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Question

Physics Question on Gravitation

An object weighs 200N at the surface of earth. Find the weight at a depth of R2\frac{R}{2} , where R is radius of earth

A

100N

B

300N

C

50N

D

150N

Answer

100N

Explanation

Solution

Given the weight of the object on the surface of the Earth, mg = 200 N
Where, m is the mass of the object and g is acceleration due to gravity on the surface of the earth.
The variation of acceleration due to gravity (g) with the depth (d) from the surface of the Earth is given by
gd=g1-dR
Where
gd is the acceleration due to gravity at depth d
R is the radius of the Earth
At depth d = R2\frac{R}{2}, acceleration due to gravity is given by
gd=g1-R2R\frac{R}{2R}=g2
Multiplying both sides by mass of the object, then we get
mgd = mg2\frac{mg}{2}
But, mg = 200 N
Therefore, Weight of the object at depth R/2 is given by
mgd = 2002\frac{200}{2} = 100 N
Answer. A