Answer
(a) The mass of Planet Z is $3.0\times 10^{24}~kg$
(b) The free-fall acceleration at a distance of 10,000 km above the north pole is $0.89~m/s^2$
Work Step by Step
(a) We can find the mass $M_z$ of Planet Z.
$\frac{G~M_z}{R_z^2} = 8.0~m/s^2$
$M_z = \frac{(8.0~m/s^2)~R_z^2}{G}$
$M_z = \frac{(8.0~m/s^2)(5.0\times 10^6~m)^2}{6.67\times 10^{-11}~m^3/kg~s^2}$
$M_z = 3.0\times 10^{24}~kg$
The mass of Planet Z is $3.0\times 10^{24}~kg$
(b) We can find the free-fall acceleration $g_z'$ at a distance of 10,000 km above the north pole. Let $R$ be the distance from the center of the planet. Note that $R = 15,000~km$.
$g_z' = \frac{G~M_z}{R^2}$
$g_z' = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(3.0\times 10^{24}~kg)}{(1.5\times 10^7~m)^2}$
$g_z' = 0.89~m/s^2$
The free-fall acceleration at a distance of 10,000 km above the north pole is $0.89~m/s^2$
