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Gravitation MCQs for NEET — Physics Questions with Answers

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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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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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