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Gravitation MCQs for NEET — Physics Questions with Answers

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The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is v. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is

$\frac{3}{2} v$

$\sqrt{\frac{3}{2}} v$

$\sqrt{\frac{2}{3}} v$

$\frac{2}{3} v$

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Explanation

For first satellite $h = 0$ , $v_{1} = \sqrt{\frac{G M}{R}}$

For second satellite , $h = \frac{R}{2}, v_{2} = \sqrt{\frac{2 G M}{3 R}}$

$v_{2} = \sqrt{\frac{2}{3}} v_{1} = \sqrt{\frac{2}{3}} v$

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In a satellite if the time of revolution is T, then K.E. is proportional to

$\frac{1}{T}$

$\frac{1}{T^{2}}$

$\frac{1}{T^{3}}$

$T^{\frac{- 2}{3}}$

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Explanation

(d) $v = \sqrt{\frac{G M}{r}} ∴ K . E . \propto v^{2} \propto \frac{1}{r}$ and $T^{2} \propto r^{3} ∴ K . E . \propto T^{\frac{- 2}{3}}$

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The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is

$\frac{T}{4}$

$\frac{T}{8}$

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Explanation

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If the height of a satellite from the earth is negligible in comparison to the radius of the earth R, the orbital velocity of the satellite is

gR/2

$\sqrt{g / R}$

$\sqrt{g R}$

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Explanation

(d) Orbital velocity near to earth = $\frac{v_{e}}{\sqrt{2}}$

$\Rightarrow v_{0} = \frac{\sqrt{2 g r}}{\sqrt{2}} = \sqrt{g r}$

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Choose the correct statement from the following : The radius of the orbit of a geostationary satellite depends upon -

Mass of the satellite, its time period and the gravitational constant

Mass of the satellite, mass of the earth and the gravitational constant

Mass of the earth, mass of the satellite, time period of the satellite and thegravitational constant

Mass of the earth, time period of the satellite and the gravitational constant

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Explanation

(d)

$T = 2 π \sqrt{\frac{r^{3}}{GM}} \Rightarrow r^{3} = \frac{G M T^{2}}{4 π^{2}} \Rightarrow r = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}}$

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If a body describes a circular motion under inverse square field, the time taken to complete one revolution T is related to the radius of the circular orbit as

$T \propto r$

$T \propto r^{2}$

$T^{2} \propto r^{3}$

$T \propto r^{4}$

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Explanation

(c)

$In a circular motion : F = \frac{{mv}^{2}}{r} = {mrω}^{2} In a inverse square field; F \propto \frac{1}{r^{2}} \Rightarrow {mrω}^{2} \propto \frac{1}{r^{2}} \Rightarrow ω^{2} \propto \frac{1}{r^{3}} \Rightarrow \frac{1}{T^{2}} \propto \frac{1}{r^{3}} \Rightarrow T^{2} \propto r^{3}$

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A planet moves around the sun. At a given point P, it is closest from the sun at a distance $d_{1}$ and has a speed $v_{1}$ . At another point Q, when it is farthest from the sun at a distance $d_{2}$ , its speed will be

$\frac{{d_{1}}^{2} v_{1}}{{d_{2}}^{2}}$

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

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

$\frac{{d_{2}}^{2} v_{1}}{{d_{1}}^{2}}$

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Explanation

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A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1%, its speed will

Increase by 1%

Increase by 0.5%

Decrease by 1%

Decrease by 0.5%

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Explanation

(b) $v \propto \frac{1}{\sqrt{r}} .$

% increase in speed = $\frac{1}{2}$ (% decrease in radius) $= \frac{1}{2} (1 %) = 0.5 %$

i.e. speed will increase by 0.5%

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Orbital velocity of an artificial satellite does not depend upon

Mass of the earth

Mass of the satellite

Radius of the earth

Acceleration due to gravity

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Explanation

(b) $v = \sqrt{\frac{G M}{r}}$

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The time period of a geostationary satellite is

24 hours

12 hours

365 days

One month

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Explanation

(a) $ω_{s} = ω_{e} \Rightarrow \frac{2 π}{T_{s}} = \frac{2 π}{T_{e}} o r, T_{s} = T_{e} (24 h o u r s)$

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