(a) The value of the gravitational acceleration is given by:
[tex] g=\frac{GM}{r^2} [/tex]
where
[tex] G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} [/tex] is the gravitational constant
M is the Earth's mass
[tex] r=6371 km=6.371 \cdot 10^6 m [/tex] is the Earth's radius at the pole
If we use the value for g given by the problem, [tex] g=9.830 m/s^2 [/tex], and we rearrange the equation above, we find the value of the Earth's mass:
[tex] M=\frac{gr^2}{G}=\frac{(9.830 m/s^2)(6.371 \cdot 10^6 m)^2}{6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}}=5.982 \cdot 10^{24} kg [/tex]
(b) The value we found for the Earth's mass is [tex] 5.982 \cdot 10^{24} kg [/tex], and we see that this value is slightly larger than the accepted value of [tex] 5.979 \cdot 10^{24} kg [/tex].
