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The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 \%$ of its value at the surface is
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Verified Answer
The correct answer is:
$9 R$
Acceleration due to gravity at a height $h$ from the earth's surface,
$g^{\prime}=\frac{G M}{(R+h)^2}$
Given, $\quad g^{\prime}=1 \%$ of $g=\frac{g}{100}$
$\begin{aligned}
& \Rightarrow \quad \frac{g}{100}=\frac{G M}{(R+h)^2} \\
& \frac{(R+h)^2}{100}=\frac{G M}{g} \text { or } \frac{(R+h)^2}{100}=R^2 \quad\left(\because g=\frac{G M}{R^2}\right) \\
& \text { or } \begin{aligned}
R+h & =10 R \\
h & =9 R
\end{aligned}
\end{aligned}$
where, $R$ is the radius of the earth.
$g^{\prime}=\frac{G M}{(R+h)^2}$
Given, $\quad g^{\prime}=1 \%$ of $g=\frac{g}{100}$
$\begin{aligned}
& \Rightarrow \quad \frac{g}{100}=\frac{G M}{(R+h)^2} \\
& \frac{(R+h)^2}{100}=\frac{G M}{g} \text { or } \frac{(R+h)^2}{100}=R^2 \quad\left(\because g=\frac{G M}{R^2}\right) \\
& \text { or } \begin{aligned}
R+h & =10 R \\
h & =9 R
\end{aligned}
\end{aligned}$
where, $R$ is the radius of the earth.
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