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Q.
A body weighs W newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be :
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Solution:
Weight on earth is $ \omega =mg=\frac{GmM}{{{R}^{2}}} $ ...(i) $ \because $ $ g=\frac{GM}{{{R}^{2}}} $ Now, at a height h above the earth surface, the value of g will become $ g=\frac{GM}{{{(R+h)}^{2}}} $ Now, weight of the body at this height is given by $ \omega =mg=\frac{GnM}{{{(R+h)}^{2}}} $ (ii) Now, dividing equation (ii) by (i) we obtam $ \frac{\omega }{\omega }=\frac{R}{R+h} $ $ h=\frac{R}{2} $ given, $ \omega =\omega \frac{{{R}^{2}}}{{{\left( R+\frac{R}{2} \right)}^{2}}}=\frac{4\omega }{9} $