Question

A body weighs 500 N at the surface of earth. How
much will it weigh at a height equal to the radius
of earth​

Answer 1

$F_1$ = 124.99 N

Explanation:

The initial force exerted is F = mg

                                               = 500 N

The mass of the body m,

m =$\frac{F}{g}$

     = $\frac{500}{9.8} kg$

    = 51.02 kg

Now, the variation in g due to height is

$g_1 = \frac{g}{(1 + (h/R))^2}$

Here,R = the radius of Earth

g is acceleration due to gravity close to Earth's surface (or on it)

h = R

So,

$g_1=\frac{g}{(1 + R/R)^2}$

$g_1 = \frac{g}{(2)^2}$

$g_1$ = 0.25g

     = $2.45 m/s^2$

Now, the force experienced by the body of mass 51.02 kg at height equal to the radius of Earth would be

$F_1$= 2.45 x 51.02

$F_1$ = 124.99 N