Question 298: Gravitation

Question

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)

Accepted Answer

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

Answer

$\frac{1}{3} R$

Answer

$\frac{2}{3} R$

Answer

$\frac{3}{2} R$

Answer

$\mathrm{2 R}$

Gravitation

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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