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Gravitation MCQs for NEET — Physics Questions with Answers

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For the moon to cease to remain the earth's satellite, its orbital velocity has to increase by a factor of -

$\sqrt{2}$

1/ $\sqrt{2}$

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Explanation

(b) $v_{e} = \sqrt{2} v_{0,}$ i.e. if the orbital velocity of moon is increased by factor of $\sqrt{2}$ then it will escape out from the gravitational field of earth

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The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R)

10r

20r

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Explanation

(b)

$g = \frac{G M}{R^{2}} g' = \frac{G M}{(R + h)^{2}} & g' = 1 % o f g' = \frac{1}{100} \times g \frac{g'}{g} = {(\frac{R}{R + h})}^{2} \Rightarrow \frac{1}{100} = {(\frac{R}{R + h})}^{2} \Rightarrow h = 9 r$

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Escape velocity on a planet is $v_{e}$ . If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

$4 v_{d}$

$2 v_{d}$

$v_{d}$

$\frac{1}{2} v_{d}$

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Explanation

(b) $v_{e} = \sqrt{\frac{2 G M}{R}} \Rightarrow v_{e} \propto \sqrt{M} i f R = c o n s \tan t$

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The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will be

0.2

2.57

4.81

0.39

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Explanation

$v_{e} = \sqrt{\frac{2 G M}{R}} s o, \frac{v_{e}}{v_{m}} = \sqrt{\frac{M_{e} R_{m}}{M_{m} R_{e}}} = \sqrt{\frac{81}{3.5}} = 4.81$

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If radius of earth is R then the height h’ at which value of ‘g’ becomes one-fourth is

$\frac{R}{4}$

$\frac{3 r}{4}$

$\frac{R}{8}$

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Explanation

By solving h = R

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The escape velocity from the surface of earth is $V_{e}$ . The escape velocity from the surface of a planet whose mass and radius are 3 times those of the earth will be

$V_{d}$

$3 V_{d}$

$9 V_{d}$

$27 V_{d}$

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Explanation

(a) $v_{e} = \sqrt{\frac{2 G M}{R}} s o, v_{e} \propto \sqrt{\frac{M}{R}}$

If mass and radius of the planet are three times than that of earth then escape velocity will be same

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How much energy will be necessary for making a body of 500 kg escape from the earth ?

[g = 9.8 m / s^{2}, r a d i u s o f e a r t h = 6.4 \times 10^{6} m]

$A b o u t 9.8 \times 10^{6} J$

$A b o u t 6.4 \times 10^{8} J$

$A b o u t 3.1 \times 10^{10} J$

$A b o u t 27.4 \times 10^{12} J$

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Explanation

$P E = - \frac{G M m}{R} = \frac{g R^{2} m}{R} = - m g R = - 500 \times 9.8 \times 6.4 \times 10^{6} = - 3.1 \times 10^{10} J$

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Two planets have the same average density but their radii are $R_{1}$ and $R_{2}$ . If acceleration due to gravity on these planets be $g_{1}$ and $g_{2}$ respectively, then

$\frac{g_{1}}{g_{2}}$ = $\frac{R_{1}}{R_{2}}$

$\frac{g_{1}}{g_{2}}$ = $\frac{R_{2}}{R_{1}}$

$\frac{g_{1}}{g_{2}}$ = $\frac{{R_{1}}^{2}}{{R_{2}}^{2}}$

$\frac{g_{1}}{g_{2}}$ = $\frac{{R_{1}}^{3}}{{R_{2}}^{3}}$

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Explanation

(a) g= $\frac{4}{3} π ρ G R$ . If $ρ$ = constant then $\frac{g_{1}}{g_{2}}$ = $\frac{R_{1}}{R_{2}}$

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The escape velocity for the earth is 11.2 km/sec. The mass of another planet is 100 times that of the earth and its radius is 4 times that of the earth. The escape velocity for this planet will be

112.0 km/s

5.6 km/s

280.0 km/s

56.0 km/s

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Explanation

(d) $E s c a p e v e l o c i t y v = \sqrt{\frac{2 G M}{R}} \Rightarrow \sqrt{\frac{v_{p}}{v_{e}}} = \sqrt{\frac{M_{p}}{M_{e}} \times \frac{R_{e}}{R_{p}}} \Rightarrow v_{p} = 5 v_{e} = 5 \times 11.2 = 56 k m / s$

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Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If $R_{E}$ is the maximum range of a projectile on the earth’s surface, what is the maximum range on the surface of the moon for the same velocity of projection ?

0.2 $R_{ε}$

2 $R_{ε}$

0.5 $R_{ε}$

5 $R_{ε}$

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Explanation

(d) Range of projectile $R = \frac{u^{2} \sin 2 θ}{g}$

if u and $θ$ are constant then $R \propto \frac{1}{g}$

$\frac{R_{m}}{R_{E}} = \frac{g_{E}}{g_{m}} \Rightarrow \frac{R_{m}}{R_{E}} = \frac{1}{0.2} \Rightarrow R_{m} = \frac{R_{E}}{0.2} \Rightarrow R_{m} = 5 R_{E}$

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