Physics, asked by Nikki57, 1 year ago

A body weights 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?

Answers

Answered by Suryavardhan1
7
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/s2

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

Answered by Anonymous
3
Hi

Here is your answer,

Acceleration due to gravity g and height h is given by 

          g = gR²/( R + h)² = gR²/(R +R/2)² = 4/9g

 Gravitational force on body at height h is 

  F = mg

 = m4/9g = mg4/9 = 63 × 4/9 = 28 N [ ∵ mg = 63]



Hope it helps you !
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