (1/ âˆž - 1/r]

= - G M m [ (1/r - 1/ âˆž ]

Therefore Gravitational potential energy W = - G M m / r

Initial total energy = kinetic energy + potential energy

= Â½ m v

^{2}- G M m / R

Final total energy = 0 - G M m / (R + h)

Initial total energy = Final total energy, since no other external force is involved,

Â½ m v

^{2}- G M m / R = G M m / (R + h)

G M m / (R + h) = G M m / R - Â½ m v

^{2}

G M / ( R + h ) = ( G M - Â½ R v

^{2}) / R

R + h) / R = G M / (G M - Â½ R v

^{2})

(R + h ) = (G M R) / (G M - Â½ R v

^{2})

h = [(G M R) / (G M - Â½ R v

^{2})] - R

h = Â½ R

^{2}v

^{2}/ ( G M - Â½ R v

^{2})

With the given values, the height reached by the rocket = h

h = { Â½ Ã— (6.4 Ã— 10^{6})^{2} Ã— (5 Ã— 10^{3})^{2}} ____________________________________________________ m

{(6.67 Ã— 10 ^{-11} Ã— 6 Ã— 10^{24}) - ( Â½ Ã— 6.4 Ã— 10^{6} Ã— (5 Ã— 10^{3})^{2})}

= (512 Ã— 10^{6}) / (40 Ã— 10^{13}) - (80 Ã— 10^{12})

= 1.6 Ã— 10^{6} m

[For such long calculations as above, first evaluate the product groups in small units individually. Convert them to scientific notation i.e, with powers of 10 separated. Long chain calculations in one step may lead to large errors due to rounding off of significant digits].

# Question-20

**The escape speed of a projectile on the Earth's surface is 11.2 km 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 Sun and other planets?**Solution:**

Initial kinetic energy = Â½ m v

_{1}

^{2}[v

_{1}= initial speed of the body]

Final kinetic energy = Â½ m v

_{2}

^{2}[ v

_{2}= final speed of the body]

Initial potential energy = -G M m / R

= Â½ m V

^{2}[ V = escape velocity]

[When the body is projected with the escape velocity its K.E is Â½ m V

^{2}. Its potential energy at that time is - G M m / R. The body will then escape with zero energy from earth's gravitational influence. Hence this equation]

Final potential energy = 0

Initial kinetic energy + Initial potential energy = final kinetic energy + Final potential energy

Â½ m v

_{1}

^{2}- G M m / R = Â½ m v

_{2}

^{2}+ 0

Â½ m v

_{1}

^{2}- Â½ m V

_{2}

^{2}= Â½ m v

_{2}

^{2}

Â½ m ( 3 V)

^{2}- Â½ m V

^{2}= Â½ m v

_{2}

^{2}+ 0

Therefore 8 V

^{2}= v

_{2}

^{2}

Speed of the body after escape = âˆš 8 V = âˆš 8 11.2 km/s

= 31.68 kms

^{-1}

# Question-21

**A satellite orbits the Earth at a height of 400 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 kg, mass of the Earth = 6.4 Ã— 10**

^{6}kg, radius of the earth = 6.4 Ã— 10^{6}m, G = 6.67 Ã— 10^{-11}N m^{2}kg^{-2}.**Solution:**

Potential energy at a height h above earth's surface = G M m / R

Orbital velocity of the satellite at a height h = ( G M / R+h)

^{1/2}

(Centripetal force = gravitational force)

Therefore mass Ã— centripetal acceleration = G M m / (R+h)

= m Ã— (v

^{2}/ R+h)

= G M m / (R+h)

v

^{2}= G M / R+h

Potential energy of the satellite at height (h) above earth = G M m / (R+h)

Kinetic energy of the satellite in orbit = Â½ m v

^{2}= Â½ m ( G M / R+h )

Total energy of the satellite in orbit = - (G M m / R+h ) + Â½ m ( G M / R+h )

= - (Â½ G M m / R+h)

Energy to be expended to launch the satellite = - total energy of the satellite

= Â½ G M / R+h

[This is similar to the example of a spring on which the work done by the spring is taken as positive and the energy spent to compress it, is negative.]

We know that G = 6.67 Ã— 10

^{-11}N m

^{2}kg

^{-2}.

Mass of the satellite m = 200 kg

Mass of the earth M = 6.0 x 10

^{24}kg

Radius of the earth R = 6.4 Ã— 10

^{6}m

Height above earth h = 200 m

Energy to be expended to launch the satellite = Â½ G M m / (R+h)

= Â½ Ã— (6.67 Ã— 10

^{-11}Ã— 6.0 Ã— 10

^{24}Ã— 200) / (6.4 Ã— 10

^{6}+ 0.4 Ã— 10

^{6})

= 5.89 Ã— 10

^{9}J.

# Question-22

**Two stars, each of one solar mass (2 Ã— 10**

^{30}kg) is approaching each other for a head-on collision. When they are at a distance 10^{9}km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10^{4}km. Assume the stars to remain undistorted until they collide. (Use the known value of G)**Solution:**

The principle of conservation of energy is applied to the collision. The data indicates that the initial kinetic energy is nearly zero. The decrease in potential energy due to the approach from 9 km should represent the increase in kinetic energy.

Mass of each star M = 2

**Ã—**10

^{30}kg

Initial distance R = 10

^{12}m

Radius of each star â€˜râ€™ = 10

^{7}m

distance at collision = 2

**Ã—**10

^{7}m = 4

**Ã—**10

^{7}m

Initial potential energy = - G M

^{2 }/ R

= - {6.67

**Ã—**10

^{-11}

**Ã—**(2

**Ã—**10

^{30})

^{2}} / 10

^{12}J

= - 26.68

**Ã—**10

^{35}J

Potential energy at collision = - G M

^{2}/ 2r

= - 6.67

**Ã—**10

^{-11}

**Ã—**(2

**Ã—**10

^{30})

^{2}/ 2

**Ã—**10

^{7}J

= - 13.34

**Ã—**10

^{42}J

Decrease in potential energy = - 26.68

**Ã—**10

^{35}- (- 13.34

**Ã—**10

^{42}) J

= 10

^{36}(1.334

**Ã—**10

^{7}- 1.334) J

= 1.334

**Ã—**10

^{43}J

= Increase in kinetic energy

= 2

**Ã—**Â½ M v

^{2}[since initial kinetic energy = 0]

M v

^{2}= 1.334

**Ã—**10

^{43}

v

^{2}= 1.334

**Ã—**10

^{43}/ 2

**Ã—**10

^{30}

= 6.67

**Ã—**10

^{12}

v = 2.58

**Ã—**10

^{6}ms

^{-1}

Relative velocity of collision = 2.58

**Ã—**10

^{6}ms

^{-1}.

# Question-23

**Two heavy spheres, each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational field and potential at the midpoint 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?**

**Solution:**

Gravitational Field intensity= Gravitational force on unit mass

= G M m / R

^{2}where m = 1

= G M / R

^{2}

Gravitational potential = Work done to bring unit mass from infinity

= - G M / R

Mass of each sphere M = 100 kg

Distance between the spheres = 1.0 m

Gravitational field at the mid point due to the other sphere is given by,

F

_{1}= G M / (R/2)

^{2}

= (6.67 Ã— 10

^{-11}Ã— 100) / (0.5)

^{2}N

= 2.668 Ã— 10

^{-8}N acting towards the other sphere

The two equal fields in opposite directions give a net field at the centre as zero.

Considering the potentials,

Potential at the centre due to one mass = - G M /(R/2)

= - (6.67 Ã— 10

^{-11}Ã— 100) / (0.5) J kg

^{-1}

= - 1.334 Ã— 10

^{-8}J kg

^{-1}

Potential at the centre due to the other mass = - G M /(R/2)

= - (6.67 Ã— 10

^{-11}Ã— 100) / (0.5) J kg

^{-1}

= - 1.334 Ã— 10

^{-8}J kg

^{-1}

Total gravitational potential at the centre

= - 1.334 Ã— 10

^{-8}+ (- 1.334 Ã— 10

^{-8}) J kg

^{-1}

= - 2.668 10

^{-8}J kg

^{-1}

[It is to be noted here that the Gravitational field at a point is the net effect of forces and therefore the vector resultant is to be found. In the case of Gravitational potential, the figure is the energy spent in bringing unit mass from infinity. Hence the net effect is the scalar addition of the magnitudes].

# Question-24

**As you have learnt in the text, a geostationary satellite orbits the Earth at a height of nearly 36,000 km from the surface of the Earth. What is the potential due to Earth's gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the Earth = 6.0 Ã— 10**

^{24}kg, radius = 6400 km.**Solution:**

Mass of the earth M = 6.0 Ã—10

^{24}kg.

Radius R = 6400 km = 6.4 Ã— 10

^{6}m

Height above earth = 36000 km = 36 Ã— 10

^{6}m

Potential at the satellite = -G M / (R + h)

= -(6.67 Ã— 10

^{-11}Ã— 6 Ã— 10

^{24})/ (6.4 Ã— 10

^{6}+ 36 Ã— 10

^{6})

= - 9.44 10

^{8 }J kg

^{-1}.

# Question-25

**A star 2.5 times the mass of the Sun and collapsed to the size of 12 km rotates with a speed of 1.2 rev/s ( Extremely compact stars of this kind are known as neutron stars. Certain observed stellar objects called pulsars are believed to belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = 2 Ã—10**

^{30}kg).**Solution:**

After compaction,

Mass of star M = 2.5 Ã— 2 Ã— 10

^{30}= 5 Ã— 10

^{30}kg

Radius R = 12 km = 1.2 Ã— 10

^{4}m

Acceleration due to gravity = g = G M / R

^{2}

= (6.67 Ã— 10

^{-11}Ã— 5 Ã— 10

^{30}) / (1.2 Ã— 10

^{4})

^{2}ms

^{-2}

= 23.16 Ã— 10

^{11}ms

^{-2}

= 2.316 Ã— 10

^{12}ms

^{-2}

Gravitational force on a mass of m at the equator = m g

= (m Ã— 2.316 Ã— 10

^{12}) N

Reaction to centripetal force on the body = m v

^{2}/R = mÏ‰

^{2}R

= m Ã— (2Ï€ Ã— 1.5)

^{2}Ã— 1.2 Ã— 10

^{4}N

= m Ã— 107 Ã— 10

^{4}N

= m Ã— 1.07 Ã— 10

^{6}N

The inward gravitational force is 2 Ã— 10

^{6 }times greater than the outward reaction due to centripetal force. Hence the mass will remain stuck to the surface.

# Question-26

**A space ship is stationed on Mars. How much energy must be expended on the space ship to rocket it out of the solar system? Mass of the space ship = 1000 kg, mass of the sun = 2 Ã—10**

^{30}kg, mass of Mars = 6.4 Ã— 10^{ 23}kg, radius of Mars = 3395 km, radius of the orbit of Mars = 2.28 Ã— 10^{8}km, G = 6.67 Ã— 10^{-11}N m^{2}kg^{-2}.**Solution:**

Potential energy = work done to bring the mass from infinity to within the range of gravitation

= - G M m / R

Consider the system - sun and space ship.

Potential energy due to sun's gravitational force = - G M m / R

[ M = mass of the sun , m = mass of the space ship, R = Radius of orbit]

Consider the system - Mars and space ship.

Potential energy due to gravitational force of mars = - G M

_{m}m / R

_{m}

[ M

_{m}= mass of the Mars, m = mass of the space ship, R

_{m}= Radius of Mars]

Total energy of the space ship stationed at surface Mars =

= Potential energy due to sun's gravitational force + Potential energy due to gravitational force of mars

= (- G M m / R) + (-G M

_{m}m / R

_{m})

Energy required to rocket the space ship out of solar system is given as,

= -5.97 Ã— 1011 J

# Question-27

**A rocket is fired 'vertically' from the surface of Mars with a speed of 2 km s**

^{-1}. If 20 % of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of Mars before returning to it? Mass of Mars = 6.4 Ã— 10^{23}kg, radius of Mars = 3395 km, G = 6.67 Ã— 10^{-11}N m^{2}kg^{-2}.**Solution:**

Mass of Mars M = 6.4 x 10

^{23}kg,

Radius of Mars R = 3395 km = 3.395 Ã— 10

^{6}m

G = 6.67 x10

^{-11}N m

^{2}kg

^{-2}.

Initial speed v = 2 km/s = 2 Ã— 10

^{3}ms

^{-1}

Initial potential energy = - G M m / R

Initial kinetic energy = Â½ m v

^{2}

Initial total energy = Â½ m v

^{2}- G M m / R

Available initial energy = 0.8 (Â½ m v

^{2}) - G M m / R [ since 20% K.E. is lost]

= 0.4 mv

^{2}- G M m /R

On reaching the highest point total energy = P. E [ since v =o and K.E. =0]

= - G M m / R+h

By conservation of energy

0.4 mv

^{2 }- G M m /R = - G M m / R+h

G M / R+h = (G M - 0.4 v

^{2}R) / R

(R+ h - R) / R = {G M - ( G M - 0.4 v

^{2}R) } / (G M - 0.4 v

^{2}R)

h / R = 0.4 v

^{2}R / (G M - 0.4 v

^{2}R)

h = 0.4 v

^{2}R

^{2}/ (G M - 0.4 v

^{2}R)

With the given values,

Maximum height reached â€˜hâ€™

= 4.95 x 10

^{5}m

= 495 Km.