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The escape velocity on earth is 11.2 km/s. On another planet having twice radius and 8 times mass of the earth, the escape velocity will be

3.7 km/s

11.2 km/s

22.4 km/s

43.2 km/s

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Explanation

(c) $\frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{p}}{R_{e}}} = \sqrt{8 \times \frac{1}{2}} = 2 s o, v_{p} = 2 \times v_{e} = 22.4 k m / s$

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When a body is taken from the equator to the poles, its weight

Remains constant

Increases

Decreases

Increases at N-pole and decreases at S-pole

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Explanation

(b) Because acceleration due to gravity increases

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The escape velocity of a body on the surface of the earth is 11.2 km/s. If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become

5.6 km/s

11.2 km/s (remain unchanged)

22.4 km/s

44.8 km/s

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Explanation

If M becomes double and R becomes half then escape velocity becomes two times

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A body of mass m is taken to the bottom of a deep mine. Then

Its mass increases

Its mass decreases

Its weight increases

Its weight decreases

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Explanation

d) Because acceleration due to gravity decreases

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Given mass of the moon is 1/81 of the mass of the earth and corresponding radius is 1/4 of the earth. If escape velocity on the earth surface is 11.2 km/s, the value of same on the surface of the moon is

0.14 km/s

0.5 km/s

2.5 km/s

5 km/s

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Explanation

on moon $v_{m} = \sqrt{\frac{2 G M \times 4}{81 \times R}} = \frac{2}{9} \sqrt{\frac{2 G M}{R}} = \frac{2}{9} \times 11.2 = 2.5 k m / s$

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The angular velocity of rotation of star (of mass M and radius R) at which the matter start to escape from its equator will be

$\sqrt{\frac{2 G M^{2}}{R}}$

$\sqrt{\frac{2 G M}{g}}$

$\sqrt{\frac{2 G M}{R^{3}}}$

$\sqrt{\frac{2 G R}{M}}$

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Explanation

If star rotates with angular velocity $ω$

then $ω$ = $\frac{V}{R} = \frac{1}{R} \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{2 G M}{R^{3}}}$

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A body weighs 700 gm wt on the surface of the earth. How much will it weigh on the surface of a planet whose mass $\frac{1}{7}$ of earth's mass and radius is half that of the earth ?

200 gm wt

400 gm wt

50 gm wt

300 gm wt

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Explanation

(b) We know that $g = \frac{G M}{R^{2}}$

On the planet $g, = \frac{G M / 7}{R^{2} / 4} = \frac{4 g}{7} = \frac{4 g}{7}$

Hence weight on the planet = 700 $\times \frac{4}{7} = 400 g m w t$

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What will be the acceleration due to gravity at height h if h >> R where R is radius of earth and g is acceleration due to gravity on the surface of earth ?

$\frac{g}{{(1 + \frac{h}{R})}^{2}}$

$g (1 - \frac{2 h}{R})$

$\frac{g}{{(1 - \frac{h}{R})}^{2}}$

$g (1 - \frac{h}{R})$

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Explanation

(a) g'=g ${(\frac{R}{R + h})}^{2} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

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How many times is escape velocity, of orbital velocity for a satellite revolving near earth?

$\sqrt{2}$ times

2 times

3 times

4 times

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Explanation

(a)

$E s c a p e v e l o c i t y = \sqrt{\frac{2 G M}{R}} O r b i t a l v e l o c i t y = \sqrt{\frac{G M}{R}} H e n c e, i t' s \sqrt{2} t i m e s .$

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Escape velocity on earth is 11.2 km/s. What would be the escape velocity on a planet whose mass is 1000 times and radius is 10 times that of earth ?

112 km/s

11.2 km/s

1.12 km/s

3.7 km/s

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Explanation

(a) $\frac{v_{p}}{v_{e}} = \sqrt{\frac{M_{p}}{M_{e}} \frac{R_{e}}{R_{p}}} = \sqrt{(1000) \times (\frac{1}{10})} = 10 v_{p} = 10 \times 11.2 = 112 k m / s$

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