Solution:
Acceleration due to gravity at height \( h \) is \( g' = g\left(\frac{R}{R+h}\right)^2 = g\left(\frac{6400}{8000}\right)^2 = 0.64g \). The percentage difference is \( \frac{g - g'}{g} \times 100 = (1 - 0.64) \times 100 = 36% \).
By what percentage will the acceleration due to gravity at a height of \( 1600\text{ km} \) from the surface of the Earth differ from that on the surface of the Earth? (Take radius of Earth to be \( 6400\text{ km} \))
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