A body weight 45 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half of the radius of the earth?

Question

A body weight 45 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half of the radius of the earth? (A)  20 N  (B)  45 N  (C)  40 N  (D)  90 N

Tardigrade Admin · Accepted Answer

Acceleration due to gravity at h=(R/2) g'=(g/(1+(h)Re)2)=(g/(1+(Re)2 Re)2) =(g/((3)2)2)=(4/9) g ⇒ So w'=(4/9) w=(4/9) × 45=20 N

Q. A body weight $45 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half of the radius of the earth?

$20 N$

$45 N$

$40 N$

$90 N$

Acceleration due to gravity at $h = \frac{R}{2}$
$g^{'} = \frac{g}{( 1 + \frac{h}{R _{e}} ) ^{2}} = \frac{g}{( 1 + \frac{R _{e}}{2 R _{e}} ) ^{2}}$
$= \frac{g}{( \frac{3}{2} ) ^{2}} = \frac{4}{9} g$
$\Rightarrow$ So $w^{'} = \frac{4}{9}$
$w = \frac{4}{9} \times 45 = 20 N$

Q. A body weight 45N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half of the radius of the earth?

20N

45N

40N

90N