Question

A body weigh 63N on the surface of the earth. what is the gravitational force on it due to earth at a height = half the radius of the earth

Answer 1

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



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Answer 2

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