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 determine the distance above and below the Earth's surface where a body has the same weight, we need to understand the concept of gravitational force variation above and below the Earth's surface.

The gravitational force experienced by a body at a distance \( x \) above or below the Earth's surface can be expressed using the formula for gravitational force: 

1. Above the Earth's surface:

The gravitational force at a height \( h \) above the Earth's surface:

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

2. Below the Earth's surface:

The gravitational force at a depth \( d \) below the Earth's surface:

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

We are given that these forces are equal, so:

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

By simplifying, we cancel out \(GMm\) from both sides and solve for \( h \) in terms of \( d \):

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

Simplifying further, we obtain:

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

Let \( d = h \) because they need to be symmetric for the weight to be the same. Hence the equation becomes:

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

Solving this quadratic equation for \( h \), we find:

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

This means that the body will have the same weight at a distance \( (\sqrt5 - 1)\frac{R}{2} \) both above and below the Earth's surface. Therefore, the correct answer is \((\sqrt5 - 1)\frac{R}{2}\).

This comprehensive solution confirms that the option \((\sqrt5 - 1)\frac{R}{2}\) is correct.

Answer 2

Step 1: Define Gravitational Acceleration Above and Below: - Above the earth’s surface at height h, gravitational acceleration g_p is:

\( g_{p} = \frac{gR^2}{(R + h)^2} \)

- Below the earth’s surface at depth h, gravitational acceleration g_q is:

\( g_{q} = g \left(1 - \frac{h}{R}\right) \)

Step 2: Set g_p = g_q:

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

Step 3: Simplify the Equation:

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

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

Step 4: Let \( \frac{h}{R} = x \):

\( (1 - x)(1 + x) = 1 \)

\( 1 - x^2 = 1 \)

\( x = \frac{\sqrt{5} - 1}{2} \)

Step 5: Calculate h:

\( h = \frac{R}{2}(\sqrt{5} - 1) \)

So, the correct answer is: \( h = \frac{R}{2}(\sqrt{5} - 1) \)