Gravitation

JEE Main level, single correct, medium

198 of 223

Question 198 of 223

The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

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Answer: (C) 40 hours

T2∝R3⇒(T1 T2)2=(R1R2)3\mathrm{T}^2 \propto \mathrm{R}^3 \Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3
⇒(T1 T2)=(R1R2)3/2=(14)3/2⇒T2 T1=(4)3/2=8\Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^{3 / 2}=\left(\frac{1}{4}\right)^{3 / 2} \quad \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=(4)^{3 / 2}=8
⇒T2=8×T1=8×5=40=40\Rightarrow \mathrm{T}_2=8 \times \mathrm{T}_1=8 \times 5=40=40 hours

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