Gravitation — Page 5 | JEE Physics

Q41

A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be :

A in the same circular orbit of radius R

B such that it escapes to infinity

C in a circular orbit of a different radius

D in an elliptical orbit

Correct Answer

Option D

Solution

mv

\widehat i

+ mv

\widehat j

= 2m

{\overrightarrow v ^1}

\overrightarrow v

=

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

Q42

A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx2 , is given by :

A

Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) - BL} \right]

B

Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]

C

Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) + BL} \right]

D

Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) - BL} \right]

Correct Answer

Option B

Solution

dm = (A + Bx2)dx dF =

{{GMdm} \over {{x^2}}}

F =

\int_a^{a + L} {{{GM} \over {{x^2}}}}

(A + Bx2)dx = GM

\left[ { - {A \over x} + Bx} \right]_a^{a + L}

= GM

\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]

Q43

A satellite is revolving in a circular orbit at a height h form the earth surface, such that h < < R where R is the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is :

A

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

B

\sqrt {2gR}

C

\sqrt {gR}

D

{{\sqrt {gR} } \over 2}

Correct Answer

Option A

Solution

v0 =

\sqrt {g(R + h)} \approx \sqrt {gR}

ve =

\sqrt {2g(R + h)} \approx \sqrt {2gR}

\Delta

v=ve $-$ v0 =

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

Q44

Two stars of masses 3

\times

1031 kg each, and at distance 2

\times

1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is - (Take Gravitational constant; G = 6.67

\times

10–11 Nm2 kg–2)

A 2.4

\times

104 m/s

B 1.4

\times

105 m/s

C 3.8

\times

104 m/s

D 2.8

\times

105 m/s

Correct Answer

Option D

Solution

By energy convervation between 0 & $\infty$ .

- {{GMm} \over r} + {{ - GMm} \over r} + {1 \over 2}m{V^2} = 0 + 0

[M is mass of star m is mass of meteroite) $\Rightarrow$ v

= \sqrt {{{4GM} \over r}} = 2.8 \times {10^5}

m/s

Q45

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%

Q46

Two planets have masses M and 16 M and their radii are

a

and 2

a

, respectively. The separation between the centres of the planets is 10

a

. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :

A

2\sqrt {{{GM} \over a}}

B

\sqrt {{{G{M^2}} \over {ma}}}

C

{3 \over 2}\sqrt {{{5GM} \over a}}

D

4\sqrt {{{GM} \over a}}

Correct Answer

Option C

Solution

Let at point P, net gravitational force = 0 $\therefore$

{{G\left( M \right)\left( m \right)} \over {{{\left( {10a - x} \right)}^2}}} = {{G\left( {16M} \right)\left( m \right)} \over {{x^2}}}

$\Rightarrow$ x = 8a By Conservation of Mechanical Energy, Ui + Ki = Uf + Kf $\Rightarrow$

- {{GMm} \over {8a}} - {{16GMm} \over {2a}}

+

{1 \over 2}m{v^2}

=

- {{16GMm} \over {8a}} - {{GMm} \over {2a}}

+ 0 $\Rightarrow$

{1 \over 2}m{v^2}

=

GMm\left[ {{1 \over {8a}} + {{16} \over {2a}} - {1 \over {2a}} - {{16} \over {8a}}} \right]

$\Rightarrow$

{1 \over 2}m{v^2}

=

GMm\left[ {{{1 + 64 - 4 - 16} \over {8a}}} \right]

$\Rightarrow$

{1 \over 2}m{v^2} = GMm\left[ {{{45} \over {8a}}} \right]

$\Rightarrow$ v =

\sqrt {{{90GM} \over {8a}}}

=

{3 \over 2}\sqrt {{{5GM} \over a}}

Q47

A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth’s radius Re . By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become

\sqrt {{3 \over 2}}

times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is :

A 2Re

B 3Re

C 4Re

D 2.5Re

Correct Answer

Option B

Solution

V0 =

\sqrt {{{GM} \over {{R_e}}}}

Applying Conservation of Angular Momentum, mVRe = mV'R $\Rightarrow$ m

\sqrt {{3 \over 2}} {V_0}{R_e}

= mV'R $\Rightarrow$ V' =

\sqrt {{3 \over 2}} {{{V_0}{R_e}} \over R}

Applying Conservation of Energy,

{ - {{GMm} \over {{R_e}}}}

+

{1 \over 2}m{V^2}

=

{ - {{GMm} \over R_{max}}}

+

{1 \over 2}mV{'^2}

$\Rightarrow$

{ - {{GMm} \over {{R_e}}}}

+

{1 \over 2}m\left( {{3 \over 2}V_0^2} \right)

=

{ - {{GMm} \over R_{max}}}

+

{1 \over 2}m \times

{\left( {V'} \right)^2}

$\Rightarrow$

{ - {{GMm} \over {{R_e}}} + }

{1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}

=

- {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{V_0}{{R_e^2} \over {{R^2}}}

$\Rightarrow$

{ - {{GMm} \over {{R_e}}} + }

{1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}

=

- {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}{{R_e^2} \over {{R^2}}}

$\Rightarrow$

- {1 \over {{R_e}}} + {3 \over {4{R_e}}}

=

- {1 \over R} + {{3{R_e}} \over {4{R^2}}}

$\Rightarrow$

{1 \over {4{R_e}}}

=

- {1 \over R} + {{3{R_e}} \over {4{R^2}}}

$\Rightarrow$ R = 3Re

Q48

On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by

{{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}

in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:

A

{A{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}

B

{A{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}

C

{A \over {{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}}

D

{A \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}

Correct Answer

Option C

Solution

Given

{E_x} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}

$\therefore$

{{ - dV} \over {dx}} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}

$\Rightarrow$

\int\limits_0^V {dV} = - \int\limits_\infty ^x {{{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}dx}

$\Rightarrow$

V = {A \over {{{({x^2} + {a^2})}^{1/2}}}}

Q49

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 }}

Q50

The value of the acceleration due to gravity is g1 at a height h =

{R \over 2}

(R = radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth. The ratio

\left( {{d \over R}} \right)

equals :

A

{5 \over 9}

B

{1 \over 9}

C

{7 \over 9}

D

{4 \over 9}

Correct Answer

Option A

Solution

Given,

{g_{at\,high}} = {g_{at\,depth}}

We know,

{g_{depth}} =

g\left( {1 - {d \over R}} \right)

$\therefore$

g\left( {1 - {d \over R}} \right) = {{GM_e} \over {{{(R + h)}^2}}}

$\Rightarrow$

g\left( {1 - {d \over R}} \right) =

{{G{M_e}} \over {{{\left( {R + {R \over 2}} \right)}^2}}}

=

{4 \over 9}{{G{M_e}} \over {{R^2}}}

=

{4 \over 9}g

$\Rightarrow$

{d \over R} = 1 - {4 \over 9} = {5 \over 9}