Question

A body weight 63 Newton on the surface of earth what is the gravitational force on it at the height half of radius of earth?​

Answer 1

Answer:

The initial force exerted is F = mg = 63 N

so the mass of the body would be m = F/g = 63/10 kg = 6.3 kg

Now, the variation in g due to height is

g1 = g / (1 + (h/R))^2

R is the radius of Earth

g is acceleration due to gravity close to Earth's surface (or on it)

here h is half the radius of the Earth or h = R/2

so,

g1= g(1 + R/2R)^2

g1 = g / (3/2)^2

or g1

= (4/9)g = 4.44 m/s^2

Now, the force experienced by the body of mass 6.3 kg at height half the radius of Earth would be

F1 = 6.3 x 4.4 = 27.72 N