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I want to create a story in which only the Moon differs from reality. The surface gravity of the Moon in this story is greater than in reality, about 0.6 G. In my story, the United States in the 1960s, as in reality, will successfully complete the Apollo program. My concern is that the surface gravity would be too great, making the Apollo program impossible.

Would it be possible for the United States in the 1960s to send astronauts to the moon, whose surface gravity is 0.6 G, and return them to earth?

Details

In my story, in 20,000 BC, an almost omnipotent alien suddenly replaced the moon as we know it with a nearly perfect sphere made of pure gold and left for another galaxy. This new moon has the same mass, momentum, angular momentum, and amount of heat as the old one, and is already in equilibrium from the moment it appeared. Time passes and the United States in the 1960s, as in reality, begins the Apollo program. American astronauts lands on the moon, inscribes “USA” on the golden ground, and returns to earth.

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    $\begingroup$ uh, it has same mass and distance, btu the moon is much smaller in diameter? I don't think that you get to 0.6 g at the smaller diameter... $\endgroup$
    – Trish
    Commented Jul 6 at 9:55
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    $\begingroup$ @Trish For the same mass, the smaller the diameter, the greater the surface gravity. $\endgroup$
    – Hexirp
    Commented Jul 6 at 10:22
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    $\begingroup$ Just doing some base calculations here: Source for Moon numbers To calculate the new Moon radius, we use the gravity formula F=GMm/r^2 with Newton's formula F=ma. Desired acceleration a is 5.88m/s^2 (0.6g), and the mass of the subject m can be cancelled from both sides leaving us with a=GM/r^2 which is rearranged to r=√(GM/a). Solving with the Gravitational constant G and the mass of the Moon M we get an (equatorial) radius r of 936,365 meters. $\endgroup$
    – chai_tea
    Commented Jul 6 at 11:20
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    $\begingroup$ Using the escape velocity formula of 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. $\endgroup$
    – chai_tea
    Commented Jul 6 at 11:21
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    $\begingroup$ @JBH up to a point. The Apollo program was about 6% of USA GDP at the time. Simplistically, if they had needed more than 16 times as many launches to put fuel on the moon, they could not have done it, even with 100% tax rate and martial law to suppress the revolts. $\endgroup$ Commented Jul 7 at 13:44

4 Answers 4

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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.

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    $\begingroup$ I wonder if this would push NASA to use the "lunar surface rendezvous" architecture: launch an unmanned fuel tanker on its own rocket, land it remotely, then launch the lunar module with astronauts on a second rocket. $\endgroup$
    – Cadence
    Commented Jul 6 at 15:35
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    $\begingroup$ @Cadence nah, not enough tech for that - automatic landing is hard. It needs an altimeter that works, and in 2024, a modern 2025 altimeter failed to see the moon, resulting in japan's lander crashing. So... very tricky. $\endgroup$
    – Trish
    Commented Jul 6 at 19:53
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    $\begingroup$ @Trish The US landed several robotic spacecraft successfully prior to Apollo, in the Surveyor program. The stakes here are higher, and not all of the Surveyors were successful, but it would be doable. $\endgroup$ Commented Jul 7 at 2:15
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    $\begingroup$ Also, delta-V produced is not dependent on either thrust or fuel consumption rate. I think the delta-v requirement is different for landing on Gold Moon than Moon Classic, but it shouldn't be 4x. $\endgroup$ Commented Jul 7 at 2:18
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    $\begingroup$ I tried calculating backwards from lander tonnage and arrived at 7,000 tons of fuel and oxidizer at launch (2 and 1/3 original mass). Feel free to use it, if you want. $\endgroup$ Commented Jul 7 at 13:59
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With so much more gravity, and still plenty of time to be struck by meteors to add all kinds of different materials to the surface, I'd expect the moon to have a meaningful atmosphere to throw off some calculations that take only the increased gravity into account.

There's a number of variables to consider how much of an atmosphere this construct of a moon would have, what gases can be found in this atmosphere, and in what proportions. By being bombarded with light gases ejected from solar radiation I'd expect some large proportion of the atmosphere to be of hydrogen and helium. But then maybe the solar winds could drag this away. That is unless there's a magnetic field like that on Earth to protect the atmosphere.

If we can assume the moon is struck by comets and such with an abundance of water ice then there might be a reasonable amount of water vapor in the atmosphere, perhaps even pools of liquid water on the surface with the atmosphere holding in some heat. Because of radiation breaking molecules apart, and radiation leading to varied nuclear reactions, there could be carbon, nitrogen, and varied other heavy gases in the atmosphere on the moon.

The existence of even a relatively thin atmosphere could help in making landing on the moon a bit easier as like when landing on Mars aerodynamic braking could be possible, meaning less fuel burned on landing. Taking off from the moon though would mean having to fight air resistance, and likely meaning more fuel burned in total than in the near vacuum seen on the moon as we know it.

What this might mean in practice is that to land men on the moon and return them safely to Earth could mean a need for multiple launches per mission, with many unmanned missions to bring fueled up rockets to the moon, with some kind of means to assemble them into a multiple stage return vehicle.

With the higher gravity and a potential for some kind of atmosphere a mission to the moon could look like how people expect a mission to Mars to go. There could be weather to contend with. If there's an atmosphere containing oxygen and carbon like Mars then there could be in-situ rocket fuel synthesis like has been proposed for missions to Mars, which could mean a considerable savings on mass launched to the moon for a successful manned mission.

All of the required technology for a mission to a Mars-like moon likely existed in the 1960s. Maybe it would have delayed a moon landing but the people involved should still have been able to make it work.

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Your escape velocity has changed from 2.4 km/s to 3.3 km/s.

For a 5 ton payload (the lunar module dry weight is 4-and-change tons) you need 40.8 tons of fuel (45.8 tons total). This is up from 25 tons for the landing on the old Moon.

You have to keep fuel on hand to slow down 4 km/s before entering Earth's atmosphere = 175.4 tons payload+fuel+oxidizer (95 tons original)

The intial boost to Earth escape velocity (11.2 km/s, plus 1 km/s lost to atmospheric drag) is 7,336 tons of fuel+oxidizer (3,000 tons original)

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  • $\begingroup$ +1 for a simple and concise answer. $\endgroup$
    – JBH
    Commented Jul 8 at 0:32
  • $\begingroup$ roughly... 2.5 times the launch mass. $\endgroup$
    – Trish
    Commented 2 days ago
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There were proposals for manned missions to Mars. Von Braun's Marsprojekt was written in 1952 with a proposed launch date of 1965. A lot was learned about Mars from the Mariner probes which would have modified this design: the atmosphere was thinner, which made aerobraking much more difficult. You could use Marsprojeckt as a blueprint for an Apollo-era project for landing on a larger moon and returning. Gravity on Mars is about 40% of Earth's gravity, so that is a bit less than the gravity of the moon in your story.

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