Question

A body weighs 72 N on the surface of the earth. What is the gravitational force on it due
to the earth at a height equal to half the radius of the earth from the surface?​

Answer 1

Explanation:

According to Newton's gravitational law,

W_f=\frac{W}{\left(1+\frac{h}{r}\right)^2}

here W is the weight of body on he surface of earth, h is the height of body from the surface of earth, r is the radius of earth.

here, weight to body , W = mg = 72N on the surface of earth. we have to find out weight of body at a height equal to half the radius of the earth above the earth's surface.

so, h = r/2

so, weight of body = \frac{72}{\left(1+\frac{r/2}{r}\right)^2}

= 72/(1 + 1/2)² = 72/(3/2)² = 72 × 4/9 = 32N

Answer 2

Answer:32 N

Explanation:

I have attached a solution