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 \(\frac{g - 0.64g}{g}\times 100% = 36%\).
By what percentage will the acceleration due to gravity at a height of 1600 km from the surface of the Earth differ from that on the surface of the Earth? (Take radius of Earth to be 6400 km)
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