Please Help on this question on physics...Gold has a density of 19.3X10^3...?

Question

Gold has a density of 19.3X10^3 kg/m^3. What would the radius of a solid gold sphere have to be if the acceleration due to gravity at its surface is to be 9.81 m/s^2? Check the answer against the radius of the Earth, which has a mean density of 5.5X10^3 kg/m^3. radius of a gold sphere =? Please help as I have put in my fair share of helping others on here...I am by no means a physics wiz.

Answer 1

F = -GMm/R^2
a = F/m = -GM/R^2
M = (4/3)πρR^3
a = -G((4/3)πρR^3)/R^2
a = -G(4/3)πρR
a(g) = -G(4/3)πρ(g)R(g) = -G(4/3)πρ(e)R(e)
ρ(g)R(g) = ρ(e)R(e)
R(g) = (ρ(e)/ρ(g))R(e)
R(g) = (5.5X10^3/19.3X10^3)R(e)
R(g) = (0.2849741)R(e)
Using the volumetric radius rather than equatorial radius of the Earth,
R(g) = (0.2849741)(6,370.998685023)
R(g) ≈ 1,815.57 km

Calculating from the Universal Gravitational Constant:
-G(4/3)πρR = a
R = 3a/(4πGρ)
R = (3)(9.80665 m/s^2)/[(4)(3.14159)(6.67428*10^-11 m^3/kg/s^2)(19.3*10^3 kg/m^3)
R ≈ 1,817,482.906547823234397576468 m
R ≈ 1,817.483 km

Which agrees to 3 significant figures.