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A body is taken to a height of $n R$ from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is
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Verified Answer
The correct answer is:
$(n+1)^{2}$
Aceleration due to gravity at a height $h$ above
the earth surface $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$
$$
\begin{array}{l}
\frac{g}{g^{\prime}}=\left(\frac{R+h}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=\left(\frac{R+n R}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=(1+n)^{2}
\end{array}
$$
the earth surface $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$
$$
\begin{array}{l}
\frac{g}{g^{\prime}}=\left(\frac{R+h}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=\left(\frac{R+n R}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=(1+n)^{2}
\end{array}
$$
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