Gravitation — Page 7 | NEET Physics

Q61

The masses and radii of the earth and moon are (M1, R1) and (M2, R2) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses :

A

V = {1 \over 2}\sqrt {{{4G({M_1} + {M_2})} \over r}}

B

V = \sqrt {{{4G({M_1} + {M_2})} \over r}}

C

V = {1 \over 2}\sqrt {{{2G({M_1} + {M_2})} \over r}}

D

V = {{\sqrt {2G} ({M_1} + {M_2})} \over r}

Correct Answer

Option B

Solution

{1 \over 2}m{V^2} - {{G{M_1}m} \over {r/2}} - {{G{M_2}m} \over {r/2}} = 0

{1 \over 2}m{V^2} = {{2Gm} \over r}({M_1} + {M_2})

V = \sqrt {{{4G({M_1} + {M_2})} \over r}}

Option (b)

Q62

A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is _________. 'P' has the time period of 24 hours.

A 3

B 5

C

6\sqrt 2

D

{6 \over {\sqrt 2 }}

Correct Answer

Option A

Solution

From Kepler's law

{T^2} \propto {R^3}

{\left( {{{24} \over T}} \right)^2} = {\left( {{{12R} \over {3R}}} \right)^3}

T = 3 sec

Q63

If

g

is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass

m

raised from the surface of the earth to a height equal to the radius

R

of the earth is

A

{1 \over 4}mgR

B

2mgR

C

{1 \over 2}mgR

D

mgR

Correct Answer

Option C

Solution

Gravitational potential energy on the earth surface of a body U =

-{{GmM} \over R}

And at the height h from the earth surface the potential energy

{U_h} = - {{GmM} \over {R + h}}

=

- {{GmM} \over {2R}}

[ as h = R ] So the gain in the potential energy

\Delta U = {U_h} - U

$\therefore$

\Delta U = {{ - GmM} \over {2R}} + {{GmM} \over R};

$\Rightarrow$

\Delta U = {{GmM} \over {2R}}

Now

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

\,\,\,

$\therefore$

{\mkern 1mu} {{GM} \over R} = gR

$\therefore$

\Delta U = {1 \over 2}mgR

Q64

If suddenly the gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the satellite will

A continue to move in its orbit with same velocity

B move tangentially to the original orbit with the same velocity

C become stationary in its orbit

D move towards the earth

Correct Answer

Option B

Solution

When gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the centripetal force becomes zero.

So the satellite will move tangentially to the original orbit with the same velocity as it has at the instant when gravitational force becomes zero.

Q65

A particle of mass

10

g

is kept on the surface of a uniform sphere of mass

100

kg

and radius

10

cm.

Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take

G

= 6.67 \times {10^{ - 11}}\,\,N{m^2}/k{g^2}

)

A

3.33 \times {10^{ - 10}}\,J

B

13.34 \times {10^{ - 10}}\,J

C

6.67 \times {10^{ - 10}}\,J

D

6.67 \times {10^{ - 9}}\,J

Correct Answer

Option C

Solution

We know, Work done = Difference in potential energy $\therefore$

W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right]

$\Rightarrow$

W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}

= 6.67 \times {10^{ - 10}}J

Q66

Two bodies of masses

m

and

4

m

are placed at a distance

r.

The gravitational potential at a point on the line joining them where the gravitational field is zero is:

A

- {{4Gm} \over r}

B

- {{6Gm} \over r}

C

- {{9Gm} \over r}

D zero

Correct Answer

Option C

Solution

Let the gravitational field at

P,

distant

x

from mass

m,

be zero. $\therefore$

{{Gm} \over {{x^2}}} = {{4Gm} \over {{{\left( {r - x} \right)}^2}}}

\Rightarrow 4{x^2} = {\left( {r - x} \right)^2}

\Rightarrow 2x = r - x

\Rightarrow x = {r \over 3}

Gravitational potential at point

P,

V = - {{Gm} \over {{r \over 3}}} - {{4Gm} \over {{{2r} \over 3}}} = -{{9Gm} \over r}

Q67

A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is :

A 1.5

B 2.0

C 0.7

D 3.0

Correct Answer

Option D

Solution

{T^2} \propto {R^3}

T = k{R^{3/2}}

{{dT} \over T} = {3 \over 2}{{dR} \over R}

= {3 \over 2} \times 0.02 = 0.03

% Change = 3%

Q68

A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:

A 2

B 1

C

\sqrt 2

D

{1 \over {\sqrt 2 }}

Correct Answer

Option D

Solution

{V_0} = \sqrt {{{GM} \over r}}

{V_e} = \sqrt {{{2GM} \over r}}

$\therefore$

{{{V_0}} \over {{V_e}}} = \sqrt {{{GM} \over r} \times {{2GM} \over r}} = {1 \over {\sqrt 2 }}

Q69

Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

A

\sqrt {{G \over 2}(1 + 2\sqrt 2 )}

B

\sqrt {{G \over 2}(2\sqrt 2 - 1)}

C

\sqrt {G(1 + 2\sqrt 2 )}

D

{1\over2}\sqrt {G(1 + 2\sqrt 2 )}

Correct Answer

Option D

Solution

Given, m = 1 kg, R = 1 m We know that,

F = {{G{m_1}{m_2}} \over {{r^2}}}

$\because$

{F_1} = {{Gmm} \over {{{(2R)}^2}}} = {{G{m^2}} \over {4{R^2}}}

and

{F_2} = {{Gmm} \over {{{(\sqrt 2 R)}^2}}} = {{G{m^2}} \over {2{R^2}}}

Net force on one particle,

{F_{net}} = {F_1} + {F_2}\cos 45^\circ + {F_2}\cos 45^\circ

= {F_1} + 2{F_2}\cos 45^\circ

= {{G{m^2}} \over {4{R^2}}} + 2\left( {{{G{m^2}} \over {2{R^2}}}} \right).{1 \over {\sqrt 2 }}

= {{G{m^2}} \over {4{R^2}}} + {{G{m^2}} \over {\sqrt 2 {R^2}}}

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

As the gravitational force provides the necessary centripetal force, so

{F_{net}} = {F_C} = {{m{v^2}} \over R}

Here, FC = centripetal force.

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

\Rightarrow v = {1 \over 2}\sqrt {{{Gm} \over R}(1 + 2\sqrt 2 )}

\Rightarrow v = {1 \over 2}\sqrt {G(1 + 2\sqrt 2 )}

Q70

A planet revolving in elliptical orbit has : A. a constant velocity of revolution. B. has the least velocity when it is nearest to the sun. C. its areal velocity is directly proportional to its velocity. D. areal velocity is inversely proportional to its velocity. E. to follow a trajectory such that the areal velocity is constant. Choose the correct answer from the options given below :

A D only

B E only

C C only

D A only

Correct Answer

Option B

Solution

According to Kepler’s second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.