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The ratio of the K.E. required to be given to the satellite to escape earth's gravitational field to the K.E. required to be given so that the satellite moves in a circular orbit just above earth atmosphere is

One

Two

Half

Infinity

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Explanation

(b) K.E. required for satellite to escape from earth's gravitational field

$\frac{1}{2} m v_{e}^{2} = \frac{1}{2} m {(\sqrt{\frac{2 G M}{R}})}^{2} = \frac{G M m}{R}$

K.E. required for satellite to move in circular orbit

$\frac{1}{2} m v_{o}^{2} = \frac{1}{2} m {(\sqrt{\frac{G M}{R}})}^{2} = \frac{G M m}{2 R}$

The ratio between these two energies = 2

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An astronaut orbiting the earth in a circular orbit 120 km above the surface of earth, gently drops a spoon out of space-ship. The spoon will

Fall vertically down to the earth

Move towards the moon

Will move along with space-ship

Will move in an irregular way then fall down

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Explanation

out of the space-ship.

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The period of a satellite in a circular orbit around a planet is independent of

The mass of the planet

The orbital radius of satellite around the planet

The mass of the satellite

All the three parameters (a), (b) and (c)

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Explanation

$T = 2 π \sqrt{\frac{R^{3}}{GM}}$

so, it is independent of mass of satellite.

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Two identical satellites A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of the satellite A is 3V, the speed of the satellite B will be ?

12 V

6 V

$\frac{4}{3} v$

$\frac{3}{2} v$

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Explanation

(b) $v = \sqrt{\frac{G M}{R}} \Rightarrow \frac{v_{A}}{v_{B}} = \sqrt{\frac{R_{B}}{R_{A}}} = \sqrt{\frac{R}{4 R}} = \frac{1}{2} ∴ \frac{v_{A}}{v_{B}} = \frac{3 V}{v_{B}} = \frac{1}{2} ∴ v_{B} = 6 V$

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A geostationary satellite

Revolves about the polar axis

Has a time period less than that of the near-earth satellite

Moves faster than a near-earth satellite

Is stationary in the space

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Explanation

(a) A geosstationary satellite revolved in the same direction the earth rotates about the polar axis i.e. west to east.

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A small satellite is revolving near earth's surface. Its orbital velocity will be nearly

8 km/sec

11.2 km/sec

4 km/sec

6 km/sec

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Explanation

(a) Orbital velocity $v_{0} = \sqrt{g R_{e}}$

Here, g=9.8 m/ $s^{2}$ , $R_{e}$ = $6.4 \times 10^{6} m$

$H e n c e, v_{0} = \sqrt{9.8 \times 6.4 \times 106} = 7.91 \times 10^{3} m / s = 7.91 K m / s = 8 K m / s (a p p r o x)$

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The distance of neptune and saturn from sun are nearly $10^{13}$ and $10^{12}$ meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

$\sqrt{10}$

100

$10 \sqrt{10}$

$1 / \sqrt{10}$

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Explanation

(c) $\frac{T_{1}}{T 2} = {(\frac{R_{1}}{R_{2}})}^{3 / 2} = {(\frac{10^{13}}{10^{12}})}^{3 / 2} = {(1000)}^{1 / 2} = 10 \sqrt{10}$

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The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is v. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is

$\frac{3}{2} v$

$\sqrt{\frac{3}{2}} v$

$\sqrt{\frac{2}{3}} v$

$\frac{2}{3} v$

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Explanation

For first satellite $h = 0$ , $v_{1} = \sqrt{\frac{G M}{R}}$

For second satellite , $h = \frac{R}{2}, v_{2} = \sqrt{\frac{2 G M}{3 R}}$

$v_{2} = \sqrt{\frac{2}{3}} v_{1} = \sqrt{\frac{2}{3}} v$

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In a satellite if the time of revolution is T, then K.E. is proportional to

$\frac{1}{T}$

$\frac{1}{T^{2}}$

$\frac{1}{T^{3}}$

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

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Explanation

(d) $v = \sqrt{\frac{G M}{r}} ∴ K . E . \propto v^{2} \propto \frac{1}{r}$ and $T^{2} \propto r^{3} ∴ K . E . \propto T^{\frac{- 2}{3}}$

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The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is

$\frac{T}{4}$

$\frac{T}{8}$

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Explanation

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