## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The free-fall acceleration toward the sun at the distance of the earth's orbit is $5.9\times 10^{-3}~m/s^2$
Let $g_s$ be the free-fall acceleration toward the sun; $g_s = \frac{G~M_s}{r^2}$ $g_s = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(2.0\times 10^{30}~kg)}{(1.50\times 10^{11}~m)^2}$ $g_s = 5.9\times 10^{-3}~m/s^2$ The free-fall acceleration toward the sun at the distance of the earth's orbit is $5.9\times 10^{-3}~m/s^2$.