Answer
Please see the work below.
Work Step by Step
We know that in circular motion $a_r=\frac{v^2}{r}$
We plug in the known values to obtain:
$a_r=\frac{(1.02\times 10^3)^2}{0.3844\times 10^6\times 10^3}$
$a_r=2.7\times 10^{-3}\frac{m}{s^2}$
According to Newton's law of gravitation
$F=\frac{GMm}{r^2}$
This can be rearranged as:
$\frac{F}{m}=\frac{GM}{r^2}$
$\implies a=\frac{GM}{r^2}$
We plug in the known values to obtain:
$a=\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{0.3844\times 10^9}$
$a=2.7\times 10^{-3}\frac{m}{s^2}$
Thus, it is verified that both cases give the same answer.
