Question

a body weighs 72N on the surface of the earth. what will be the force of vravity acting on the body at a height equal to half the radius of the earth above the earth's surface.answer it only if you know.answer with the ezplanation. ill mark it as the brainliest who answers first and will thank you. and the points you get are 50p.

Answer 1

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:According to Newton's gravitational law,

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 =

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

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Explanation: