Question

A body weighs 63 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?

Answer 1

PLEASE READ CAREFULLY :-

EARTH g=9.8m/s(square) [on the surface of Earth]
EARTH g=0 [at the core]
NOW,
THE g at the height half the radius=9.8/2=4.9m/s(square)
SO,THE g at the height equal to half the radius =9.8+4.9=14.7 m/s(square)

THE MASS OF THE OBJECT AT THE SURFACE OF EARTH---->
W=MXg
63n=mx9.8
m=6.428kg

SO, THE GRAVITATIONAL FORCE ACTING at the height equal to half the radius of earth----->
g=14.7
W=MXg
W=6.248X14.7
W OR GRAVITATIONAL FORCE ACTING=94.4976N

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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.