#### Answer

(a) The free-fall acceleration at the surface of the moon is $1.62~m/s^2$
(b) The free-fall acceleration at the surface of Jupiter is $25.9~m/s^2$

#### Work Step by Step

(a) We can find the free-fall acceleration $g_m$ at the surface of the moon.
$g_m = \frac{G~M_m}{R_m^2}$
$g_m = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(7.35\times 10^{22}~kg)}{(1.74\times 10^6~m)^2}$
$g_m = 1.62~m/s^2$
The free-fall acceleration at the surface of the moon is $1.62~m/s^2$.
(b) We can find the free-fall acceleration $g_j$ at the surface of Jupiter.
$g_j = \frac{G~M_j}{R_j^2}$
$g_j = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.90\times 10^{27}~kg)}{(6.99\times 10^7~m)^2}$
$g_j = 25.9~m/s^2$
The free-fall acceleration at the surface of Jupiter is $25.9~m/s^2$.