Question 309: Gravitation

Question

The height ' $h$ ' above the earth's surface at which the value of acceleration due to gravity $(\mathrm{g})$ becomes $\left(\frac{\mathrm{g}}{3}\right)$ is ( $\mathrm{R}=$ radius of the earth)

Accepted Answer

$$\begin{aligned}
&	ext { We know that, }\
&\begin{aligned}
& g_h=g\left(\frac{R}{R+h}ight)^2 \
& 	ext { For } g_h=\frac{g}{3} \
& \frac{g}{3}=g\left(\frac{R}{R+h}ight)^2 \
& \frac{1}{\sqrt{3}}=\frac{R}{R+h} \
& \sqrt{3} R=R+h \
	herefore\quad& h=(\sqrt{3}-1) R
\end{aligned}
\end{aligned}$$

Answer

$(\sqrt{3}+1) R$

Answer

$(\sqrt{3}-1) R$

Answer

$\sqrt{3} \mathrm{R}$

Answer

$3 \sqrt{R}$

Gravitation

The height ' h ' above the earth's surface at which the value of acceleration due to gravity (\mathrm{g}) becomes \left(\frac{\mathrm{g}}{3}\right) is ( \mathrm{R}= radius of the earth)

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