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

The free-fall acceleration on the surface of Titan is $1.35~m/s^2$
We can find the free-fall acceleration $g_T$ at the surface of Titan. $g_T = \frac{G~M}{R^2}$ $g_T = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.35\times 10^{23}~kg)}{(2.58\times 10^6~m)^2}$ $g_T = 1.35~m/s^2$ The free-fall acceleration on the surface of Titan is $1.35~m/s^2$.