Question

At what distance above and below the surface of the earth a body will have the same weight (take radius of earth as R)?

• A. \(\frac{R}{2}\)
• B. \((\sqrt5-1)\frac{R}{2}\)
• C. \((\sqrt3-1)\frac{R}{2}\)
• D. \((\sqrt5-1)R\)

Answer 1

The Correct Option is B

To find the distances above and below Earth's surface where an object's weight is constant, we analyze gravitational force variations.

Gravitational force at a distance \( x \) from Earth's center (where \( R \) is Earth's radius) is:

1. Above the surface (at height \( h \)):

\(F_{\text{above}} = \frac{GMm}{(R+h)^2}\)

2. Below the surface (at depth \( d \)):

\(F_{\text{below}} = \frac{GMm}{R^2} \cdot \left(1 - \frac{d}{R}\right)\)

Setting these forces equal:

\(\frac{GMm}{(R+h)^2} = \frac{GMm}{R^2} \cdot \left(1 - \frac{d}{R}\right)\)

Canceling \( GMm \) and solving for \( h \) in terms of \( d \):

\(\frac{1}{(R+h)^2} = \frac{1}{R^2} \left(1 - \frac{d}{R}\right)\)

Further simplification yields:

\(R^2 = (R+h)^2 \left(1 - \frac{d}{R}\right)\)

For symmetric weight, \( d = h \). Substituting this:

\((R+h)^2 \left(1 - \frac{h}{R}\right) = R^2\)

Solving this quadratic equation for \( h \):

\(h = (\sqrt5 - 1)\frac{R}{2}\)

Thus, an object has the same weight at a distance of \( (\sqrt5 - 1)\frac{R}{2} \) above and below Earth's surface.

This confirms \( (\sqrt5 - 1)\frac{R}{2} \) as the correct distance.