Class 11 - Physics - Gravitation

**Question1.**

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:

- No we cannot shield a body from the gravitational influence of nearby matter.
- As electrical forces depend upon the nature of the intervening medium while the gravitational forces don’t depend upon the nature of the intervening medium.
- So, such shielding acts are not possible in case of gravitation i.e., gravity screens are not possible.
- Yes, if the size of space station is large enough, then the astronaut will be able to detect the change in earth’s gravity.
- Distance between earth and moon is very less as compared to distance between earth and sun.
- Tidal effect is inversely proportional to the cube of the distance between this shows it is not governed by the inverse square law which is obeyed by gravitational force. Hence tidal effect of moon is larger than that due to the sun.

**Question2.**

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:

. Acceleration due to gravity at depth h is given as :__Decreases__

g = (1-(2h)/ (R_{e})) g Where,

R_{e}= 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.

. Acceleration due to gravity at depth d is given as :__Decreases__

g_{d} = (1-(d/R_{e})) g

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

- Acceleration due to gravity is independent of the
.__mass of the body__

Acceleration due to gravity of body of mass m is given by the relation:

g = (GM)/ (R^{2})

Where,

G = Universal gravitational constant

M = Mass of the Earth

R = Radius of the Earth

It is clear that acceleration due to gravity is independent of the mass of the body.

- The formula –G Mm (1/r
_{2}– 1/r_{1}) is more accurate than the formula mg (r_{2}– r_{1}).

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}) = - (GMm)/ (r_{1})

V (r_{2}) = - (GMm)/ (r_{2})

Therefore difference in potential energy, V = V (r_{2}) - V (r_{1})

=-GmM ((1/ r_{2}) – (1/ r_{1}))

Hence, this formula is more accurate than the formula mg (r_{2}– r_{1}).

**Question 3.**

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

Answer:

Let,

Radius of orbit of earth = r_{e} = 1 AU

Radius of orbit of planet = r_{p}

Also Time taken by the earth to complete one revolution around the sun,

T_{e }= 1 year

Time taken by the planet to complete one revolution around the sun,

T_{p} = (1/2) T_{e}

= (1/2) year

From Kepler’s third law of planetary motion,

(r_{p}/ r_{e})^{3} =( T_{p}/ T_{e})^{2}

(r_{p}/ r_{e}) = (T_{p}/ T_{e}) (^{2/3)}

= ((1/2)/ (1))^{ (2/3)}

= (0.5)^{ (2/3)}

=0.63

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

**Question 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:

Given:

Orbital period of Io, T_{Io }= 1.769 days = 1.769 x 24 x 60 x 60 sec

Orbital radius of Io, R_{lo} = 4.22 x 10^{8} m

Satellite Io is revolving around the Jupiter

Mass of the satellite is given as:

M_{J} = (4 π^{2} R_{Io}^{3})/ (GT_{Io}^{2})

Where,

M_{J} = Mass of Jupiter

G = Universal gravitational constant

Orbital period of the Earth, T_{e} = 365.25 days = 365.25 x 24 x 60 x60s

Orbital radius of the Earth,

R_{e} = 1 AU = 1.496 x 10^{11} m

Mass of the Sun M_{s} = (4π ^{2} R_{e}^{3})/ (GT_{e}^{2}) … (ii)

Therefore, (M_{s}/ M_{J}) = (4 π^{2} R_{e}^{3})/ (GT_{e}^{2}) x (GT_{Io}^{2})/ (4π ^{2} R_{Io}^{3})

= ((1.769 x 24 x 60 x 60)/ (365.25 x 24 x 60 x60))^{ 2} x ((1.496 x 10^{11})/ (4.22 x 10^{8}))^{3}

= 1045.04

Therefore, (M_{s}/ M_{J}) ≈ 1000

M_{s} ≈ 1000 M_{J}

This shows the mass of Jupiter is (1/1000^{th}) the mass of the sun.

**Question 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:

Given:

Mass of the galaxy, M = 2.5 × 10^{11} solar mass

=2.5 × 10^{11} x (2 x 10^{30})

=5 x 10^{41}kg

Radius of the orbit of star, r= 50000ly

=50000 x 9.46 x 10^{15}m

=4.73 x 10^{20}m

Therefore time period of the star T =2 π √ (r^{3})/ (GM)

=2 π √ (4.73 x 10^{20})^{3}/ (6.67 x 10^{-11} x 5 x 10^{41})

=1.12 x 10^{16} sec

**Question 6.**

Choose the correct alternative:

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

(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy

required to project a stationary object at the same height (as the satellite) out of earth’s influence.

Answer:

. Total mechanical energy of a satellite is sum of its kinetic energy (which is always positive) and potential energy (may be negative).__Kinetic energy__- At infinity gravitational potential energy of the satellite is 0. As the satellite is bound to earth therefore total energy of the satellite is negative.
- The total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.
. An orbiting satellite acquires a certain amount of energy that enables it to revolve around the earth. This energy is provided by its orbit.__Less__- So it requires lesser energy to move out of the influence of earth’s gravitational field than a stationary object on the earth’s surface which does not have any energy initially.

**Question7.**

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 of a body from earth does not depend on the mass of the body because V__No___{esc}= √ (2GM/R).. The value of g depends on the latitude which is different at different locations on earth. Escape speed is given as :-√( 2gR).__Yes__. It does not depend on the direction of projection.__No__. As escape velocity Vesc = √ (2gR).and the value of ‘g’ depends upon the height of location from where body is projected.__Yes__

**Question 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:

- Linear speed (v= is not constant because it covers the distance R of the comet from the sun changes due to its elliptical orbit around sun.
- The angular speed is not constant as it covers different angles in equal interval of time.
- The angular momentum remains constant.
- As the linear speed varies, the kinetic energy ((1/2) mv
^{2}) is not constant. - As the distance of the comet from the sun changes therefore the potential energy is not constant.
- Total energy throughout the orbit remains constant.

**Question 9.**

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

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

In the following two exercises, choose the correct answer from among the given ones:

Answer:

- On earth, the weight of the body of a person is carried by his legs. But in space the astronaut is at the state of weightlessness.
- Therefore, the feet of the astronaut do not get swollen.
- In free space, the face of the astronaut is expected to get more blood supply. Therefore, the face of the astronaut may get swollen.
- In the gravity free space, the swollen face of the astronaut may cause headache.
- Space has different orientations. Therefore, orientational problem can affect an astronaut in space.

**Question 10.**

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) 0.

Answer:

Correct option: __(iii).__

Gravitational potential (V) is constant at all points inside a spherical shell.

Therefore gravitational potential gradient (-dV/dr) at all points inside a spherical shell is zero.

Since gravitational intensity E (-dV/dr), it shows gravitational intensity is 0 at all points inside the spherical shell.

When we remove the upper part of the spherical shell, we will get hemispherical shell.

As the gravitational intensity at centre C is 0, the direction of gravitational intensity must be downwards. Therefore option (iii) c is correct.

**Question 11.**

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.

Answer:

Correct option: - __(ii) e__

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.

**Question 12.**

A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero?

Mass of the sun = 2 × 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:

Given:

Mass of Sun M = 2× 10^{30} kg

Mass of the earth m =6× 10^{24} kg

Distance between Sun and Earth r = 1.5 × 10^{11} m

Consider a point R, where the gravitational force on the rocket due to earth = gravitational force on the rocket due to sun.

Distance of point R from the Earth = x

Therefore, (Gm)/ (r^{2}) = (GM)/(r-x)^{ 2}

(r-x)^{ 2}/(x^{2}) = (M/m)

= (2 x 10^{30})/ (6 x 10^{24})

= (10^{6})/ (3)

=(r-x)/(x) = ((10)^{3}/ (√3)

(r/x) = ((10)^{3}/ (√3) + 1

≈ (10)^{3} / (√3)

x = (√3) r/ (10)^{3}

= (1.732 x 1.5 x 10^{11})/ (10^{3}) m

=2.6 x 10^{8} m

**Question 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:

Given:

Radius of the earth orbit R= 1.5 × 10^{11}

Time period of the earth around the sun = 365 years

= 365 x 24 x 60 x 60 s

Gravitational constant G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

Time period of the satellite = T = 2π√(R+h)^{ 2}/ (GM)

Or Mass of the sun, M = 4 π^{2}(R+h)^{ 3} / (GT^{2})

= (4 π^{2} x (1.5 × 10^{11})^{3})/ (6.67 x 10^{-11}) x (365 x 24 x 60 x60)^{2}

=2.01 x 10^{30} kg

**Question 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:

Given:

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 r^{3}/GM)^{ 1/3}

For Saturn and Sun, it can be written as:

(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} x (29. 5)^{ (2/3)}

= 1.5 × 10^{11} x 9.55

=14.32 x 10^{11}m

Hence, the distance between Saturn and the Sun = 14.32 x 10^{11}m

**Question 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:

Given:

Weight of the body = 63N

Height h = (R_{e}/2) where Radius of the earth = R_{e}

Acceleration due to gravity at height h from the Earth’s surface is as:

g’= (g)/((1+h)/R_{e})^{2}

Where,

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

g’ = (g)/(1+ (R_{e}/2xR_{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’

=mx (4/9) g

= (4/9) mg

= (4/9) x 63

=28N

**Question 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:

Given:

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

Where g = acceleration due to gravity on earth’s surface, mass of the body = m and weight of the body = W

Depth, d = (1/2) R_{e}

Where, R_{e} = Radius of the Earth

Acceleration due to gravity at depth =g’ and is given as:

g' = (1- (d/ R_{e}))g

= (1- (R_{e})/ (2x R_{e})) g

= (1/2) g

Weight of the body at depth d,

W’ = mg’

= m x (1/2) g

= (1/2) W

= (1/2) x 250

=125N

Q**uestion 17.**

A rocket is fired 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} = 6.0 x 10^{24 }kg

Radius of the Earth, R_{e} = 6.4 x 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} + (-GM_{e}m)/ (R_{e})

At highest point h, v =0

And Potential energy = - (GM_{e}m) (R_{e} +h)

Total energy of the rocket = 0 + (- GM_{e}m) (R_{e} +h)

= (- GM_{e}m) (R_{e} +h)

By applying law of conservation of energy,

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

(1/2) mv^{2} + (-GM_{e}m)/ (R_{e}) = (-GM_{e}m)/ (R_{e }+ h)

(1/2) v^{2} = GM_{e} ((1/ R_{e}) – (1/ R_{e }+ h))

= GM_{e} ((R_{e }+ h - R_{e})/ (R_{e} (R_{e }+ h)))

(1/2) v^{2} = ((GM_{e} h)/ (Re (R_{e }+ h))) x (R_{e}/ R_{e})

(1/2) v^{2} = (g R_{e} h)/ (R_{e }+ h)

Where g (acceleration due to gravity on earth’s surface)

= (GM_{e} /R_{e}^{2})=9.8m/s^{2}

Therefore, (v^{2} R_{e}) = h (2gR_{e}h)

h= (R_{e} v^{2})/ (2gR_{e} - v^{2})

= (6.4 x 10^{6} x (5 × 10^{3})^{2})/ (2 x 9.8 x 6.4 x 10^{6} – (5x 10^{3})^{2})

h = (6.4 x 25 x 10^{12})/ (100.44 x 10^{6})

h = 1.6 x 10^{6} m

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

= (R_{e }+ h)

= (6.4 x 10^{6} + 1.6 x 10^{6})

=8.0 x 10^{6} m

**Question 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:

Given:

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) m v^{2}_{esc}

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) m v^{2}_{esc} = (1/2) mv_{f}^{2}

v_{f} = √ (v_{p}^{2} - v^{2}_{esc})

=√ (3v_{esc})^{ 2} – (v^{2}_{esc})

=√ (8) v_{esc}

=√ (8) x 11.2

=31.68km/s

**Question 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 m2 kg^{–2}.

Answer:

Given:

Mass of the Earth, M_{e} = 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

Potential energy of the satellite when revolving = (-GM_{e}m)/ (R_{e} + h)

Kinetic energy of the satellite when revolving = (1/2) mv^{2}

Orbital velocity of the satellite, v=√ (GM_{e})/ (R_{e} +h)

Total energy of the orbiting satellite = (1/2) mv^{2} + (-GM_{e}m)/ (R_{e} + h)

=- (1/2) (GM_{e}m)/ (R_{e} +h)

(-) ive sign indicates that the satellite is bound to the earth. This energy is known as Bound energy.

Energy to be imparted to put the satellite out of earth’s gravitational influence

= – (Bound energy)

= (1/2) (GM_{e}m)/ (R_{e} +h)

= (1/2) x (6.67 × 10^{–11} x 6.0 × 10^{24} x 200)/ (6.4 × 10^{6} + 0.4 ×10^{6})

= (1/2) x (6.67 x6 x 2 x 10)/ (6.8 x 10^{6})

= (5.9 x 10^{9}) J

**Question 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 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).

Answer:

Given:

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 v = 0 total energy of two stars separated at distance r

= (-GMm)/(r) + (1/2) mv^{2}

= (-GMm)/(r) + 0 … (i)

When the stars are about to collide:

Velocity of the stars = v

Distance between the centres of the stars will be twice the radius of a star i.e.

= 2R

Total kinetic energy of both stars = (1/2) Mv^{2} + (1/2) Mv^{2} = Mv^{2}

Total potential energy of both stars = (-GMm)/ (2R)

Total energy of the two stars = Mv^{2} - (GMm)/ (2R)... (ii)

Using the law of conservation of energy,

Mv^{2} - (GMm)/ (2R) = - (GMM)/(r)

v^{2} = - (GM)/(r) + (GM)/ (2R)

=GM ((-1/r) + (1/2R))

=6.67 x 10^{-11} x 2 x 10^{30}[(-1/10^{12}) + (1/2) x10^{7})]

=13.34 x 10^{19} [-10^{-12} + (5 x 10^{-8})]

=-6.67 x 10^{12}

v= √ (6.67 x 10^{12})

v= (2.58 x 10^{6}) m/s

**Question 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 centres of the spheres?

Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Answer:

Given:

Gravitational Constant G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

Mass of spheres = 100kg

Radius of each sphere = 0.10m

Distance between two spheres d = 1.0m

Let mid-point between two spheres = T

Therefore TA = TB =r = (d/2) = 0.5m

Force due to gravitational field at T due to sphere A = (GM)/ (0.5)^{2}

Force due to gravitational field at T due to sphere B = (GM)/ (0.5)^{2}

At mid- point gravitational force will be zero because gravitational force exerted by each sphere will act in opposite directions.

Gravitational potential at mid-point due to two spheres will be:

= (-GM/(r) + (GM)/(r) = - (2GM/r)

= - (6.67 x 10^{-11} x 100 x2)/ (0.5)

=-2.668 x 10^{-8} Jkg^{-1}

Any object placed at mid-point will be in equilibrium state, but the equilibrium is unstable.

If the objects are displaced from mid-point towards either sphere, the object will not return to its initial position of equilibrium.

.