Gravitation

JEE Main level, single correct, medium

98 of 223

Question 98 of 223

The height at which the acceleration due to gravity becomes g9\frac{\mathrm{g}}{9} (where g=\mathrm{g}= the acceleration due to gravity on the surface of the earth) in terms of RR, the radius of the earth is

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Answer: (A) 2R2 R

g′=GM(R+h)2g^{\prime}=\frac{G M}{(R+h)^2}, acceleration due to gravity at height hh
⇒g9=GMR2⋅R2(R+h)2=g(RR+h)2⇒19=(RR+h)2⇒RR+h=13⇒3R=R+h⇒2R=h\begin{aligned} & \Rightarrow \frac{g}{9}=\frac{G M}{R^2} \cdot \frac{R^2}{(R+h)^2}=g\left(\frac{R}{R+h}\right)^2 \\ & \Rightarrow \frac{1}{9}=\left(\frac{R}{R+h}\right)^2 \Rightarrow \frac{R}{R+h}=\frac{1}{3} \\ & \Rightarrow 3 R=R+h \Rightarrow 2 R=h \end{aligned}

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