Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

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

$$\sqrt {gR/2} $$

B

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

C

$$\sqrt {2gR} $$

D

$$\sqrt {gR} $$

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

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

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

= $$\sqrt {{{GM} \over R}} $$ [ 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

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

2

MCQ (Single Correct Answer)

From a solid sphere of mass $$M$$ and radius $$R,$$ a spherical portion of radius $$R/2$$ is removed, as shown in the figure. Taking gravitational potential $$V=0$$ at $$r = \infty ,$$ the potential at the center of the cavity thus formed is:

($$G=gravitational $$ $$constant$$)

($$G=gravitational $$ $$constant$$)

A

$${{ - 2GM} \over {3R}}$$

B

$${{ - 2GM} \over R}$$

C

$${{ - GM} \over {2R}}$$

D

$${{ - GM} \over R}$$

Before removing the spherical portion, potential at point $$P$$(Center of cavity)

$${V_{sphere}}\,\, = {{ - GM} \over {2{R^3}}}\left[ {3{R^2} - {{\left( {{R \over 2}} \right)}^2}} \right]$$

$$ = {{ - GM} \over {2{R^3}}}\left( {{{11{R^2}} \over 4}} \right) = - 11{{GM} \over {8R}}$$

Mass of removed part = $${M \over {{4 \over 3}\pi {R^3}}} \times {4 \over 3}\pi {\left( {{R \over 2}} \right)^2}$$ = $${M \over 8}$$

Due to cavity part potential at point $$P$$

$${V_{cavity}}\,\,$$ = $$ - {{G{M \over 8}} \over {2{{\left( {{R \over 2}} \right)}^3}}}\left[ {3{{\left( {{R \over 2}} \right)}^2} - {0^2}} \right]$$

= $$ - {{3GM} \over {8R}}$$

So potential at P due to remaining part,

$$ = {V_{sphere}}\,\, - \,\,{V_{cavity}}\,\,$$

$$ = \,\, - {{11GM} \over {8R}} - \left( { - {3 \over 8}{{GM} \over R}} \right)$$

$$ = {{ - GM} \over R}$$

V = $$ - {{GM} \over {2{R^3}}}\left[ {3{R^2} - {r^2}} \right]$$

Here M = mass of the sphere, r = distance from the center of the sphere

3

MCQ (Single Correct Answer)

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

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

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 F_{cp} = $${{M{v^2}} \over R}$$

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

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

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

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 F

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

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

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

4

MCQ (Single Correct Answer)

What is the minimum energy required to launch a satellite of mass $$m$$ from the surface of a planet of mass $$M$$ and radius $$R$$ in a circular orbit at an altitude of $$2R$$?

A

$${{5GmM} \over {6R}}$$

B

$${{2GmM} \over {3R}}$$

C

$${{GmM} \over {2R}}$$

D

$${{GmM} \over {3R}}$$

Energy of the satellite on the surface of the planet

E_{i} = K.E + P.E = 0 + $$\left( { - {{GMm} \over R}} \right)$$ = $${ - {{GMm} \over R}}$$

Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v

E_{f} = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$

In the orbital of planet, the centripetal force is provided by the gravitational force

$$\therefore$$ $${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}$$

$$ \Rightarrow {v^2} = {{GM} \over {3R}}$$

$$\therefore$$ E_{f} = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$

$$ = {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$$

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

$$\therefore$$ Minimum energy required required to launch the satellite

= E_{f} - E_{i}

= $$ - {{GMm} \over {6R}}$$ - $$\left( { - {{GMm} \over R}} \right)$$

= $${{5GMm} \over {6R}}$$

E

Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v

E

In the orbital of planet, the centripetal force is provided by the gravitational force

$$\therefore$$ $${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}$$

$$ \Rightarrow {v^2} = {{GM} \over {3R}}$$

$$\therefore$$ E

$$ = {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$$

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

$$\therefore$$ Minimum energy required required to launch the satellite

= E

= $$ - {{GMm} \over {6R}}$$ - $$\left( { - {{GMm} \over R}} \right)$$

= $${{5GMm} \over {6R}}$$

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