The height ' $h$ ' from the surface of the earth at which the value of ' g ' will be reduced by $64 \%$ than the value at surface of the earth is ( $\mathrm{R}=$ radius of the earth)

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Aicharya · Accepted Answer

Since, $g=\frac{G M}{R^2}$ $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}$
When value of $g$ is reduced by $64 \%$, $\mathrm{g}_{\mathrm{h}}=36 \%$ of g $\begin{array}{ll} \therefore & \frac{g_h}{g}=\frac{R^2}{(R+h)^2}=\frac{36}{100} \\ \therefore & \frac{R}{(R+h)}=\frac{6}{10} \\ \therefore & 4 R=6 h \\ \therefore & h=\frac{4}{6} R=\frac{2}{3} R \end{array}$

The height ' $h$ ' from the surface of the earth at which the value of ' g ' will be reduced by $64 \%$ than…

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