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Q.
An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, $R =6400 km$. The percentage decrease in the weight of the object will be
$g _{\text {eff }}=\frac{ g }{\left(1+\frac{ h }{ R }\right)^2} ; g _{\text {eff }}=\frac{ g }{\left(1+\frac{1}{4}\right)^2}=\frac{16 g }{25} $
change $=\frac{ g _{\text {eff }}- g }{ g } \times 100=\frac{\frac{16}{25}-1}{1} \times 100$
$ =\frac{-9}{25} \times 100=-36 \%$
