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Gravitation MCQs for NEET — Physics Questions with Answers

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If acceleration due to gravity on the surface of a planet is two times that on surface of earth and its radius is double that of earth. Then escape velocity from the surface of that planet in comparison to earth will be -

2 v_e

3 v_e

4 v_e

None of these

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Explanation

(a) v= $\sqrt{2 g R} .$ If acceleration due to gravity and radius of the planet, both are double that of earth then escape velocity will be two times. i.e. $v_{p} = 2 v_{e}$

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A body weight W newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be

$\frac{w}{2}$

$2 \frac{w}{3}$

$4 \frac{w}{9}$

$\frac{8 w}{27}$

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Explanation

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The escape velocity of a rocket launched from the surface of the earth

Does not depend on the mass of the rocket

Does not depend on the mass of the earth

Depends on the mass of the planet towards which it is moving

None of the above

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Explanation

(a)

$v_{e} = \sqrt{\frac{2 G M}{R}} \Rightarrow T h e e s c a p e v e l o c i t y o f a r o c k e t l a u n c h e d f r o m t h e s u r f a c e o f t h e e a r t h D o e s n o t d e p e n d o n t h e m a s s o f t h e r o c k e t .$

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Which of the following is the evidence to show that there must be a force acting on earth and directed towards the sun?

Deviation of the falling bodies towards east

Revolution of the earth around the sun

Phenomenon of day and night

The apparent motion of the sun around the earth

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Explanation

(b)

The earth revolves around the sun due to the gravitation pull of the sun. Due to this gravitational attraction between this celestial body, the centripetal force is generated which binds the solar system together. Hence revolution of the earth around the sun is the evidence to show that there must be a force acting on earth and directed towards the sun.

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A mass of $6 \times 10^{24} k g$ is to be compressed in a sphere in such a way that the escape velocity from the sphere is $3 \times 10^{8} m / s$ . Radius of the sphere should be $(G = 6.67 \times 10^{- 11} N - m^{2} / k g^{2})$

9 km

9 m

9 cm

9 mm

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Explanation

$(d) v_{e} = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{R}} = 3 \times 10^{8} B y s o l v i n g R = 9 m m$

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The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

$9.8 m / s e c^{2}$

$4.9 m / s e c^{2}$

$980 m / s e c^{2}$

$19.6 m / s e c^{2}$

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Explanation

(b) $\frac{g'}{g} = \frac{M'}{M_{2}} {[\frac{R}{R'}]}^{2} = [\frac{2 M}{M}] {[\frac{R}{2 R}]}^{2} = \frac{1}{2}$

$g^{1} = \frac{g}{2} = \frac{9.8}{2} = 4.9 m / s^{2}$

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Force of gravity is least at

The equator

The poles

A point in between equator and any pole

None of these

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Explanation

(a)

The equator has the least gravitational pull, that is there will be the least weight for an object there.

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The velocity with which a projectile must be fired so that it escapes earth’s gravitation does not depend on:

Mass of the earth

Mass of the projectile

Radius of the projectile’s orbit

Gravitational constant

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Explanation

(b) We know that, $V_{e} = \sqrt{\frac{2 G M}{r}}$

From above expression we can see, mass of projectile does not matter.

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The gravitational force between two stones of mass 1 kg each separated by a distance of 1 metre in vacuum is

Zero

6.675 $\times 10^{‐ 5} n e w t o n$

$6.675 \times 10^{‐ 11} n e w t o n$

$6.675 \times 10^{‐ 8} n e w t o n$

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Explanation

(c) $F = G \frac{m_{1} m_{2}}{r^{2}} = 6.675 \times \frac{1 \times 1}{1^{2}} \times 10^{‐ 11} = 6.675 \times 10^{‐ 11} N$

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The radius of a planet is $\frac{1}{4}$ of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surface ?

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

$\sqrt{2}$

$2 \sqrt{2}$

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Explanation

$(a) v = \sqrt{2 g R} \Rightarrow \frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{e}}{R_{p}}} = \sqrt{2 \times \frac{1}{4}} = \frac{1}{\sqrt{2}} s o, v_{p} = \frac{v_{e}}{\sqrt{2}}$

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