Question
At depth $x$ from the earth's surface the value of $g^{\prime}$ remains half of that of at the surface. What will be the value of $g ^{\prime}$ obtained at the same height.
✓
Answer
Value of $g ^{\prime}$ at depth $x$,
$g^{\prime}=g\left(1-\frac{x}{R_c}\right)$
According to question, $g^{\prime}=\frac{g}{2}$
$\begin{aligned}
\therefore \quad \frac{g}{2} & =g\left(1-\frac{x}{R_c}\right) \\
\frac{1}{2} & =1-\frac{x}{R_c} \\
\frac{x}{R_c} & =1-\frac{1}{2}=\frac{1}{2} \\
R_e & =2 x \\ & \\
x & =\frac{1}{2} R_c
\end{aligned}$
Formula for height,
$g^{\prime}=\frac{g}{\left(1+\frac{x}{R_c}\right)^2}$
Put the values,
$\begin{array}{l}
g^{\prime}=\frac{g}{\left(1+\frac{\frac{1}{2} R_{\varepsilon}}{R_e}\right)^2}=\frac{g}{\left(\frac{3}{2}\right)^2} \\
\therefore g^{\prime}=\frac{4}{9} g
\end{array}$
Hence, value of $g ^{\prime}$ will be $=\frac{4 g}{9}$.
$g^{\prime}=g\left(1-\frac{x}{R_c}\right)$
According to question, $g^{\prime}=\frac{g}{2}$
$\begin{aligned}
\therefore \quad \frac{g}{2} & =g\left(1-\frac{x}{R_c}\right) \\
\frac{1}{2} & =1-\frac{x}{R_c} \\
\frac{x}{R_c} & =1-\frac{1}{2}=\frac{1}{2} \\
R_e & =2 x \\ & \\
x & =\frac{1}{2} R_c
\end{aligned}$
Formula for height,
$g^{\prime}=\frac{g}{\left(1+\frac{x}{R_c}\right)^2}$
Put the values,
$\begin{array}{l}
g^{\prime}=\frac{g}{\left(1+\frac{\frac{1}{2} R_{\varepsilon}}{R_e}\right)^2}=\frac{g}{\left(\frac{3}{2}\right)^2} \\
\therefore g^{\prime}=\frac{4}{9} g
\end{array}$
Hence, value of $g ^{\prime}$ will be $=\frac{4 g}{9}$.
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