Q133 / 160 · AP EAMCET 2024 (20 May Shift 2) · Physics

What is the height from the surface of earth, where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth? $\left(R_E=6400 \mathrm{~km}\right)$

  • Correct answer
    A
    6400 km
  • B
    3200 km
  • C
    1600 km
  • D
    640 km

Solution

Acceleration due to gravity at height $h$ is $\begin{aligned} & g_h=\frac{g}{(R+h)^2} \Rightarrow g_h \propto \frac{1}{(R+h)^2} \\ & \frac{g_{h_2}}{g_{h_1}}=\left(\frac{\mathrm{R}+\mathrm{h}_1}{R+h_2}\right)^2 \Rightarrow \frac{\frac{g}{4}}{g}=\left(\frac{6400+0}{6400+h_2}\right)^2 \\ & \Rightarrow \frac{1}{2}=\frac{6400}{6400+h_2} \Rightarrow h_2=6400 \mathrm{~km} \end{aligned}$

Hence, the correct answer is option: 'A'.

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