NEET Class neet Answered
A body weight 72 N on the surface of earth. What is the gravitation force on it, at a height equal to half the radius of the earth
Asked by honeysap7 | 31 Jul, 2021, 09:44: AM
Expert Answer
Let m be the mass of the body . Gravitational force F acting on the body is given as
F = m g = m × [ (G M ) / R2 ] = 72 N ..........................(1)
where g is acceleration due to gravity on earth's surface , G is gravitatioal constant , M is mass of earth and R is radius of earth.
Gravitational force F' acting on the body at a height equals half of earths radius R is given as
F' = m × [ (G M ) / (R+0.5R) 2 ] = (1/2.25) m × [ (G M ) / R2 ] .....................(2)
Using eqn.(1) , Gravitational force F' acting on the body at a height equals half of earths radius R is given as
F' = ( 1 / 2.25 ) × 72 = 32 N
Answered by Thiyagarajan K | 31 Jul, 2021, 12:47: PM
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