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Gravitation MCQs for NEET — Physics Questions with Answers

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A satellite is to revolve round the earth in a circle of radius 8000 km. The speed at which this satellite be projected into an orbit, will be

$3 k m / s$

16 km/s

7.15 km/s

8 km/s

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Explanation

(c) $v_{0} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g R^{2}}{r}} = \sqrt{\frac{10 \times {(64 \times 10^{5})}^{2}}{8000 \times 10^{3}}} = 71.5 \times 10^{2} m / s = 7.15 k m / s$

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If mass of a body is M on the earth surface, then the mass of the same body on the moon surface is

M/6

Zero

None of these

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Explanation

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Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to -

$\frac{1}{R}$

$\frac{1}{\sqrt{R}}$

$\frac{1}{R^{3 / 2}}$

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Explanation

(a) $K . E = \frac{G M m}{2 R}$

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A particle falls towards earth from infinity. It’s velocity on reaching the earth would be -

Infinity

$\sqrt{2 g R}$

$2 \sqrt{g R}$

Zero

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Explanation

$(b) K_{i} + U_{i} = K_{f} + U_{f} \Rightarrow 0 + 0 = \frac{1}{2} {mv}^{2} - \frac{GMm}{R^{2}} \Rightarrow \frac{1}{2} v^{2} = \frac{GM}{R^{2}} \Rightarrow v^{2} = \sqrt{\frac{2 GM}{R^{2}}} \Rightarrow v = \sqrt{2 gR}$

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Two satellite A and B, ratio of masses 3 : 1 are in circular orbits of radii r and 4r. Then ratio of total mechanical energy of A to B is

1 : 3

3 : 1

3 : 4

12 : 1

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Explanation

(d) Total mechanical energy of satellite

$E = \frac{- G M m}{2 r} \Rightarrow \frac{E_{A}}{E_{B}} = \frac{m_{A}}{m_{B}} \times \frac{r_{B}}{r_{A}} \Rightarrow \frac{3}{1} \times \frac{4 r}{r} = \frac{12}{1}$

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The orbital velocity of a planet revolving close to earth's surface is

$\sqrt{2 g R}$

$\sqrt{g R}$

$\sqrt{\frac{2 g}{R}}$

$\sqrt{\frac{g}{R}}$

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Explanation

$\frac{m v^{2}}{R} = \frac{G M m}{R^{2}} \Rightarrow v = \sqrt{\frac{G M}{r}} \sin c e, g = \frac{G M}{R^{2}} H e n c e, v = \sqrt{g R}$

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If the gravitational force between two objects were proportional to 1/R (and not as $1 / R^{2}$ ) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

$1 / R^{2}$

$R^{\circ}$

$R^{1}$

1/R

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Explanation

(b) Gravitational force provides the required centripetal force for orbiting the

satellite $\frac{m v^{2}}{R} = \frac{K}{R}$

because $(F \propto \frac{1}{R})$
$∴ v \propto R^{\circ}$

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Potential energy of a satellite having mass ‘m’ and rotating at a height of $6.4 \times 10^{6} m$ from the earth surface is -

$- 0.5 m g R_{e}$

$- m g R_{e}$

$- 2 m g R_{e}$

$4 m g R_{e}$

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Explanation

(a) Potential energy = $= \frac{- G M m}{r} = \frac{G M m}{R_{e} + h} = \frac{- G M m}{2 R_{e}}$

$= - \frac{g R^{2} m}{2 R_{e}} = - \frac{1}{2} m g R_{e} = - 0.5 m g R_{e}$

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When a satellite going round the earth in a circular orbit of radius r and speed v loses some of its energy, then r and v change as

r and v both with increase

r and v both will decrease

r will decrease and v will increase

r will decrease and v will decrease

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Explanation

(c) $E_{T o t a l} = - \frac{G M m}{2 r} w h e r e r i s t h e r a d i u s o f o r b i t S o i f e n e r g y d e c r e a s e s, r a l s o d e c r e a s e s . A l s o, K . E . = \frac{G M m}{2 r} S o, i f r a d i u s d e c r e a s e s, K . E . i n c r e a s e s$ If B.E. decreases then r also decreases and v increases as

$v \propto \frac{1}{\sqrt{r}}$

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Which of the following quantities does not depend upon the orbital radius of the satellite ?

$\frac{T}{R}$

$\frac{T^{2}}{R}$

$\frac{T^{2}}{R^{2}}$

$\frac{T^{2}}{R^{3}}$

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Explanation

(d) $T^{2} \propto R^{3} ∴ \frac{T^{2}}{R^{3}}$ = constant

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