Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$a = 4.92\times 10^{-21}~m/s^2$
The gravitational force between the earth and the meteorite will cause both objects to accelerate toward each other. We can find the acceleration of the earth just before impact. Let $M_E$ be the mass of the earth. Then; $F_G = M_E~a$ $\frac{G~M_E~M_M}{R^2} = M_E~a$ $a = \frac{G~M_M}{R^2}$ $a = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(3000~kg)}{(6.38\times 10^6~m)^2}$ $a = 4.92\times 10^{-21}~m/s^2$