what is the height at which the value of g is same as depth of R2?

Dear Student,

 

Please find below the solution to the asked query:

Suppose $\rho$ is the density of earth, and M is the total mass, so we have
$M=\frac{4}{3}{\mathrm{\pi R}}^3$
Now the mass for the sphere of radius $\frac{R}{2}$ is
 $M\text{'}=\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{R}}{2}\right)}^3=\frac{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}{8}=\frac{\mathrm{M}}{8}$
So g at the the depth of $\frac{R}{2}$ is 
$g_{r/2}=\frac{M\text{'}G}{{\left(R/2\right)}^2}=\frac{GM/8}{R^2/4}=\frac{1}{2}\frac{GM}{R^2}$
Now suppose the g is same at height h so we have
$$\begin{gathered} g_h=\frac{GM}{(R+h{)}^2}=\frac{1}{2}\frac{GM}{R^2} \\ 2R^2=(R+h{)}^2=R^2+h^2+2Rh \\ h=-2.4R \text{and} 0.4R \end{gathered}$$
Sine negative value of h is not possible, the answer is 0.4R. 
 

Hope this information will clear your doubts about gravitation.

 

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

 

Regards