NCERT Solutions for Class 11th: Ch 8 Gravitation Physics Science

Page No: 201

**Excercises**

**8.1. 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?

**Answer**

(a) We cannot shiwld a body from the gravitational influence of nearby matter, because the gravitational force on a body due to nearby matter is independent of the presence of other matter, whereas it is not so in case of electric forces. It means the gravitational screens are not possible.

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

(c) 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.

**8.2. 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 –G* Mm*(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}distance away from the centre of the earth.

**Answer**

(a) decreases

(b) decreases

(c) mass of the body

(d) more.

8.3. Suppose

there existed a planet that went around the sun twice as fast as the

earth.What would be its orbital size as compaorange to that of the earth?

**Answer**

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} = ½*T*_{e} = ½ year

Orbital radius of the

planet = *R*_{p }

From Kepler’s

third law of planetary motion, we can write:

(*R*_{p} / *R*_{e})^{3} = (*T*_{p} / *T*_{e})^{2}

(*R*_{p} / *R*_{e}) = (*T*_{p} / *T*_{e})^{2/3}

= (½ / 1)^{2/3} = 0.5^{2/3} = 0.63

Hence, the orbital

radius of the planet will be 0.63 times smaller than that of the

Earth.

8.4. Io, one of the satellites of Jupiter, has an orbital period of

1.769 days and the radius of the orbit is 4.22 × 10^{8}

m. Show that the mass of Jupiter is about one-thousandth that of the

sun.

**Answer**Orbital period of

I

_{0},

*T*

_{I0}= 1.769 days = 1.769 × 24 × 60 × 60 s

I

_{0},

*R*

_{I}

_{0}= 4.22 × 10

^{8}m

_{0}is

revolving around the Jupiter

*M*

_{J}= 4π

^{2}

*R*

_{I0}

^{3}/ G

*T*

_{I0}

^{2}…..

**(i)**

*M*

_{J}= Mass of Jupiter

_{e }= 365.25 days = 365.25 × 24 × 60 × 60 s

*R*

_{e }= 1 AU = 1.496 × 10

^{11 }m

*M*

_{s}= 4π

^{2}

*R*

_{e}

^{3}/ G

*T*

_{e}

^{2 }……

**(ii)**

*M*

_{s}/

*M*

_{J}= (4π

^{2}

*R*

_{e}

^{3}/ G

*T*

_{e}

^{2}) × (G

*T*

_{I0}

^{2}/ 4π

^{2}

*R*

_{I0}

^{3}) = (

*R*

_{e}

^{3}×

*T*

_{I0}

^{2}) / (

*R*

_{I0}

^{3}×

*T*

_{e}

^{2})

^{2}× (1.496 × 10

^{11}/ 4.22 × 10

^{8})

^{3}

*M*

_{s}/

*M*

_{J}~ 1000

*M*

_{s }~ 1000 ×

*M*

_{J}

one-thousandth that of the Sun.8.5. Let us assume that our galaxy consists of 2.5 × 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}ly.

**Answer**

Mass of our galaxy

Milky Way, *M* = 2.5 × 10^{11} solar mass

Solar mass = Mass of

Sun = 2.0 × 10^{36}

kg

Mass of our galaxy, *M*

= 2.5 × 10^{11} ×

2 × 10^{36} = 5

× 10^{41} kg

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

Radius of Milky Way, *r*

= 5 × 10^{4} ly

1 ly = 9.46 ×

10^{15} m

∴*r* = 5 ×

10^{4} × 9.46 ×

10^{15}

=

4.73 ×10^{20} m

Since a star revolves

around the galactic centre of the Milky Way, its time period is given

by the relation:*T *= ( 4π^{2}r^{3}/ G*M*)^{1/2}

= [ (4 × 3.14^{2} × 4.73^{3} × 10^{60}) / (6.67 × 10^{-11} × 5 × 10^{41}) ]^{1/2}

= (39.48 × 105.82 × 10^{30}/ 33.35 )^{1/2}

= 1.12 × 10^{16} s

1 year = 365 × 324 × 60 × 60 s

1s = 1 / (365 × 324 × 60 × 60) years

∴ 1.12 × 10^{16} s = 1.12 × 10^{16} / (365 × 24 × 60 × 60) = 3.55 × 10^{8} years.

**8.6. 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 requiorange to launch an orbiting satellite out of earth’s

gravitational influence is more/less than the energy requiorange to

project a stationary object at the same height (as the satellite) out

of earth’s influence.

**Answer**

(a) Kinetic energy

(b) Less

**8.7. 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?

**Answer**

The escape velocity is indpendent of the mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since, this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.

8.8. 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.

**Answer**

A comet while going on elliptical orbit around the Sun has constant angular momentum and totaal energy at all locations but other quantities vary with locations.

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

**Answer**

(a) 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.

(b) 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.

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

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

8.10. 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 8.12) (i) a, (ii) b, (iii) c, (iv) O.

**Answer**

Gravitational potential (*V*) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (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 arrow **c**.

**8.11. Choose the correct answer from among the given ones:**

For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

**Answer**

Gravitational potential (*V*) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (*dV*/*dR*) 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 arrow **e**.

rocket is fiorange 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 ×10

^{30}

kg, mass of the earth = 6 × 10

^{24}

kg. Neglect the effect of other planets etc. (orbital radius = 1.5

× 10

^{11}

m).

**Answer**Mass of the Sun,

*M*

_{s}

= 2 × 10

^{30}kg

Mass of the Earth, *M*_{e}

= 6 × 10 ^{24} kg

Orbital radius,* r*

= 1.5 × 10^{11} m

Mass of the rocket = *m*

*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:

G*mM*_{s} / (*r* – *x*)^{2} = G*mM*_{e} / *x*^{2}

[ (*r* – *x*) / *x* ]^{2} = *M*_{s} / *M*_{e}

(*r* – *x*) / *x* = [ 2 × 10^{30} / 60 × 10^{24}]^{1/2} = 577.35

1.5 × 10^{11} – *x* = 577.35*x*

578.35*x* = 1.5 × 10^{11}*x* = 1.5 × 10^{11} / 578.35 = 2.59 × 10^{8} m.

8.13.How

will you ‘weigh the sun’, that is estimate its mass? The

mean orbital radius of the earth around the sun is 1.5 × 10^{8}

km.

**Answer**

Orbital radius of the Earth around the Sun, *r* = 1.5 × 10^{11} m

Time taken by the Earth to complete one revolution around the Sun,*T* = 1 year = 365.25 days

= 365.25 × 24 × 60 × 60 s

Universal gravitational constant, G = 6.67 × 10^{–11} Nm^{2} kg^{–2}

Thus, mass of the Sun can be calculated using the relation,*M* = 4π^{2}*r*^{3} / G*T*^{2}

= 4 × 3.14^{2} × (1.5 × 10^{11})^{3}/ [ 6.67 × 10^{-11} × (365.25 × 24 × 60 × 60)^{2}]

= 2 × 10^{30} kg

Hence, the mass of the Sun is 2 × 10^{30} kg.

8.14. A

Saturn year is 29.5 times the earth year. How far is the Saturn from

the sun if the earth is 1.50 ×10^{8}

km away from the sun?

**Answer**

Distance of the Earth

from the Sun, *r*_{e} = 1.5 ×

10^{8} km = 1.5 ×

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* = (4π^{2}*r*^{3} / G*M*)^{1/2}

For Saturn and Sun, we

can write*r*_{s}^{3} / *r*_{e}^{3} = *T*_{s}^{2} / *T*_{e}^{2}*r*_{s} = *r*_{e}(*T*_{s} / T_{e})^{2/3}

= 1.5 × 10^{11} (29.5 *T*_{e} / *T*_{e})^{2/3}

= 1.5 × 10^{11} (29.5)^{2/3}

= 14.32 × 10^{11} m

Hence, the distance

between Saturn and the Sun is 1.43 × 10^{12} m.

8.15. 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?

**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*‘ = *g* / [1 + ( *h* /* R*_{e}) ]^{2}

Where,*g* = Acceleration due to

gravity on the Earth’s surface

*R*e = Radius of

the Earth

For *h* = *R*_{e} / 2*g*‘ = *g* / [(1 + (*R*_{e} / 2*R*_{e}) ]^{2}

= *g* / [1 + (1/2) ]^{2} = (4/9)*g*

Weight of a body of

mass *m* at height *h* is given as:*W*‘ = *mg*

= *m* × (4/9)*g* = (4/9)*mg*

= (4/9)*W*

= (4/9) × 63 = 28 N.

8.16. 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

N on the surface?

**Answer**

Weight of a body of

mass *m* at the Earth’s surface, *W* = *mg* =

250 N

Body of mass *m *is located at depth, *d* = (1/2)*R*_{e}

Where,*R*_{e} = Radius of the Earth

Acceleration due to

gravity at depth *g *(*d*) is given by the relation:

*g*‘ = (1 – (*d* / *R*_{e})g

= [1 – (*R*_{e} / 2*R*_{e}) ]*g* = (1/2)*g*

Weight of the body at

depth *d*,

*W*‘ = *mg*‘

= *m* × (1/2)*g* = (1/2) *mg* = (1/2)*W*

= (1/2) × 250 = 125 N

8.17. A

rocket is fiorange vertically with a speed of 5 km 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 ×

10^{24}

kg; mean radius of the earth = 6.4 ×

10^{6}*m*; G=

6.67 ×

10^{–11}

N *m*^{2}

kg^{–}^{2.}^{}

**Answer**

Velocity of the rocket,*v* = 5 km/s = 5 × 10^{3}

m/s

Mass of the Earth, *M*_{e} = 6 × 10^{24} kg

Radius of the Earth, *R*_{e} = 6.4 × 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

= (1/2)*mv*^{2} + (-G*M*_{e}*m* / *R*_{e})

At highest point *h*,*v* = 0

And, Potential energy = -G*M*_{e}*m* / (*R*_{e} + *h*)

Total energy of the

rocket = 0 + [ -G*M*_{e}*m* / (*R*_{e} + *h*) ]

= -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*

(1/2)*mv*^{2} + (-G*M*_{e}*m* / *R*_{e}) = -G*M*_{e}*m* / (*R*_{e} + *h*)

(1/2)*v*^{2} = G*M*_{e} [ (1/*R*_{e}) – 1 / (*R*_{e} + *h*) ]

= G*M*_{e}[ (*R*_{e} + *h* – *R*_{e}) / *R*_{e}(*R*_{e}+ *h*) ]

(1/2)*v*^{2} = *gR*_{e}*h* / (*R*_{e} + *h*)

Where *g* = G*M* / *R*_{e}^{2} = 9.8 ms^{-2}

∴ *v*^{2} (*R*_{e} + *h*) = 2*gR*_{e}h*v*^{2}*R*_{e} = *h*(2*gR*_{e} – *v*^{2})*h* = *R*_{e}*v*^{2} / (2*gR*_{e} – *v*^{2})

= 6.4 × 10^{6} × (5 × 10^{3})^{2} / [ 2 × 9.8 × 6.4 × 10^{6} – (5 × 10^{3})^{2}*h* = 1.6 × 10^{6} m

Height achieved by the

rocket with respect to the centre of the Earth = *R*_{e} + *h*

= 6.4 × 10^{6} + 1.6 × 10^{6} = 8 × 10^{6} m.

8.18. 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 the sun and

other planets.

**Answer**

Escape velocity of a

projectile from the Earth, *v*_{esc }= 11.2 km/s

Projection velocity of

the projectile, *v*_{p} = 3v_{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 = (1/2)*mv*_{p}^{2} – (1/2)*mv*_{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 = (1/2)*mv*_{f}^{2}

From the law of

conservation of energy, we have

(1/2)*mv*_{p}^{2} – (1/2)*mv*_{esc}^{2} = (1/2)*mv*_{f}^{2}*v*_{f} = ( *v*_{p}^{2} – *v*_{esc}^{2 })^{1/2}

= [ (3*v*_{esc})^{2} – *v*_{esc}^{2}]^{1/2}

= √8 *v*_{esc}

= √8 × 11.2 = 31.68 km/s.

8.19. 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 = 6.0 ×10^{24} kg; radius of the earth = 6.4 ×10^{6} m; G = 6.67 × 10^{–11} N m^{2} kg^{–2}.

**Answer**

Mass of the Earth, *M*

= 6.0 × 10^{24} kg

Mass of the satellite,*m* = 200 kg

Radius of the Earth, *R*_{e}

= 6.4 × 10^{6} m

Universal gravitational

constant, G = 6.67 × 10^{–11} Nm^{2}kg^{–2}

Height of the

satellite, *h* = 400 km = 4 × 10^{5} m = 0.4 ×10^{6}

m

Total energy of the

satellite at height *h* = (1/2)*mv*^{2} + [ -G*M*_{e}*m* / (*R*_{e} + *h*) ]

Orbital velocity of the

satellite, *v* = [ G*M*_{e} / (*R*_{e} + *h*) ]^{1/2}

Total energy of height,*h* = (1/2)G*M*_{e}*m* / (*R*_{e} + *h*) – G*M*_{e}*m* / (*R*_{e} + *h*) = -(1/2)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 requiorange to send

the satellite out of its orbit = – (Bound energy)

= (1/2) G*M*_{e}*m* / (*R*_{e} + *h*)

= (1/2) × 6.67 × 10^{-11} × 6 × 10^{24} × 200 / (6.4 × 10^{6} + 0.4 × 10^{6})

= 5.9 × 10^{9} J.

8.20. Two stars each of one

solar mass (= 2× 10^{30} kg) are approaching each other

for a head on collision. When they are a distance 109 km, their

speeds are negligible. What is the speed with which they collide? The

radius of each star is 104 km. Assume the stars to remain undistorted

until they collide. (Use the known value of G).

**Answer**

Mass of each star, *M* = 2 × 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*

= [ -G*MM* / *r* ] + (1/2)*mv*^{2}

= [ -G*MM* / *r* ] + 0 ….**(i)**

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 = (1/2) *Mv*^{2} + (1/2)*Mv*^{2} = *Mv*^{2}

Total potential energy of both stars = -G*MM* / 2*R*

Total energy of the two stars = *Mv*^{2} – G*MM* / 2*R* ….**(ii)**

Using the law of conservation of energy, we can write:*Mv*^{2} – G*MM* / 2*R* = -G*MM* / *r**v*^{2} = -G*M* / *r* + G*M* / 2*R*

= G*M* [ (-1/*r*) + (1/2*R*) ]

= 6.67 × 10^{-11} × 2 × 10^{30}[ (-1/10^{12} ) + (1 / 2 × 10^{7}) ]

~ 6.67 × 10^{12}

v = ( 6.67 × 10^{12})^{1/2} = 2.58 × 10^{6} m/s.

8.21. 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 force and potential at the mid point of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium

stable or unstable?

**Answer**

Grvitational field at the mid-point of the line joining the centres of the two spheres

= G*M*/(*r*/2)^{2} (alog negative *r*) + G*M*/(*r*/2) (along *r*) = 0

Gravitational potential at the midpoint f the line joining the centres of the two spheres is

*V* = – G*M*/*r*/2 + (-G*M*/*r*/2) = -4G*M*/*r* = -4 × 6.67 × 10^{-11} × 100/1.0

= -2.7 × 10^{-8} J/Kg

As the effective force on the body placed at mid-point is zero, sso the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its inital position of equilibrium. Hence, the body is in unstable equilibrium.

8.22. 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

× 10^{24}

kg, radius = 6400 km.

**Answer**

Mass of the Earth, *M*

= 6.0 × 10^{24} kg

Radius of the Earth, *R*

= 6400 km = 6.4 × 10^{6} m

Height of a

geostationary satellite from the surface of the Earth,

*h* = 36000 km =

3.6 × 10^{7} m

Gravitational potential

energy due to Earth’s gravity at height *h,*

= -G*M* / (*R* + *h*)

= – 6.67 × 10^{-11} × 6 × 10^{24} / (3.6 × 10^{7} + 0.64 × 10^{7})

= -9.4 × 10^{6} J/kg.

8.23. A star 2.5 times the

mass of the sun and collapsed to a size of 12 km rotates with a speed

of 1.2 rev. per second. (Extremely compact stars of this kind are

known as neutron stars. Certain stellar objects called pulsars belong

to this category). Will an object placed on its equator remain stuck

to its surface due to gravity? (Mass of the sun = 2 × 10^{30}

kg).

**Answer**

A body gets stuck to

the surface of a star if the inward gravitational force is greater

than the outward centrifugal force caused by the rotation of the

star.

Gravitational force, *f*_{g} = – GMm / R^{2}

Where,*M *= Mass of the

star = 2.5 × 2 × 10^{30} = 5 × 10^{30 }kg*m *= Mass of the

body*R =* Radius of

the star = 12 km = 1.2 ×10^{4} m

∴ *f*_{g} = 6.67 × 10^{-11} × 5 × 10^{30} × m / (1.2 × 10^{4})^{2} = 2.31 × 10^{11} m N

Centrifugal force, *f*_{c}^{}= *mr**ω*^{2}*ω*

= Angular speed = 2π*ν*

ν = Angular

frequency = 1.2 rev s^{–1}*f*_{c} =*mR* (2π*ν*)^{2}

= *m* ×

(1.2 ×10^{4}) × 4 × (3.14)^{2} ×

(1.2)^{2} = 1.7 ×10^{5}*m* N

Since *f*_{g}

> *f*_{c}, the body will remain stuck to the surface

of the star.

8.24. A

spaceship is stationed on Mars. How much energy must be expended on the

spaceship to launch it out of the solar system? Mass of the space ship =

1000 kg; mass of the Sun = 2 × 10^{30} kg; mass of mars = 6.4 × 10^{23} kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10^{8}kg; G= 6.67 × 10^{–11} m^{2}kg^{–2}.

**Answer**

Mass

of the spaceship, *m*_{s}= 1000 kg

Mass

of the Sun, *M*

= 2 × 10^{30}

kg

Mass

of Mars, *m*_{m}

= 6.4 × 10 ^{23}

kg

Orbital

radius of Mars, *R*

= 2.28 × 10^{8 }kg

=2.28 × 10^{11}m

Radius

of Mars, *r *=

3395 km = 3.395 × 10^{6}

m

Universal

gravitational constant, G = 6.67 × 10^{–11}

m^{2}kg^{–2}

Potential

energy of the spaceship due to the gravitational attraction of the

Sun = -G*Mm*_{s} / *R*

Potential

energy of the spaceship due to the gravitational attraction of Mars = -G*M*_{m}*m*_{s} / *r*

Since

the spaceship is stationed on Mars, its velocity and hence, its

kinetic energy will be zero.

Total

energy of the spaceship = -G*Mm*_{s}/ *R* **–**

G*M*_{m}*m*_{s} / *r*

= -G*m*_{s}[ (*M* / *R*) + (*m*_{m} / *r*) ]

The negative sign

indicates that the system is in bound state.

Energy requiorange for

launching the spaceship out of the solar system

= – (Total energy

of the spaceship)

= G*m*_{s}[ (*M* / *R*) + (*m*_{m} / *r*) ]

= 6.67 × 10^{-11 }× 10^{3} × [ (2 × 10^{30}/ 2.28 × 10^{11}) + (6.4 × 10^{23}/ 3.395 × 10^{6} ) ]

= 596.97 × 10^{9} = 6 × 10^{11} J.

8.25. A rocket is fiorange

‘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× 1023 kg;

radius of mars = 3395 km; G = 6.67× 10^{-11} N m^{2}

kg^{–2}.

**Answer**

Initial velocity of the

rocket, *v* = 2 km/s = 2 × 10^{3} m/s

Mass of Mars, *M*

= 6.4 × 10^{23} kg

Radius of Mars, *R*

= 3395 km = 3.395 × 10^{6} m

Universal gravitational

constant, G = 6.67× 10^{–11} N m^{2} kg^{–2}

Mass of the rocket = *m*

Initial kinetic energy

of the rocket = (1/2)*mv*^{2}

Initial potential

energy of the rocket = -G*Mm* / *R*

Total

initial energy = (1/2)*mv*^{2}– G*Mm* / *R*

If

20 % of initial kinetic energy is lost due to Martian atmospheric

resistance, then only 80 % of its kinetic energy helps in reaching a

height.

Total

initial energy available = (80/100) × (1/2) *mv*^{2} – G*Mm */ *R* = 0.4*mv*^{2} – G*Mm* / *R*

Maximum

height reached by the rocket = *h*

At

this height, the velocity and hence, the kinetic energy of the rocket

will become zero.

Total

energy of the rocket at height *h = *-G*Mm* / (*R* + *h*)

Applying

the law of conservation of energy for the rocket, we can write:

0.4*mv*^{2} – G*Mm* / *R* = -G*Mm* / (*R* + *h*)

0.4*v*^{2} = GM / R – G*M* / (*R* + *h*)

= G*Mh* / *R*(*R* + *h*)

(*R* + *h*) / *h* = G*M* / 0.4*v*^{2}*R**R* / *h* = ( G*M* / 0.4*v*^{2}*R* ) – 1

h = R / [ (GM / 0.4v^{2}*R*) – 1 ]

= 0.4*R*^{2}*v*^{2}/ (GM – 0.4*v*^{2}R)

= 0.4 × (3.395 × 10^{6})^{2} × (2 × 10^{3})^{2}/ [ 6.67 × 10^{-11} × 6.4 × 10^{23} – 0.4 × (2 × 10^{3})^{2} × (3.395 × 10^{6}) ]

= 18.442 × 10^{18} / [ 42.688 × 10^{12} – 5.432 × 10^{12} ]

= 18.442 × 10^{6 }/ 37.256

= 495 × 10^{3} m = 495 km.