Animated Solution for Physics - Gravitation: The height $h$ at which the weight of a body will be the same as that at the same depth $h$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected)

Equating Weights

Imagine you are holding an object at a certain height $h$ above the Earth's surface. Now, imagine taking that exact same object deep underground to a depth $h$. The problem presents a fascinating scenario: the weight of the object is identical in both locations.

Since weight is the product of mass and the acceleration due to gravity ($W=mg$), equating the weights directly implies that the acceleration due to gravity at height $h$ ($g_h$) is equal to the acceleration due to gravity at depth $h$ ($g_d$).

To solve this, we must deploy the precise formulas for gravity variation. For depth, the formula is straightforward and linear:

For height, we must be cautious. A common trap is to use the binomial approximation $g_h\approx g(1-2h/R)$. However, this is only valid when $h\ll R$. Since the problem does not state this, and the options are in terms of $R$, we must use the exact inverse-square formula:

Equating the two gives us our master equation:

First, we can elegantly cancel out $g$ from both sides. Taking the LCM on the right side yields:

Now, we cross-multiply to eliminate the fractions, setting the stage for some algebraic expansion:

Let's carefully expand the right-hand side. We know $(R+h{)}^2=R^2+2Rh+h^2$. Multiplying this by $(R-h)$ gives:

Notice how the $R^3$ terms beautifully cancel out on both sides. Grouping the remaining terms leaves us with:

Since we are looking for a non-zero height, we can safely divide the entire equation by $h$. This reduces our cubic equation into a much friendlier quadratic equation:

This is a standard quadratic equation in the form $a{x}^2+bx+c=0$. Applying the quadratic formula to solve for $h$:

Since $h$ represents a physical distance, it must be a positive value. Therefore, we discard the negative root, leaving us with our final, elegant answer: