How much does a moon’s gravity effect the "free-fall acceleration" of an object on the planets surface?

Question

The Planet

The Planet I'm speaking of has the same size as earth but is much lighter. That's because the "Upper Mantle" is about 1000 km wider and mostly consist out of organic material. Such a huge mass of carbon on one planet isn't very realistic but I need it to have an excuse for lower gravity on the surface. Picture from Wikipedia

The Moon

I haven't put much thought in the moon neither. The Moon is supposed to be close and big enough to affect the habits of animals on the surface by lowering their "Free-fall acceleration". (big enough difference for them to orient themselves on it) As an example: In these phases some of them would be hunting because it needs them less energy.

The Questions

To the physicists among us:

• If my approach is realistic, how big/heavy and close would the moon have to be? (The right equations and some pointers might do it too.)

• If it isn't realistic, what would be the best explanation for periodical gravity reductions?

Sorry if I wasted your time. Have a great day!

Answer 1

I answered this question (How to get periodic low gravity) by positing a binary planet. If Earth were a moon of Jupiter, in the orbit now occupied by the actual Jovian moon Metis, Jupiter overhead would effectively decrease weight on things on Earth by 30%.

Which astronomical or cosmological event would explain periodical low-grav-effect on earth-like planet?

So first: how much am I lifted by our own Moon overhead? Moon mass = 0.01 Earth weights My mass = a svelte 100 kg Moon distance from Earths center: 6371 km core to surface + 384,400 km to the moon = 390771 km

https://www.omnicalculator.com/physics/gravitational-force

$0.002$ Newtons. In contrast the Earth's gravity on me is $981$ N. So just a little bit of gravity from the moon. But enough to move the oceans!

How to get more gravity out of the moon? Let us try making it more massive. If instead of $0.01$ the "moon" has $1000$ times the mass of the Earth that would produce a pull of $260$ N or about $30\%$ Earths gravity. That would improve your vertical leap.

This "moon" is $6\times 10^{27}$ kg (Earth being $6\times 10^{24}$). Uranium is $52 cc/kg$ and so this pure uranium moon (I like pure uranium moons!) is approx $3\times 10^{29}$ cc. This sphere calculator gave me a radius of $4207480000\space\text{cm}$ or $42074\space\text{km}$. Earth is only $6371\space\text{km}$ radius but Jupiter is $69,911\space\text{km}$. So if you want it at the orbit of the moon, a largish planet sized chunk of uranium.

I am delighted by this prospect but can understand that you might have problems with a Jupiter sized chunk of uranium being your world's moon. You could make this more reasonable by bringing the dense moon into a tighter orbit than our moon is, and so it would be closer (as in the linked Metis example) where a Jupiter sized chunk of Jupiter stuff can accomplish the same pull as a Jupiter sized chunk of uranium at Luna distance. Or you could assert your moon is made of superdense matter - maybe the stuff of neutron stars, or the theoretical superheavy elements which start at about 3 times the mass of uranium.

Answer 2

Let's set some things right: gravity has an infinite range.

This means that any body in the Universe feels the gravity of any other body in the Universe. So the crow which is now in my garden is attracted by the Andromeda Galaxy millions of light years away. BUT...

But the magnitude of this attraction is negligible when compared to the gravitation attraction of Earth on that crow.

This concept I tried to explain you with the crow in my garden is generalized by the Hill sphere

An astronomical body's Hill sphere is the region in which it dominates the attraction of satellites.

If a moon was so massive to have a significant influence on beings living on the surface of the planet, the roles would be swapped and what you call planet would be the actual moon.

Something similar happens on Io, a Jupiter satellite, where the tidal forces are so strong that the dissipated energy keep the moon warmer than it would be thanks to the Sun.

Unlike Earth and the Moon, Io's main source of internal heat comes from tidal dissipation rather than radioactive isotope decay [...]. Such heating is dependent on Io's distance from Jupiter, its orbital eccentricity, the composition of its interior, and its physical state. [...] The vertical differences in Io's tidal bulge, between the times Io is at periapsis and apoapsis in its orbit, could be as much as 100 m.

Answer 3

The wonderful world of tidal forces

We'll call the planet Earth and the satellite we'll call the Moon.

In a first approximation, the gravitational force of the Moon on a human walking on Earth is zero. At a closer inspection there is indeed a force, but it's tiny.

Why is this so?

The gravitational force of the Moon on a human walking on Earth is close to zero because the Moon attracts not only the human, but also the Earth; while the human is indeed falling towards the Moon with a certain acceleration, Earth is falling towards the Moon too with just about the same acceleration, and thus the net effect on the human relative to Earth is almost zero. Look at the nice picture:

A human walking on Earth falls towards the Moon with an acceleration just a tiny little higher than the acceleration with which Earth itself falls towards the Moon. The mass of the human is assumed to be negligible with respect to the masses of the Earth and the Moon. Own work, available on Flickr under the CC Attribution license.

• The gravitational acceleration of Earth towards the Moon is

  $$a_{\text{E-M}} = \frac{G \cdot m_{\text{Moon}}}{d^2_{\text{E-M}}}$$

• The gravitational acceleration of the human towards the Moon is

  $$a_{\text{H-M}} = \frac{G \cdot m_{\text{Moon}}}{(d_{\text{E-M}} - r_{\text{E}})^2}$$

What we are really interested in is the difference between the acceleration of the human towards the Moon, $a_{\text{H-M}}$, and the acceleration of the Earth towards the Moon, $a_{\text{E-M}}$:

$$a_\text{H-M} - a_\text{E-M} = G \cdot m_\text{Moon} \frac {2d_\text{E-M}r_\text{E} - r^2_\text{E}}{d^2_\text{E-M}(d_\text{E-M} - r_\text{E})^2} \simeq G \cdot m_\text{Moon} \frac{2r_\text{E}}{d^3_\text{E-M}}$$

We notice that the relative acceleration of the human with respect to Earth is just about proportional with the radius of the Earth and inversely proportional with the cube of the distance between the Earth and the Moon. (The approximation holds if the radius of Earth is small with respect to the distance between Earth and the Moon.)

The problem is that pesky cube, which makes the relative acceleration of the human with respect to Earth decrease very very quickly as the distance between the Earth and the Moon increases.

Also, look the the Wikipedia article on Tidal forces.

Answer 4

Look into the Roche Limit

One of the ways that planets end up with rings is that another smaller (or with lower surface gravity) celestial body comes close enough to it that the planets gravity exceeds the other body's gravity at its surface ripping it apart. So, if you want to have tides significantly effect gravity then you have to look for a system where the planet is approaching the limit for how close it could reach the moon without being destroyed by it. Now in reality that Roche limit does not exist because the smaller body would be ripped apart first. As a result if you want want a planet with significant fluctuation in free-fall acceleration due to tidal forces of another celestial body then you would probably be looking at something a lot closer to a binary system or a system where your "planet" is actually the moon than an earth/luna system.

Now if we say for example that the two bodies are of identical mass and radius, assuming perfectly ridged spheres, then with gravity diminishing by a power of two, the distance between the surfaces of the bodies for a 30% reduction in gravity due to tidal effects when compared to the same body in isolation gives ~83% planetary radius. If we compare between peak gravity and minimum gravity they will be allowed to be a little further apart since both gravities will add when on the opposite side of the planet, but the difference wouldn't be much. At this distance you end up the other body being very big in the sky, you will have very frequent eclipses. Also if the bodies have atmospheres than their outer bounds could be touching leading to orbital decay over extreme lengths of time, making such a configuration unlikely.

In summary, you may want your "planet" to actually be the moon.

Answer 5

The Moon is supposed to be close and big enough to affect the habits of animals on the surface by lowering their "Free-fall acceleration".

This isn't going to be a long-term stable situation.

What you're talking about is "gravitational tidal effects". Our Moon raises tides on Earth in the same way you're planet's satellite does. Most people know about the ocean tides, but there are also "Earth tides": The surface of the Earth moves up and down about 1/3 of a meter.

Over the longer term, the resulting "tidal acceleration" effect (a bit of a misnomer) slows down the planetary rotation and speeds up the satellite's orbit, until they end up being tide-locked as our Moon is now.

It's hard to estimate the speed of this process for your world without more information on its geology. If there's a sensible effect (maybe 10% of normal gravity? More?), and if your planet's mass and moment of inertia are smaller (to give smaller g), this is going to be geologically fast: The planet's rotation will slow to match the satellite's orbit period, and the variations will disappear.

Any process that periodically changes g at the surface is going to have the same result: The planet itself is going to respond, and that's going to result in lots of energy dissipation.

Maybe you can turn that into a story point? "This world can't be very old, and that's a part of it's mystery!"

Answer 6

Some additional information. First of all, though, let me note AlexP's answer which is excellent. It's good and I shan't repeat it.

But several additional points to consider:

First, the distance at which the satellite can provide a meaningful improvement in a jump is perilously close to the point where the world begins to break up as the satellite's tidal forces overwhelm the planet's cohesion due to its own gravity.

Second, "So what?" you say. Just set up the satellite's orbit far enough away not to disrupt the planet by close enough to allow for a 10% local diminution of effective gravity. But those enormous tidal forces are going to raise tremendous tides, not only in the planet's oceans but in the solid rock. And this will release vast amounts of heat. Interestingly, that heat started out as the energy of rotation of the planet. (If the planet and satellite mass anything like Earth, and the planet rotates in 24 hours, the satellite would probably have to be inside geosynch to raise a 10% tide, so the effect of the tides would be to steal the planet's rotational energy to drive the satellite out and slow the planet's rotation down. Eventually, there'd be a stable situation where the planet and the satellite were tidelocked to each other. At this point, the energy loss from tides would greatly diminish.

Finally, long before that happened, the tides themselves would have renderedthe planet uninhabitable. The rock tides would be tens or hundred of miles high, would completely disrupt the planet's surface, probably melt it, and certainly stir up massive volcanic activity to saturate the rubble in lava. Io on steroids!

Bottom line: Better jump fast, but it's not going to be very long before no one is jumping anywhere on the planet.