Gravitation | JEE Main previous year questions with Solutions

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Topic Question Set

Q 1 :
Two satellite A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of A is 3v, the speed of B will be [2024]

$6 v$
$12 v$
$\frac{4}{3} v$
$3 v$

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4

(A)

Q 2 :
A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is

(Given = Radius of geo-stationary orbit for earth is $4.2 \times 10^{4}$ km) [2024]

$1.05 \times 10^{4}$ km
$1.4 \times 10^{4}$ km
$1.68 \times 10^{5}$ km
$8.4 \times 10^{4}$ km

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(A) By Kepler's law, $T^{2} = \frac{4 π^{2} R^{3}}{G M} \Rightarrow T^{2} \propto \frac{R^{3}}{M}$

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3} (\frac{M_{2}}{M_{1}})$

${(\frac{24}{6})}^{2} = {(\frac{4.2 \times 10^{4}}{R_{2}})}^{3} \times (\frac{\frac{M_{E}}{4}}{M_{E}})$

$4^{2} \times 4 = {(\frac{4.2 \times 10^{4}}{R_{2}})}^{3} \Rightarrow R_{2} = 1.05 \times 10^{4}$ km

Q 3 :
Correct formula for height of a satellite from Earth's surface is _____. [2024]

${(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R$
${(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 2} - R$
${(\frac{T^{2} R^{2}}{4 π^{2} g})}^{1 / 3} - R$
${(\frac{T^{2} R^{2} g}{4 π^{2}})}^{- 1 / 3} + R$

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(1)

$\frac{m v^{2}}{(R + h)} = \frac{G M m}{{(R + h)}^{2}} \Rightarrow v = \sqrt{\frac{G M}{R + h}}$

$Now T = \frac{2 π (R + h)}{\sqrt{\frac{G M}{R + h}}}$

$\Rightarrow T = \frac{2 π {(R + h)}^{3 / 2}}{\sqrt{G M}} = \frac{2 π {(R + h)}^{3 / 2}}{R \sqrt{g}}$

$\Rightarrow T^{2} R^{2} g = 4 π^{2} (R + h^{3})$

$\Rightarrow {(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R = h$

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