1.

Find the height from the surface of earth at which weight of a body of mass m reduced to 36% of its weight on the surface. (Re = 6400 km.)

Answer»

h = ?, R = 6400 km

g' = g\(\big(1-\frac{2h}{R}\big)\)= g - \(\frac{2gh}{R}\)

or g - g' = \(\frac{2gh}{R}\)

Percentage decrease in weight

= \(\frac{mg-mg'}{mg}\) x 100

= \(\frac{g-g'}{g}\) x 100

\(\frac{g-g'}{g}\) x 100 = \(\frac{2gh}{gR}\) x 100

= \(\frac{2h}{R}\times100\)

36 = \(\frac{2\times h}{6400}\times100\)

or h = 1.152 km


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