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Gravitation

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Question : 15 of 25
Marks: +1, -0
A body weighs 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?
Solution:  
Acceleration due to gravity g at height h is given by
g′=gR2(R+h)2g' = \frac{g R^{2}}{(R+h)^{2}}
=gR2(R+R2)2= \frac{g R^{2}}{\left(R+\frac{R}{2}\right)^{2}}
=49g= \frac{4}{9} g
Gravitational force on body at height h is
F=mg′F = m g'
=m49g= m \frac{4}{9} g
=mg49= m g \frac{4}{9}
=63×49=28N [∵mg=63]= 63 \times \frac{4}{9} = 28 \mathrm{N} \ [\because m g = 63]
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