Question

Find the percentage decrease in the acceleration due to gravity when a body is taken from the surface of earth to a height of $64km$ from its surface. [Take $R_e=6.4\times {10}^{6}m]$.

Answer 1

$g=g(1-\frac{2h}{R})$

$\frac{g}{g}=1-\frac{2\times 64}{6400}=1-\frac{128}{6400}$

for $\%R$ change

$1-\frac{g}{g}=1-1+\frac{128}{6400}$

$\frac{g-g}{g}=\frac{128}{6400}=0.02$

$\%$ change $=0.02\times 100=2\%$