Answer
The free-fall acceleration on the surface of Titan is $1.35~m/s^2$
Work Step by Step
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 $1.35~m/s^2$.
