Acceleration due to gravity at a height \(\mathrm{H}\) from the surface of a planet is the same as that at a depth of \(\mathrm{H}\) below of surface. If \(\mathrm{R}\) be the radius of the planet, then \(\mathrm{H}\) vs. \(\mathrm{R}\) graph for different planets will be

• A

  
  
• B

  
  
• Correct answer
  C

  
  
• D

  
  

Solution

Hint :
\(\begin{aligned} & \frac{G M}{(R+H)^2}=\frac{G M(R-H)}{R^3} \\ & (R+H)^2(R-H)=R^3 \\ & R^3-R H^2+H R^2-H^3=R^3 \\ & H^2-R^2+R H=0 \\ & H=\frac{(\sqrt{5}-1)}{2} R\end{aligned}\)

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