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If the radius of a planet is four times that of earth and the value of g is the same for both, the escape velocity on the planet will be:

11.2 km/s

5.6 km/s

22.4 km/s

None

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Explanation

$v_{e} = \sqrt{\frac{2 G M}{R}} a n d g = \frac{G M}{R^{2}} \Rightarrow v_{e} = \sqrt{2 g R} (F o r e a r t h) v_{P} = \sqrt{2 \times g \times 4 R} = 2 v_{e} \Rightarrow v_{p} = 2 \times 11.2 k m / s = 22.4 k m / s$

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The earth (mass = $6 \times 10^{24} k g$ ) revolves round the sun with angular velocity $2 \times 10^{‐ 7} r a d / s$ in a circular orbit of radius $1.5 \times 10^{8} k m$ . The force exerted by the sun on the earth in Newtons, is

$18 \times 10^{25}$

Zero

$27 \times 10^{39}$

$36 \times 10^{21}$

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Explanation

(d) $m = 6 \times 10^{24} k g, ω = 2 \times 10^{‐ 7} r a d s, R = 1.5 \times 10^{11} m$

The force exerted by the sun on the earth

$F = m ω^{2} R$

By substituting the value we can get,

$F = 36 \times 10^{21} N$

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If the radius and acceleration due to gravity both are doubled, escape velocity of earth will become

11.2 km/s

22.4 km/s

5.6 km/s

44.8 km/s

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Explanation

$(b) v = \sqrt{2 g R} . I f g a n d R b o t h a r e d o u b l e d t h e n v w i l l b e c o m e s t w o t i m e s i . e . 11.2 \times 2 = 22.4 k m / s$

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The gravitational force between two point masses $m_{1}$ and $m_{2}$ at separation r is given by $F = k \frac{m_{1} m_{2}}{r^{2}}$

The constant k

Depends on system of units only

Depends on medium between masses only

Depends on both (a) and (b)

Is independent of both (a) and (b)

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Explanation

(a) k represents gravitational constant which depends only on the system of units.

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The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be

$\frac{1}{4} R$

$\frac{1}{2} R$

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Explanation

(d) $g = \frac{4}{3} π p G R \Rightarrow \frac{R_{p}}{R_{ε}} = (\frac{g_{p}}{g_{ε}}) (\frac{p_{ε}}{p_{p}}) = (1) \times (\frac{1}{2}) \Rightarrow R_{p} = \frac{R_{ε}}{2} = \frac{R}{2}$

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If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

$4 π G / 3 g R$

$3 π R / 4 g G$

$3 g / 4 π R G$

$π R G / 12 G$

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Explanation

$g = \frac{4}{3} \frac{π R^{3} \times G ρ}{R^{2}} \Rightarrow ρ = \frac{3 g}{4 π R G}$

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A planet has twice the radius but the mean density is $\frac{1}{4} t h$ as compared to earth. What is the ratio of escape velocity from earth to that from the planet ?

3:1

1:2

1:1

2:1

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Explanation

$(c) v = R \sqrt{\frac{8}{3} πρG} \Rightarrow \frac{v_{p}}{v_{e}} = \frac{R_{p}}{R_{e}} \sqrt{\frac{ρ_{p}}{ρ_{e}}} = 2 \sqrt{\frac{1}{4}} = 1$

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For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

$\frac{1}{2}$

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

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Explanation

$(b) F o r a m o v i n g s a t e l l i t e K i n e t i c e n e r g y = \frac{G M m}{2 r} P o t e n t i a l e n e r g y = \frac{- G M m}{r} s o, \frac{K i n e t i c e n e r g y}{P o t e n t i a l e n e r g y} = \frac{1}{2}$

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A body weight 500 N on the surface of the earth. How much would it weigh half way below the surface of the earth ?

125 N

250 N

500 N

1000 N

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Explanation

(b) Weight on surface of earth, mg= 500 N

and weight below the surface of earth at $d = \frac{R}{2}$

$m g' = m g (1 - \frac{d}{R}) = m g (1 - \frac{1}{2}) = \frac{m g}{2} = 250 N$

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Who among the following gave first the experimental value of G?

Cavendish

Copernicus

Brook Taylor

None of these

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Explanation

(a)

It was Lord Henry Cavendish who determined the experimental value of G(universal gravitational constant) in 1798 using a torsion balance.

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