You overestimate the gravity a little...
$$g=\frac {GM}{R^2}$$
That's surface attraction, G is a constant (6.67E-11 m³/ kg s²), M is a moon (7E22 kg), so all we need is R. We start with the volume of a sphere:
$$V=\frac M \rho=\frac 4 3 \pi R^3$$
Transform to R, and plug in gold (19300 kg/m³ ~20000 kg/m³)...
$$R=\left(\frac 3 {4\pi} \frac M \rho \right)^{\frac{1}{3}} \sim \left(\frac 3 {4 \pi} \frac {7 \times 10^{22}}{20000} \right)^{\frac{1}{3}}=\left(\frac {21} {8 \pi} \times 10^{18} \right)^{\frac{1}{3}} = \left( 0.8357 \times 10^{18} \right)^{\frac{1}{3}}=943000 [\text m]$$
Plugging that into the initial formula, I get...
$$g\sim5.25 \left[\frac {\text m} {\text s^2}\right]\sim0.535g_{Earth} = 3.23 g_{Moon} $$
What does that mean for our Moon Lander?
Basically, it means the engines need to have about (generously rounded) 3.5 times the power of the lunar module. The Aerojet AJ10-137 of the Lunar module had an ISP of 314.5 s and brought the ascent stage up. Let's be simple and use the more boosters approach. So we slap 4 times the engine to the tank. That should us get spaceborn, right? Well... not so fast. We need to have more fuel. How much?
Well, we need about 1900 dV, which is pretty much the power we carry in fuel. It's not in liters, gallons, or kilos, but how much oomph it has to push this specific spacecraft of known dry weight. So... how much mass in fuel does that correspond to, if we take 4 engines? Well, first of all, we burn 4 times the fuel per time, since our engine did not get more efficient (which is what the ISP measures, pretty much 'mile per galon' of sorts). And we need to lug up the extra fuel, so we don't just need 4 times the fuel, we need a little more, because the tyranny of rockets (you need to launch not just the craft, but also the extra fuel) is bad.
The Lunar Lander packed about 85% of its dry mass in fuel. So in the ballpark and rounding up liberally, the dry mass in fuel. We need 4 times as much fuel per time for the four engines (plus the fuel to raise that), so by clever rounding up, we can pin it to need about four times the lunar ascent module in fuel, and we might have some to spare.
So all in all, our wet mass of the lunar ascent module went from 1.85 to 5 times its dry mass. That pushes back to the start launch over other steps... so all in all, the whole launch system might just about double to triple in total weight, just to get this last stage to have enough fuel. Ballpark that is. That might be doable, and the proportions of the Saturn would be roughly similar, because volume goes to the power of 3 of the dimensions.
F=GMm/r^2
with Newton's formulaF=ma
. Desired accelerationa
is 5.88m/s^2 (0.6g), and the mass of the subjectm
can be cancelled from both sides leaving us witha=GM/r^2
which is rearranged tor=√(GM/a)
. Solving with the Gravitational constantG
and the mass of the MoonM
we get an (equatorial) radiusr
of 936,365 meters. $\endgroup$v=√(2GM/r)
we calculate the escape velocity of the new Moon is about 3235m/s. In comparison, the regular Moon's escape velocity is 2380m/s. $\endgroup$