Question

 Compute for the value of the acceleration due to gravity g of an object at an altitude equal to twice the radius of the Earth? (radius of Earth = 6.4 x 10⁶ m) 

Answer 1

$g(0)=G\frac{M}{R^2}$

$g(h)=G\frac{M}{(R+h)^2}$ . So, we have

$g(h)=g(0)\cdot (\frac{R}{R+h})^2=g(0)\cdot (\frac{R}{R+R})^2=g(0)/4=9.18/4=2.45\ (m/s^2)$ . Answer