Gravitation — Page 4 | JEE Physics

Q31

The ratio of the weights of a body on the Earth’s surface to that on the surface of a planets is 9 : 4. The mass of the planet is

{1 \over 9}

th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ? (Take the planets to have the same mass density)

A

{R \over 9}

B

{R \over 2}

C

{R \over 3}

D

{R \over 4}

Correct Answer

Option B

Solution

W = mg as m = constant everywhere $\therefore$ W $\propto$ g

{{{g_E}} \over {{g_p}}}

=

{9 \over 4}

We know,

g = {{GM} \over {{R^2}}}

$\therefore$

{{{g_E}} \over {{g_p}}} = {{{M_E}} \over {{M_p}}} \times {{R_p^2} \over {R_E^2}}

$\Rightarrow$

{9 \over 4} = {9 \over 1} \times {{R_p^2} \over {R_E^2}}

$\Rightarrow$

{{{R_p}} \over {{R_E}}} = {1 \over 2}

{R_p} = {{{R_E}} \over 2}

=

{R \over 2}

[Here RE = R ]

Q32

Four particles, each of mass

M

and equidistant from each other, move along a circle of radius

R

under the action of their mutual gravitational attraction. The speed of each particle is :

A

\sqrt {{{GM} \over R}}

B

\sqrt {2\sqrt 2 {{GM} \over R}}

C

\sqrt {{{GM} \over R}\left( {1 + 2\sqrt 2 } \right)}

D

{1 \over 2}\sqrt {{{GM} \over R}\left( {1 + 2\sqrt 2 } \right)}

Correct Answer

Option D

Solution

All those particles are moving due to their mutual gravitational attraction.

The force between each masses are repulsive force.

On mass M at C, due to mass at D the repulsive force is F in the vertical direction.

On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.

On mass M at C, due to mass at A the repulsive force is F' Net force acting on particle at C, =

2F\,\cos \,{45^ \circ } + F'

Where

F = {{G{M^2}} \over {{{\left( {\sqrt 2 R} \right)}^2}}}

and

F' = {{G{M^2}} \over {4{R^2}}}

\Rightarrow {{2 \times G{M^2}} \over {\sqrt 2 {{\left( {R\sqrt 2 } \right)}^2}}} + {{G{M^2}} \over {4{R^2}}}

\Rightarrow {{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]

This net force will balance by the centripetal force Fcp =

{{M{v^2}} \over R}

$\therefore$

{{M{v^2}} \over R} =

{{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]

$\Rightarrow$

v = \sqrt {{{Gm} \over R}\left( {{{\sqrt 2 + 4} \over {4\sqrt 2 }}} \right)}

= {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)}

Q33

A satellite is revolving in a circular orbit at a height

'h'

from the earth's surface (radius of earth

R;h < < R

). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere.)

A

\sqrt{2 g R}

B

\sqrt{g R}

C

\sqrt{g R / 2}

D

\sqrt{g R}(\sqrt{2}-1)

Correct Answer

Option D

Solution

Orbital velocity of satellite,

{v_0} = \sqrt {{{GM} \over {R + h}}}

=

\sqrt {{{GM} \over R}}

[ As

h < < R

then R + h = R ] =

\sqrt {gR}

[ As

g = {{GM} \over {{R^2}}}

] Escape velocity

{v_e} = \sqrt {{{2GM} \over R}}

=

\sqrt {2gR}

[ As

g = {{GM} \over {{R^2}}}

] $\therefore$ The minimum increase in its orbital velocity required to escape from the earth gravitational field

= \sqrt {2gR} - \sqrt {gR} = \sqrt {gR} \left( {\sqrt 2 - 1} \right)

Q34

If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh

{3 \over 4}

W. Radius of the Earth is 6400 km and g=10 m/s2.

A 1.1

\times

10−3 rad/s

B 0.83

\times

10−3 rad/s

C 0.63

\times

10−3 rad/s

D 0.28

\times

10−3 rad/s

Correct Answer

Option C

Solution

Initially when earth is not rotating then weight of the person is

w

. When earth rotares about it's axis then weight =

{{3\omega } \over 4}

Then, g' = g $-$ $\omega$ 2R cos2 $\theta$ $\Rightarrow$

\,\,\,

{{3g} \over 4}

= g $-$ $\omega$ 2R cos2 0o $\Rightarrow$

\,\,\,

$\omega$ 2R =

{g \over 4}

\,\,\,

$\omega$ =

\sqrt {{g \over {4R}}}

$\Rightarrow$

\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}}

= 0.63 $\times$ 10 $-$ 3 rad/s

Q35

Take the mean distance of the moon and the sun from the earth to be

0.4 \times {10^6}

km and

150 \times {10^6}

km respectively. Their masses are

8 \times {10^{22}}

kg and

2 \times {10^{30}}

kg respectively. The radius of the earth is

6400

km. Let

\Delta {F_1}

be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and

\Delta {F_2}

be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to

{{\Delta {F_1}} \over {\Delta {F_2}}}

is :

A

2

B

{10^{ - 2}}

C

0.6

D

6

Correct Answer

Option A

Solution

As gravitational force of attraction, F =

{{GMm} \over {{R^2}}}

\therefore\,\,\,\,

Force of attraction berween earth and moon F1 =

{{G{M_e}m} \over {r_1^2}}

Force of attraction between earth and sun, F2 =

{{GMeMs} \over {r_2^2}}

\therefore\,\,\,\,

\Delta

F1 = $-$

{{2G{M_e}m} \over {r_1^3}}

\Delta

r1

\Delta

F2 = $-$

{{2GMe\,Ms} \over {r_2^3}}

\Delta

r2

\therefore\,\,\,\,

{{\Delta {F_1}} \over {\Delta {F_2}}}

=

{{m\Delta {r_1}} \over {r_1^3}} \times {{r_2^3} \over {Ms\,\Delta {r_2}}}

=

\left( {{m \over {Ms}}} \right)\left( {{{r_2^3} \over {r_1^3}}} \right)\left( {{{\Delta {r_1}} \over {\Delta {r_2}}}} \right)

\Delta {r_1} = \Delta {r_2}

= diameter of the earth = 2 Rearth Given m = 8 $\times$ 1022 kg Ms = 2 $\times$ 1030 kg r1 = 0.4 $\times$ 106 km r2 = 150 $\times$ 106 km

\therefore\,\,\,\,

{{\Delta {F_1}} \over {\Delta {F_2}}} = \left( {{{8 \times {{10}^{22}}} \over {2 \times {{10}^{30}}}}} \right){\left( {{{150 \times {{10}^6}} \over {0.4 \times {{10}^6}}}} \right)^3} \times 1

= 2

Q36

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius

{R \over 2},

and the other mass, in a circular orbit of radius

{3R \over 2}

. The difference between the final and initial total energies is :

A

- {{GMm} \over {2R}}

B

+ {{GMm} \over {6R}}

C

{{GMm} \over {2R}}

D

- {{GMm} \over {6R}}

Correct Answer

Option D

Solution

Initially gravitational potenrial energy Ei = $-$

{{GMm} \over {2R}}

Final gravitational potential energy Ef = $-$

{{GM\left( {{m \over 2}} \right)} \over {2\left( {{R \over 2}} \right)}}

$-$

{{GM\left( {{m \over 2}} \right)} \over {2\left( {{{3R} \over 2}} \right)}}

= $-$

{{GMm} \over {2R}} - {{GMm} \over {6R}}

= $-$

{{4GMm} \over {6R}}

= $-$

{{2GMm} \over {3R}}

\therefore\,\,\,\,

Required difference in energies =

{E_f} - {E_i}

= $-$

{{GMm} \over R}\left( {{2 \over 3} - {1 \over 2}} \right)

= $-$

{{GMm} \over {6R}}

Q37

A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet? [Given ; Mass of planet = 8 × 1022 kg, Radius of planet = 2 × 106 m, Gravitational constant G = 6.67 × 10–11 Nm2 /kg2]

A 13

B 9

C 17

D 11

Correct Answer

Option D

Solution

{{m{V^2}} \over r} = {{GMm} \over {{r^2}}}

V = \sqrt {{{GM} \over r}}

n = {{VT} \over {2\pi r}} = \sqrt {{{GM} \over r}} {T \over {2\pi r}}

= \left( {\sqrt {{{GM} \over {{r^3}}}} } \right) \times {T \over {2\pi }} = \sqrt {{{6.67 \times {{10}^{ - 11}} \times 8 \times {{10}^{22}}} \over {{{\left( {202 \times {{10}^4}} \right)}^3}}}} \times {T \over {2\pi }}

= {{24 \times 3600} \over {2 \times 3.14}}\sqrt {{{6.67 \times 8 \times {{10}^{11}}} \over {{{\left( {202} \right)}^3} \times {{10}^{12}}}}}

= {{24 \times 3600} \over {2 \times 3.14 \times 1242.8}} = {{24 \times 3600} \over {74.51}} = 11

Q38

The value of acceleration due to gravity at Earth's surface is 9.8 ms–2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms–2, is close to : (Radius of earth = 6.4 × 106 m)

A 1.6 × 106 m

B 9.0 × 106 m

C 6.4 × 106 m

D 2.6 × 106 m

Correct Answer

Option D

Solution

{{GM} \over {{{\left( {R + h} \right)}^2}}} = {{GM} \over {2{R^2}}}

R + h = \sqrt 2 R

h = \left( {\sqrt 2 - 1} \right)R \simeq 2.6 \times {10^6}m

Q39

A test particle is moving in a circular orbit in the gravitational field produced by a mass density

\rho (r) = {K \over {{r^2}}}

. Identify the correct relation between the radius R of the particle's orbit and its period T

A T2/R3 is a constant

B TR is a constant

C T/R2 is a constant

D T/R is a constant

Correct Answer

Option D

Solution

For circular motion of particle:

{{m{V^2}} \over r} = mE

= m\left( {{{GM} \over {{r^2}}}} \right)

Where

M = \int\limits_0^r {\left( {4\pi {x^2}dx} \right)\left( {{k \over {{x^2}}}} \right)} = 4\pi kr

\Rightarrow {{m{V^2}} \over r} = m\left( {{{G\left( {4\pi k} \right)} \over r}} \right)

$\Rightarrow$ V = constant

T = {{2\pi r} \over V}

$\Rightarrow$

{T \over R}

= Constant

Q40

A rocket has to be launched from earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon ? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon :-

A E/32

B E/16

C E/4

D E/64

Correct Answer

Option B

Solution

Minimum energy required (E) = – (Potential energy of object at surface of earth)

= \left( { - {{GMm} \over R}} \right) = {{GMm} \over R}

Now Mearth = 64 Mmoon

\rho .{4 \over 3}\pi R_e^3 = 64.{4 \over 3}\pi R_m^3

$\Rightarrow$ Re = 4Rm Now

{{{E_{moon}}} \over {{E_{earth}}}} = {{{M_{moon}}} \over {{M_{earth}}}}.{{{R_{earth}}} \over {{R_{moon}}}} = {1 \over {64}} \times {4 \over 1}

\Rightarrow {E_{moon}} = {E \over {16}}