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

Question

A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth? (A) 48 N (B) 32 N (C) 30 N (D) 24 N

Tardigrade Admin · Accepted Answer

ws=mgs=72 N wh=mgh=(mgs/(1+(h)R)2) =(72 N/(1+(R /2)2)) =(72/9 /4) wh=32 N

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

48 N

32 N

30 N

24 N

$w_{s} = m g_{s} = 72 N$
$w_{h} = m g_{h} = \frac{m g _{s}}{( 1 + \frac{h}{R} ) ^{2}}$
$= \frac{72 N}{( 1 + \frac{R /2}{2} )}$
$= \frac{72}{9/4}$
$w_{h} = 32 N$

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

48 N

32 N

30 N

24 N

ws​=mgs​=72N wh​=mgh​=(1+Rh​)2mgs​​ =(1+2R/2​)72N​ =9/472​ wh​=32N

$w_{s} = m g_{s} = 72 N$
$w_{h} = m g_{h} = \frac{m g _{s}}{( 1 + \frac{h}{R} ) ^{2}}$
$= \frac{72 N}{( 1 + \frac{R /2}{2} )}$
$= \frac{72}{9/4}$
$w_{h} = 32 N$