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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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3 particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is

zero

$\frac{3 G M}{L^{2}}$

$\frac{9 G M}{L^{2}}$

$\frac{12}{\sqrt{3}} \frac{G M}{L^{2}}$

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The weight of an object in the coal mine, sea level, at the top of the mountain are $W_{1}$ , $W_{2}$ and $W_{3}$ respectively, then

$W_{1} \leq W_{2} \geq W_{3}$

$W_{1} = W_{2} = W_{3}$

$W_{1} \leq W_{2} \leq W_{3}$

$W_{1} \geq W_{2} \geq W_{3}$

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Explanation

(a) Because value of g decreases when we move either in coal mine or at the top of mountain

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The value of escape velocity on a certain planet is 2 km/s. Then the value of orbital speed for a satellite orbiting close to its surface is -

12km/s

1km/s

$\sqrt{2} k m / s$

$2 \sqrt{2} k m / s$

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Explanation

$(c) v_{0} = \frac{v_{e}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} k m / s$

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Four particles each of mass M, are located at the vertices of a square with side L. The gravitational potential due to this at the centre of the square is

$- \sqrt{32} \frac{G M}{L}$

$- \sqrt{64} \frac{G M}{L^{2}}$

zero

$\sqrt{32} \frac{G M}{L}$

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Explanation

$P o t e n t i a l a t t h e c e n t r e d u e t o \sin g l e m a s s = \frac{- G M}{L / \sqrt{2}} P o t e n t i a l a t t h e c e n t r e d u e t o a l l f o u r m a s s e s = - 4 \frac{G M}{L / \sqrt{2}} = - 4 \sqrt{2} \frac{G M}{L} = - \sqrt{32} \times \frac{G M}{L} .$

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The mass of the moon is $7.34 \times 10^{22} k g$ and the radius is $1.74 \times 10^{6} m$ . The value of gravitation field will be

$1.45 N / k g$

$1.55 N / k g$

$1.75 N / k g$

$1.62 N / k g$

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Explanation

(d) $g = \frac{G M}{R^{2}} = \frac{6.67 \times 10^{‐ 11} \times 7.34 \times 10^{22}}{{(1.74 \times 10^{6})}^{2}} = 1.62 N / k g$

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The centripetal force acting on a satellite orbiting round the earth and the gravitational force of earth acting on the satellite both equal F. The net force on the satellite is

Zero

$F \sqrt{2}$

2 F

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Explanation

(b) Actually gravitational force provides the centripetal force.

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