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?

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

g_{1} = 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,

g_{1} = g(1 + R/2R)^{2}

g_{1} = g / (3/2)^{2}

or g_{1} = (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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