Question

A body weighs 63 N on the surface of the earth. What is the gravitational force exerted on it due to the earth at a height equal to half the radius of the earth?

Answer 1

Answer:-

Weight of the body, W = 63 N
Acceleration due to gravity at height hfrom the Earth’s surface is given by the relation:
g‘ = g / [1 + ( h / Re) ]2
Where,
g = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth
For h = Re / 2
g‘ = g / [(1 + (Re / 2Re) ]2
= g / [1 + (1/2) ]2  =  (4/9)g
Weight of a body of mass m at height h is given as:
W‘ = mg
= m × (4/9)g  =  (4/9)mg
= (4/9)W
= (4/9) × 63  =  28 N.

Answer 2

Answer:

Weight of the body, W = 63 N

Acceleration due to gravity at height h from the Earth’s surface is given by the relation:

g' = g / [1 + ( h / Re) ]2

Where,

g = Acceleration due to gravity on the Earth’s surface

Re = Radius of the Earth

For h = Re / 2

g' = g / [(1 + (Re / 2Re) ]2

= g / [1 + (1/2) ]2 = (4/9)g

Weight of a body of mass m at height h is given as:

W' = mg

= m × (4/9)g = (4/9)mg

= (4/9)W

= (4/9) × 63 = 28 N.