Answer
(a) $6.2\times 10^{-4}m/s^2$
(b) $9.7h$
Work Step by Step
(a) We can find the acceleration due to gravity on the surface of the asteroid as follows:
$g_A=G\frac{M}{R^2}$
We plug in the known values to obtain:
$g_A=(6.67\times 10^{-11}N.m^2/Kg^2)(\frac{3.35\times 10^{15}Kg}{1.9\times 10^4})^2$
$g_A=6.2\times 10^{-4}m/s^2$
(b) We know that
$T=\sqrt{\frac{4\pi ^2R^3}{GmA}}$
We plug in the known values to obtain:
$T=\sqrt{\frac{4\pi^2(1.9\times 10^4m)^3}{(6.67\times 10^{-11}N.m^2/Kg^2)(3.35\times 10^{15}Kg)}}$
$T=3.5\times 10^4s$
$T=9.7h$
