Question

a body weighs 64 Newton on the surface of the earth what is the gravitational force on it due to the earth at a height equals to the one third of the radius of the earth​

Answer 1

Answer:

36

m=64kg

gravitational force=?

h=1/3*radius of the earth

gH=gr^2/(R+h)^2

mgh=mg* R^2/(R+h)^2

64*R^2/(R+R/3)^2

64*R^2/16R^2/9

64*9/16

4*9

=36

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.