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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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A satellite moves round the earth in a circular orbit of radius R making one revolution per day. A second satellite moving in a circular orbit, moves round the earth once in 8 days. The radius of the orbit of the second satellite is -

8 R

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Explanation

(b) Given that, $T_{1} = 1$ day and $T_{2} = 8$ days

$∴ \frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{\frac{3}{2}} \Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{2}{3}} = {(\frac{8}{1})}^{\frac{2}{3}} = 4 \Rightarrow r_{2} = 4 r_{1} = 4 R$

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To an astronaut in a spaceship, the sky appears:

Black

White

Green

Blue

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Explanation

(a)

On earth, we see the blue sky due to the presence of the atmosphere. The atmosphere refracts light rays coming from the sun and we see the color of the sky. In space, due to the absence of the atmosphere the light rays do not refract and reach the astronomer's eyes therefore it appears black.

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A satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon’s orbit. The satellite completes one revolution in

$\frac{1}{2}$ lunar month

$\frac{2}{3}$ lunar month

$2^{\frac{- 3}{2}}$ lunar month

$2^{\frac{3}{2}}$ lunar month

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Explanation

$\frac{T_{s}}{T_{m}} = {(\frac{r_{s}}{r_{m}})}^{\frac{3}{2}} = {(\frac{1}{2})}^{\frac{3}{2}} \Rightarrow T_{s} = 2^{\frac{- 3}{2}}$ lunar month.

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