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Step-by-Step Solution
1. Understand the Problem
We need to find the height $h$ above the Earth's surface at which a body has the same weight as it would at a depth $h$ below the Earth's surface. We neglect Earth’s rotation, and let $R$ be the Earth’s radius.
2. Mass Distribution
Let $M$ be the mass of the Earth. For the point located a depth $h$ below the surface, only the portion of the Earth of radius $(R - h)$ contributes to the gravitational force. If $M_1$ is the mass of the spherical region of radius $(R - h)$ inside Earth, then:
$M_1 = M \cdot \dfrac{\frac{4}{3}\pi (R - h)^3}{\frac{4}{3}\pi R^3} \;=\; M \dfrac{(R - h)^3}{R^3}.$
3. Equating Gravitational Accelerations
At depth $h$, the gravitational acceleration is
$g_P = \dfrac{G M_1}{(R - h)^2},$
and at height $h$, the gravitational acceleration is
$g_Q = \dfrac{G M}{(R + h)^2}.$
Since the weight is the same at these two points, $g_P = g_Q.$ Therefore,
$\dfrac{G M_1}{(R - h)^2} \;=\; \dfrac{G M}{(R + h)^2}.$
4. Substitute Mass Expression and Simplify
Substitute $M_1 = M \dfrac{(R - h)^3}{R^3}$ into the above equation:
$\dfrac{G \, M \bigl(\frac{(R - h)^3}{R^3}\bigr)}{(R - h)^2}
\;=\; \dfrac{G M}{(R + h)^2}.$
Cancel $G$ and $M$ on both sides, and simplify:
$\dfrac{(R - h)^3}{R^3 \,(R - h)^2} = \dfrac{1}{(R + h)^2}
\;\;\Longrightarrow\;\;
\dfrac{(R - h)}{R^3} = \dfrac{1}{(R + h)^2}.$
Rearranging, we get:
$(R - h)\,(R + h)^2 = R^3.$
5. Expand and Solve for h
Expand $(R - h)(R + h)^2$:
$(R - h)(R + h)^2 = (R - h)(R^2 + 2Rh + h^2).$
Upon expansion and simplification, it equals $R^3.$ We end up with a quadratic in $h$:
$h^2 + R\,h - R^2 = 0.$
Solving for $h$ using the quadratic formula $h = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ (with $a=1, b=R, c=-R^2$):
$h = \dfrac{-R \pm \sqrt{R^2 + 4R^2}}{2}
\;=\;
\dfrac{-R \pm \sqrt{5}\,R}{2}.$
Since $h$ must be positive (height above the surface),
$h = \dfrac{\sqrt{5}\,R - R}{2}.$
6. Final Answer
Therefore, the height $h$ at which the body’s weight is the same as at the depth $h$ inside Earth is
$\displaystyle h \;=\; \dfrac{\sqrt{5}\,R - R}{2}\,.$
