Answer
a) $g_M=3.70\frac{m}{s^2}$
b) $g_V=8.87\frac{m}{s^2}$
Work Step by Step
(a) We can find the acceleration due to gravity on the surface on Mercury as follows:
$g_M=G\frac{M }{r^2}$
We plug in the known values to obtain:
$g_M=6.67\times 10^{-11}\times \frac{(0.0553)(5.97\times 10^{24})}{(2.440\times 10^6)^2}$
$g_M=3.70\frac{m}{s^2}$
(b) We can find the acceleration due to gravity on the surface of Venus as follows:
$g_V=G\frac{M}{r^2}$
We plug in the known values to obtain:
$g_V=6.67\times 10^{-11}\times \frac{0.816\times 5.97\times 10^{24}}{(6.052\times 10^6)^2}$
$g_V=8.87\frac{m}{s^2}$
