Question
A body weighs 72 N on the surface of the Earth. What will be the force of gravity acting on the body at a height equal to half the radius of the Earth above the Earth's surface?
Answer: Option A
:
A
Weight of the body at earth's surface is:
We=GMmR2 .......(1)
where, G-universal gravitational constant, M-mass of earth and R-radius of earth, m- Mass of the body
Weight of the body at a height h above the earth's surface is given by
W′e=GMm(R+h)2
According to the question, h=R2
W′e=GMm(R+R2)2
W′e=4GMm9R2 ....(2)
Dividing equation (2) by equation(1), we get
W′eWe=4GMm/9R2GMm/R2
W′eWe=49
Given weight on earth's surface is, We=72 N
W′e72=49
We=32 N
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:
A
Weight of the body at earth's surface is:
We=GMmR2 .......(1)
where, G-universal gravitational constant, M-mass of earth and R-radius of earth, m- Mass of the body
Weight of the body at a height h above the earth's surface is given by
W′e=GMm(R+h)2
According to the question, h=R2
W′e=GMm(R+R2)2
W′e=4GMm9R2 ....(2)
Dividing equation (2) by equation(1), we get
W′eWe=4GMm/9R2GMm/R2
W′eWe=49
Given weight on earth's surface is, We=72 N
W′e72=49
We=32 N
Was this answer helpful ?
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