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Worked example 12.2: Acceleration of a rocket

Question: A rocket is located a distance 3.5 times the radius of the Earth above the Earth's surface. What is the rocket's free-fall acceleration?

Answer: Let $R_\oplus$ be the Earth's radius. The distance of the rocket from the centre of the Earth is $r_1= (3.5+1) R_\oplus= 4.5 R_\oplus$. We know that the free-fall acceleration of the rocket when its distance from the Earth's centre is $r_0=R_\oplus$ (i.e., when it is at the Earth's surface) is $g_0= 9.81 {\rm m/s^2}$. Moreover, we know that gravity is an inverse-square law (i.e., $g\propto 1/r^2$). Hence, the rocket's acceleration is

g_1 = g_0\left(\frac{r_0}{r_1}\right)^2 = \frac{9.81\times 1}{(4.5)^2} = 0.484 {\rm m/s^2}.

Richard Fitzpatrick 2006-02-02