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Gravitation MCQs for NEET — Physics Questions with Answers

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Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to -

$R^{(\frac{n + 1}{2})}$

$R (\frac{n - 1}{2})$

$R^{n}$

$R (\frac{n - 2}{2})$

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Explanation

(a) ${mω}^{2} R \propto \frac{1}{R^{n}} \Rightarrow m (\frac{4 π^{2}}{T^{2}}) R \propto \frac{1}{R^{n}} \Rightarrow T^{2} \propto R^{n + 1} ∴ T \propto R^{(\frac{n + 1}{2})}$

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The orbital speed of an artificial satellite very close to the surface of the earth is $V_{o}$ . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is

$4 V_{0}$

$2 V_{0}$

$0.5 V_{0}$

$4 V_{0}$

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Explanation

If $r = R + h = R + 3 R = 4 R$ then $v = \frac{V_{o}}{2} = 0.5 V_{0}$

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If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would -

Decrease by 2%

Remain unchanged

Increase by 2%

Increase by 1%

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Explanation

% increase in g = 2(% decrease in R) = 2x1% = 2%

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The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to -

5 R

7 R

10 R

18 R

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Explanation

(b) 6R from the surface of earth and 7R from the centre.

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In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ${ms}^{- 2}$ and radius of earth is 6400 kms)

0 rad/sec

$\frac{1}{800}$ rad/sec

$\frac{1}{80}$ rad/sec

$\frac{1}{8}$ rad/sec

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Explanation

(b) $g' = g - ω^{2} {Rcos}^{2} λ$

For weightlessness at equator $λ = 0$ and $g' = 0$

$∴ 0 = g - ω^{2} R \Rightarrow ω = \sqrt{\frac{g}{R}} = \frac{1}{800} \frac{rad}{s}$

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A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of is $T_{2} / T_{1}$ is

$\sqrt{2}$

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Explanation

(d) If acceleration due to gravity is g at the surface of earth then at height R its value becomes $g' = g {(\frac{R}{R + h})}^{2} = \frac{g}{4}$

$T_{1} = 2 π \sqrt{\frac{l}{g}}$ and $T_{2} = 2 π \sqrt{\frac{l}{g / 4}}$

$∴ \frac{T_{2}}{T_{1}} = 2$

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A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be

mgR

$\frac{1}{2}$ mgR

2 mgR

$\frac{1}{4}$ mgR

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Explanation

(b) $∆ U = \frac{mgh}{1 + \frac{h}{R}} = \frac{1}{2} mgR (∵ h = R)$

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Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is
[g is acceleration due to gravity at Earth’s surface]

$2 π \sqrt{\frac{2 R}{g}}$

$4 \sqrt{2} π \sqrt{\frac{R}{g}}$

$2 π \sqrt{\frac{R}{g}}$

$8 π \sqrt{\frac{R}{g}}$

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Explanation

(b) $T = 2 π \sqrt{\frac{{(R + h)}^{3}}{{gR}^{2}}} = 2 π \sqrt{\frac{{(2 R)}^{3}}{{gR}^{2}}} = 4 \sqrt{2 π} \sqrt{\frac{R}{g}}$

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An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy $E_{0}$ . Its potential energy is

$- E_{0}$

$1.5 E_{0}$

$2 E_{0}$

$E_{0}$

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Explanation

Because we know = $U = \frac{- GMm}{r}$ and $E_{0} = \frac{- GMm}{2 r}$

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Given radius of Earth ‘R’ and length of a day ‘T’ , the height of a geostationary satellite is [G–Gravitational Constant, M–Mass of Earth]

${(\frac{4 π^{2} GM}{T^{2}})}^{\frac{1}{3}}$

${(\frac{4 πGM}{R^{2}})}^{\frac{1}{3}} - R$

${(\frac{{GMT}^{2}}{4 π^{2}})}^{\frac{1}{3}} - R$

${(\frac{{GMT}^{2}}{4 π^{2}})}^{\frac{1}{3}} + R$

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Explanation

(c) $T = 2 π \sqrt{\frac{r^{3}}{GM}} \Rightarrow T^{2} = \frac{4 π^{2}}{G M} {(R + h)}^{3} \Rightarrow R + h = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}} \Rightarrow h = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}} - R$

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