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Question: A man weight \(W\) on the surface of the earth and his weight at a height \(R\) from the surface of ...

A man weight WW on the surface of the earth and his weight at a height RR from the surface of the earth is ( RR is the radius of the earth)
(A)W4\dfrac{W}{4}
(B)W2\dfrac{W}{2}
(C)WW
(D)4W4W

Explanation

Solution

The weight of a body is the force acting on that body due to gravitational pull. S.I. unit of weight is newton (N)(N) . The mass of a body always remains constant but the weight of a body changes with the change in gravitational force acting on it.
Formula used:
W=mgW = mg
W=GMmD2W = \dfrac{{GMm}}{{{D^2}}}
where WW is the gravitational force acting between two bodies, mm is mass of one of the body, MM is mass of another body, GG is the universal gravitational constant, and DD is the distance between the center of the two bodies.

Complete Step-by-step solution:
It is given that the weight of the man on the surface of the earth is WW.
Hence the mass (m)(m) of man will be Wg\dfrac{W}{g}.
i.e. m=Wg \Rightarrow m = \dfrac{W}{g}
where ggis the acceleration due to gravity near the earth’s surface which is equal to GMR2\dfrac{{GM}}{{{R^2}}}.
From Newton's law of gravitation,
W=GMmD2W = \dfrac{{GMm}}{{{D^2}}}
Where WW is the gravitational force acting between two bodies, mm is the mass of one of the body, MM is the mass of another body, GGis the universal gravitational constant, and DD is the distance between the center of the two bodies.
In our solution with regard to the question, it can be written as,
W=GMmR2\Rightarrow W = \dfrac{{GMm}}{{{R^2}}}
where WWis the weight of man at the surface of the earth, mm is the mass of man, MM is the mass of earth, GGis the universal gravitational constant, and RRis the radius of the earth.
Let the weight of man at a height RRfrom the surface of the earth i.e. at a distance of 2R2Rfrom the center of the earth be W0{W_0}.
So,
W0=GMm(R+R)2\Rightarrow {W_0} = \dfrac{{GMm}}{{{{(R + R)}^2}}}
W0=GMm(2R)2=GMm4R2\Rightarrow {W_0} = \dfrac{{GMm}}{{{{(2R)}^2}}} = \dfrac{{GMm}}{{4{R^2}}}
As we have seen earlier, W=GMmR2W = \dfrac{{GMm}}{{{R^2}}}.
Hence,
W0=W4\Rightarrow {W_0} = \dfrac{W}{4}

Therefore the correct answer to our question is (A) W4\dfrac{W}{4}.

Note:
In many questions with an increase in height, we did not consider the weight of the body is changing as the change of weight with small variation in height (in comparison to radius of the earth) will be very small and thus can be neglected. But in this question change in height was not small and was comparable to the radius of the earth thus the change produced can’t be neglected.