NCERT Solutions for Class 11 Physics Chapter 7 Gravitation

Q: 7.1 (a) Answer the following questions:

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?

Answer:

No, because gravitational force doesn't depend upon the material medium. It is independent of the presence of other materials.

Q: 7.1 (b) Answer the following :

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?

Answer:

Yes, if the size of the space station is large, then he can detect gravity.

Q: 7.1 (c) Answer the following:

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:

Apart from the gravitational pull, the tidal effect also depends on the cube of the distance between the two. Since the distance between Earth and the sun is much larger than the distance between the sun and the moon so it is not balanced, but is more than the effect of the gravitational force. Thus, the tidal effect of the moon’s pull is greater than the tidal effect of the sun.

Q: 7.2 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 −GMm(1/r2−1/r1) is more/less accurate than the formula mg(r2−r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.

Answer:

(a) Acceleration due to gravity decreases with increasing altitude.

The relation between the two is given by :

gh = (1 − 2hRe)g

(b) Acceleration due to gravity decreases with increasing depth.

The relation is given below:

gd = (1 − dRe)g

(c) Acceleration due to gravity is independent of the mass of the body.

g = GMR2 Here, M is the mass of the Earth.

(d) The formula −GMm(1/r2−1/r1) is more accurate than the formula mg(r2−r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.

Q:7.3 Suppose there existed a planet that went around the sun was twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Answer:

Time taken by planet to complete a revolution around the sun = 12Te

Using Kepler's law of planetary motion, we can write :

(RpRe)3 = (TpTe)2

or RpRe = (121)23

or RpRe = 0.63

Thus, the planet is 0.63 times smaller than Earth.

Q: 7.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22×108m . Show that the mass of Jupiter is about one-thousandth that of the sun.

Answer:

The orbital period in days is = 1.769×24×60×60 s

Mass is given by :

M = 4π2R3GT2

Thus, the ratio of the mass of Jupiter and the mass of the sun is :

MsMj= 4π2Re3GTe24π2Rio3GTio2

or MsMj= (1.769×24×60×60365.25×24×60×60)2×(1.496×10114.22×108)3

or MsMj≈1045

Thus, the mass of Jupiter is nearly one-thousandth that of the sun.

Q: 7.5 Let us assume that our galaxy consists of 2.5×1011 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 105ly.

Answer:

We know that one light year is 9.45×1015 m .

The time period of rotation is given by :

T = (4πr3GM)12

Putting all the values (in SI units) in the above equation, we get :

T= (4×(3.14)2×(4.73)3×10606.67×10−11×5×1041)12

or T= 1.12 ×1016 s

In years :

T= 1.12 ×1016365×24×60×60 = 3.55×108 years

Q: 7.6 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:

(a) The total energy will be negative of its kinetic energy. Since at infinity, potential energy is zero, and total energy is negative.

(b) The energy required will be less as the stationary object on earth has no energy initially, whereas the satellite has gained energy due to rotational motion.

Q: 7.7 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 from the Earth is given by :

vesc = 2gR

Since the escape velocity depends upon the reference (potential energy), it only depends upon the height of the location.

(a) No

(b) No

(c) No

(d) Yes

Q: 7.8 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:

Among all, only the angular momentum and total energy of the comet will be constant; all other factors given will vary from point to point.

Q: 7.9 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) In space, we have a state of weightlessness, so the swollen feet do not affect astronauts, as there is no gravitational pull (so they cannot stand).

(b) The swollen face will be affected as the sense organs, such as eyes, ears, etc. will be affected.

(c) It will affect the astronaut as it may cause mental strain.

(d) It will affect the astronaut as space has different orientations.

Q: 7.12 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 =2×1030 kg, mass of the earth =6×1024 kg. Neglect the effect of other planets etc. (orbital radius =1.5×1011 m).

Answer:

Let the distance where the gravitational force acting on satellite P becomes zero be x from the Earth.

Thus, we can write :

GmMs(r−x)2 = GmMer2

or (r−xx) = (2×103060×1024)12

or (r−xx)= 577.35

Hence :

x = 1.5×1011578.35 = 2.59×108 m (Since r = 1.5×1011 )

Q: 7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5×108km .

Answer:

Mass of the sun can be calculated by using the following formula:-

M = 4×π2×r3GT2

Putting the known values in the above formula, we obtain :

M= 4×(3.14)2×(1.5×1011)36.67×10−11×(365.25×24×60×60)2

or M= 133.24×106.64×104 = 2×1030 Kg

Thus, the mass of the sun is nearly 2×1030 Kg.

Q: 7.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50×108km away from the sun?

Answer:

Kepler's third law gives us the following relation :

T = (π2r3GM)12

Thus, we can write :

rs3re3 = Ts2Te2

or rs = re(TsTe)23

or rs= 1.5×1011×(29.5TeTe)23

or rs= 14.32×1011 m

Thus, the distance between the sun and Saturn is 14.32×1011 m.

Q: 7.15 A body weighs 63N 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:

Acceleration due to gravity at height h from the surface of Earth is :

g′ = g(1 + hR)2

For h = R2 we have :

g′ = g(1 + R2R)2

or g′= 49g

Thus, the weight of the body will be :

W = mg′

or W= m×49g = 49mg

or W= 28 N

Q: 7.16 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 250N on the surface?

Answer:

Position of the body is (depth) :

= 12Re

Acceleration due to gravity at depth d is given by :

g′ = (1 − dRe)g

or g′= (1 − Re2Re)g

or g′= 12g

Thus, the weight of the body is:-

W = mg′

or W= m×12g = mg2

or W= 125 N

Thus, the weight of the body is 125 N.

Q: 7.17 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 =6.0×1024 kg; mean radius of the earth =6.4×106 m;G=6.67×10−11 N m2 kg−2.

Answer:

The total energy is given by :

Total energy = Potential energy + Kinetic energy

= (−GmMeRe) + 12mv2

At the highest point, velocity will be zero.

Thus, the total energy of the rocket is :

= (−GmMeRe + h) + 0

Now we will use the conservation of energy :

Total energy initially (at Earth's surface) = Total energy at height h

(−GmMeRe) + 12mv2 = (−GmMeRe + h)

or 12v2 = gRehRe + h

or h = Rev22gRe − v2

or h= 6.4×106×(5×103)22g×6.4×106 − (5×103)2

or h= 1.6×106 m

Hence, the height achieved by the rocket from Earth's centre = R + h

R+h= 6.4×106 + 1.6×106

or R+h= 8×106 m

Q: 7.18 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:

Let us assume the speed of the body far away from the earth is vf.

Total energy on Earth is :

= 12mvp2 − 12mvesc2

And the total energy when the body is far from the earth is :

= 12mvf2

(Since the potential energy at a far distance from the earth is zero.)

We will use conservation of energy: -

12mvp2 − 12mvesc2 = 12mvf2

or vf = (vp2 − vesc2)

or vf= ((3vesc)2 − vesc2)

or vf= 8vesc

or vf= 31.68 km/s

Q: 7.19 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 =6.0×1024 kg; radius of the earth =6.4×106 m:G=6.67×10−11 N m2 kg−2

Answer:

The total energy of the satellite at height h is given by :

= 12mv2 + (−GMemRe + h)

We know that the orbital speed of the satellite is :

v = (GMeRe + h)

Thus, the total energy becomes :

E= 12m×(GMeRe + h) + (−GMemRe + h)

or E= −12(GMemRe + h)

Thu,s the required energy is the negative of the total energy :

nbsp; Ereq = 12(GMemRe + h)

or Ereq= 12(6.67×10−11×6×1024×2006.4×106 + 0.4×106)

or Ereq= 5.9×109 J

Q: 7.20 Two stars each of one solar mass (=2×1030 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:

The total energy of stars is given by :

E = −GMMr + 12Mv2

or E= −GMMr + 0

or E= −GMMr

Now, when stars are just about to collide, the distance between them is 2R.

The total kinetic energy of both stars is :

KE= 12Mv2 + 12Mv2 = Mv2

And the total energy of both the stars is :

E′= Mv2 + −GMM2r

Using conservation of energy, we get :

Mv2 + −GMM2r = −GMMr

or v2 = GM(−1r + 12R)

or v2= 6.67×10−11×2×1030(−11012 + 12×107)

or v2= 6.67×1012

Thus, the velocity is: 6.67×1012 = 2.58×106 m/s

Q: 7.21 Two heavy spheres each of mass 100kg and radius 0.10m are placed 1.0m 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:

The gravitational force at the midpoint will be zero. This is because both spheres are identical, and their forces will be equal but opposite in direction.

The gravitational potential is given by :

= −GMr2 − −GMr2 = −4GMr

= −4×6.67×10−11×1001

= −2.67×10−8 J/Kg

At the midpoint, we have equal forces in opposite directions, so it is in equilibrium, but if we move the body slightly, then the particle will move in one direction (as one force will be greater). So it is an unstable equilibrium.

Q:2 A star 2.5 times the mass of the sun and collapsed to a size of the 12 km rotates with a speed of 1.5 rev per second. (Extremely compact stars of this kind are known as neutron stars. Centain onserved steller objects called pulsars are believed to belong this category ). Will an object placed on its equator remain struck to its surface due to gravity? (Mass of the sun =2×1030 kg ).

Answer:

A body will get stuck at the star's surface if the centrifugal force of the star is less than the gravitational force.

The gravitational force is given by :

Fg = GMmr2

or Fg= 6.67×10−11×5×1030×m(1.2×104)2 = 2.31×1012 m N

The centrifugal force is given by :

Fc = mrω2

or Fc= mr(2πv)2 = m×1.2×104×(2π×1.2)2

or Fc= 6.8×105 m N

As we can see that the gravitational force is greater than the centrifugal force, thus the body will remain at the star.

Q:3 A space-ship is stationed on Mars. How much energy must be expended on the spaceship to rocket it out of the solar system ? Mass of the spaceship =1000 kg, Mass of the sun =2×1030 kg. Mass of the Mars =6.4×1023 kg, Radius of Mars =3395 km. Radius of the orbit of Mars =2.28×1011 m,G=6.67×10−11Nm2 kg−2

Answer:

Firstly, the potential energy of the spaceship due to the sun is given by :

= −GMmsr

and the potential energy of the spaceship due to Mars is given by :

= −GMmmR

It is given that the spaceship is stationary, so its kinetic energy is zero.

Thus, the total energy of the spaceship is :

= −GMmsr + −GMmmR

Thus, the energy needed to launch the spaceship is :

= GMmsr + GMmmR

= 6.67×10−11×103(2×10302.28×1011 + 6.4×10233.395×106)

= 596.97×109 J

= 6×1011 J

Q:4 A rocket is fired vertically from the surface of Mars with a speed of 2kms−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,

Answer:

The kinetic energy of the rocket is (initial):-

= 12mv2

And the initial potential energy is :

= −GMmR

Thus total initial energy is given by :

= 12mv2 + −GMmR

Further, it is given that 20 per cent of kinetic energy is lost.

So the net initial energy is :

= 0.4mv2 − GMmR

The final energy is given by :

= GMmR + h

Using the law of energy conservation, we get :

0.4mv2 − GMmR = GMmR + h

Solving the above equation, we get :

h = RGM0.4v2R − 1

or h= 0.4R2v2GM − 0.4v2R

or h= 18.442×101842.688×1012 − 5.432×1012

or h= 495×103 m

or h= 495 Km

Thus, the required distance is 495 Km.