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Gravitation

NEET Physics · 96 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
A planet moving along an elliptical orbit is closest to the sun at a distance r 1 and farthest away at a distance of r 2 . If vv 1 and vv 2 are the linear velocities at these points respectively, then the ratio v1v2{{{v_1}} \over {{v_2}}} is
A (r 1 /r 2 ) 2
B r 2 /r 1
C (r 2 /r 1 ) 2
D r 1 /r 2
Correct Answer
Option B
Solution

Angular momentum is conserved ∴\therefore L 1 = L 2 ⇒\Rightarrow mr 1 v 1 = mr 2 v 2 ⇒\Rightarrow r 1 v 1 = r 2 v 2

⇒v1v2=r2r1\Rightarrow {{{v_1}} \over {{v_2}}} = {{{r_2}} \over {{r_1}}}
Q42
The dependence of acceleration due to gravity g on the distance r from the centre of the earth, assumed to be a sphere of radius R of uniform density is as shown in figures below The correct figure is
A (4)
B (1)
C (2)
D (3)
Correct Answer
Option A
Solution

The acceleration due to gravity at a depth d below surface of earth is

g′=GMR2(1−dR)=g(1−dR)g' = {{GM} \over {{R^2}}}\left( {1 - {d \over R}} \right) = g\left( {1 - {d \over R}} \right)

g' = 0 at d = R. i.e., acceleration due to gravity is zero at the centre of earth.

Thus, the variation in value g with r is For, r > R,

g′=g(1+hR)2=gR2r2⇒g′∝1r2g' = {g \over {{{\left( {1 + {h \over R}} \right)}^2}}} = {{g{R^2}} \over {{r^2}}} \Rightarrow g' \propto {1 \over {{r^2}}}

Here, R + h = r For r < R,

g′=g(1−dR)=grRg' = g\left( {1 - {d \over R}} \right) = {{gr} \over R}

Here,

R−d=r⇒g′∝rR - d = r \Rightarrow g' \propto r

Therefore, the variation of g with distance from centre of the earth will be as shown in the figure.

Q43
(1) Centre of gravity (C.G) of a body is the point at which the weight of the body acts (2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius. (3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its C.G. (4) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis. Which one of the following pairs of statements is correct ?
A (4) and (1)
B (1) and (2)
C (2) and (3)
D (3) and (4)
Correct Answer
Option A
Solution

Centre of gravity of a body is the point at which the weight of the body acts and the radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.

Q44
The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R 1 to another of radius R 2 (R 2 > R 1 ) is
A GmM (1R12−1R22)\left( {{1 \over {R_1^2}} - {1 \over {R_2^2}}} \right)
B GmM (1R1−1R2)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)
C 2GmM (1R1−1R2)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)
D 12{1 \over 2}GmM (1R1−1R2)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)
Correct Answer
Option D
Solution

Total energy of the orbiting satellite of mass m having orbital radius r is

E=−GMm2rE = - {{GMm} \over {2r}}

where M is the mass of the planet.

Additional kinetic energy required to transfer the satellite from a circular orbit of radius R 1 to another radius R 2 is E 2 −E 1

=−GMm2RE−(−GMm2R1)=−GMm2R2+GMm2R1= - {{GMm} \over {2{R_E}}} - \left( { - {{GMm} \over {2{R_1}}}} \right) = - {{GMm} \over {2{R_2}}} + {{GMm} \over {2{R_1}}}
=GMm2(1R1−1R2)= {{GMm} \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)
Q45
A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be
A 9.9 m
B 10.1 m
C 10 m
D 20 m
Correct Answer
Option B
Solution

Since the man is in gravity free space, force on man + stone system is zero.

Therefore centre of mass of the system remains at rest.

Let the man goes x m above when the stone reaches the floor, then M man × x = M stone × 10

x=0.550×10x = {{0.5} \over {50}} \times 10

x = 0.1 m Therefore final height of man above floor = 10 + x = 10 + 0.1 = 10.1 m

Q46
A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The magnitude of the gravitational potential at a point sutuated at a/2 distance from the centre, will be :
A GMa{{GM} \over a}
B 2GMa{{2GM} \over a}
C 3GMa{{3GM} \over a}
D 4GMa{{4GM} \over a}
Correct Answer
Option C
Solution

Here, Mass of a particle = M Mass of a spherical shell = M Radius of a spherical shell = a Let O be centre of a spherical shell.

Gravitational potential at point P due to particle at O is

V1=−GMa/2{V_1} = - {{GM} \over {a/2}}

Gravitational potential at point P due to spherical shell is

V2=−GMa{V_2} = - {{GM} \over a}

Hence, total gravitational potential at point P is V = V 1 + V 2

=−GMa/2+(−GMa)=−2GMa−GMa=−3GMa= - {{GM} \over {a/2}} + \left( { - {{GM} \over a}} \right) = - {{2GM} \over a} - {{GM} \over a} = - {{3GM} \over a}
∣V∣=3GMa\left| V \right| = {{3GM} \over a}
Q47
The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3V, then the speed of satellite B will be
A 3V4{{3V} \over 4}
B 6VV
C 12VV
D 3V2{{3V} \over 2}
Correct Answer
Option B
Solution

Orbital velocity of a satellite in a circular orbit of radius a is given by

v=GMav = \sqrt {{{GM} \over a}}
⇒v∝1a\Rightarrow v \propto \sqrt {{1 \over a}}
⇒v2v1=a1a2\Rightarrow {{{v_2}} \over {{v_1}}} = \sqrt {{{{a_1}} \over {{a_2}}}}

∴\therefore

v2=v14RR{v_2} = {v_1}\sqrt {{{4R} \over R}}

= 2v 1 = 6V

Q48
The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t 1 is the time for the planet to move from C to D and t 2 is the time to move from A to B then
A t 1 = 4t 2
B t 1 = 2t 2
C t 1 = t 2
D t 1 > t 2
Correct Answer
Option B
Solution

According to Kepler’s law, the areal velocity of a planet around the sun always remains constant.

SCD : A 1 – t 1 (areal velocity constant) SAB : A 2 – t 2

A1t1=A2t2{{{A_1}} \over {{t_1}}} = {{{A_2}} \over {{t_2}}}
t1=t2.A1A2{t_1} = {t_2}.{{{A_1}} \over {{A_2}}}

(given A 1 = 2A 2 )

=t2.2A2A2= {t_2}.{{2{A_2}} \over {{A_2}}}

∴\therefore

t1=2t2{t_1} = 2{t_2}
Q49
Two satellites of earth, S 1 and S 2 are moving in the same orbit. The mass of S 1 is four times the mass of S 2 . Which one of the following statements is true ?
A The potential energies of earth and satellite in the two cases are equal.
B S 1 and S 2 are moving with the same speed.
C The kinetic energies of the two satellites are equal.
D The time period of S 1 is four times that of S 2 .
Correct Answer
Option B
Solution

Since orbital velocity of satellite is

v=GMrv = \sqrt {{{GM} \over r}}

it does not depend upon the mass of the satellite. Therefore, both satellites will move with same speed.

Q50
The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is
A 1/2
B 2\sqrt 2
C 1/2\sqrt 2
D 1/3
Correct Answer
Option C
Solution

Escape velocity of the body from the surface of earth is v =

2gR\sqrt {2gR}

For escape velocity of the body from the platform potential energy + kinetic energy = 0

−GMm2R+12mv2=0- {{GMm} \over {2R}} + {1 \over 2}m{v^2} = 0
⇒fvescape=GMR2.R=gR=fv\Rightarrow f{v_{escape}} = \sqrt {{{GM} \over {{R^2}}}.R} = \sqrt {gR} = fv

From the surface of the earth,

vescape=2gR{v_{escape}} = \sqrt {2gR}
∴fvescape=vescape2\therefore f{v_{escape}} = {{{v_{escape}}} \over {\sqrt 2 }}

, ∴\therefore

f=12f = {1 \over {\sqrt 2 }}
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