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

Answer: (B)

$(\sqrt5-1)\frac{R}{2}$

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