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I been trying to imagine what it would be like to move around a small moon, around the size of Saturn's moon Iapetus, which possess Earth-like gravity on the surface. I recall reading an xkcd article on an asteroid the size of the Little Prince and the effects of it having 1 g gravity and I wonder if this moon is small enough for tidal forces on people to be notable. I also wonder if the escape velocity have been lowered enough for a car to achieve it. This leads to me to wonder what driving on it would be like, would the curvature be noticeable?

The radius of Iapetus is 735 km (see here).

Hypnosifl
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HerbertSnick
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    Welcome to worldbuilding. Please note we enforce a "one question per post" policy, while you are asking several questions. Please narrow it down to a single question. It is fine to post multiple questions related by a common background. You can take the [tour] and visit the [help] to find out more. – L.Dutch Aug 03 '19 at 16:54
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    For reference, this is the XKCD article OP is refering to: https://what-if.xkcd.com/68/ – The Square-Cube Law Aug 04 '19 at 02:43
  • This doesn't seem like a worldbuilding question. It seems like an arithmetic question. "How big does a body need to be for people on it to feel tidal effects" is just arithmetic. There's no discussion to be had here. – Ryan_L Aug 04 '19 at 04:06
  • If you will actually write such story, you might be interested in this question: https://worldbuilding.stackexchange.com/q/76107/809 – Mołot Aug 04 '19 at 11:57
  • The mods are at it again lol. oh well. What you are asking is straight forward enough. you can get earth like gravity on the surface with a small dense moon. Math out the mass vs radius. nbd. the "escape velocity" of the moon will be low as that's tied to the mass of the moon. but you are still in a planet's gravity well, so you'll want to leverage the moons speed around it's planet to escape from the planet. Tides are tied to the mass of the planet. If it's a gas giant the tides will be huge, even if it's an earth sized world they will be much more significant than the tides on earth. – MParm Aug 04 '19 at 23:41
  • additional cool factoid: the surface gravity of Mercury is the same as Mars' due to the high metal content. Neeto. – MParm Aug 05 '19 at 00:09
  • @Ryan_L: It takes a bit more than just arithmetic, but first-year college calculus & physics should do it. If you're on a STEM track, anyway. – jamesqf Feb 03 '20 at 00:20

2 Answers2

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The force of gravity on a spherical body like a planet or moon is governed by Newton's laws. If a moon had the same surface gravity as Earth:

\begin{equation} F_{gravity} = \frac{G*M_{earth}*m_{object}}{r_{earth}^2} = \frac{G*M_{moon}*m_{object}}{r_{moon}^2} \end{equation}

Solving for the relative density of the moon and subbing in the info for Earth and the radius of Iapetus provided by Google:

\begin{equation} D_{moon} = \frac{D_{earth}*r_{earth}}{r_{moon}} = \frac{5.51 g/cm^3 * 12,742 km}{1,469 km} = 47.8g/cm^3r \end{equation}

Taking a quick look at a density table [1], this puts it well above Osmium and Iridium and into the realm of exotic matter, probably involving temperatures and pressures that would be fatally unfriendly to humans and conventional life support equipment. Odds are this moon would have a core of transuranic materials which would explode in an uncontrolled fission reaction as soon as the moon accreted. There's no liquid dense enough to comprise a significant portion of the moon's surface (at reasonable human-friendly pressures and temperatures), so no noticeable tides. And escape velocity is tied to surface gravity, so it would be the same as on earth. No driving off the planet.

[1] https://en.wikipedia.org/wiki/Talk%3AList_of_elements_by_density/Numeric_densities

Helion
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    I think the question about tidal forces was not about "tides" in the specific sense of movement of liquids on the surface, but about whether the gravitational pull would vary noticeably over short distances (in physics 'tidal forces' just refers to any effects of differing gravitational pull on different parts of a body). For a moon the size of Iapetus I don't think the difference on your feet and head would be noticeable but you might notice some difference if you climbed a mountain or something. – Hypnosifl Aug 03 '19 at 18:53
  • Also, it looks like you used the radius of the Earth's moon for $r_{moon}$ in your calculation, rather than the radius of Iapetus. If we use Iapetus' radius, your formula gives a density of 5.51 * 6378 / 735 = 47.8 g/cm^3. – Hypnosifl Aug 03 '19 at 20:53
  • @Hypnosifl, good call on tidal forces. The difference between "sea level" and the top of an Everest equivalent on this moon would be about 1.2% less gravity. Probably not noticeable. You might be right about my calculation though. Look for an edit! – Helion Aug 03 '19 at 23:50
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    One other thing, you said "escape velocity is tied to surface gravity, so it would be the same as on earth" but I don't think that's correct, surface gravity is given by g = GM/R^2 where M is the planet's mass and R is its radius, while escape velocity at the surface is given by sqrt(2GM/R) = sqrt(2Rg). So for two planets of different sizes with equal surface gravity g, the one with the smaller radius R would have a smaller escape velocity. An Iapetus with Earth gravity would have an escape velocity only sqrt(735/6378) = 0.34 times Earth's. – Hypnosifl Aug 04 '19 at 05:38
  • Escape velocity is not tied to surface gravity. Saturn, for example, has a surface gravity pretty close to Earth's, but a much higher escape velocity. – Logan R. Kearsley Aug 11 '19 at 19:38
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From this link we know that Tidal force is defined like so:

In celestial mechanics, the expression tidal force can refer to a situation in which a body or material (for example, tidal water) is mainly under the gravitational influence of a second body (for example, the Earth), but is also perturbed by the gravitational effects of a third body (for example, the Moon). The perturbing force is sometimes in such cases called a tidal force (for example, the perturbing force on the Moon): it is the difference between the force exerted by the third body on the second and the force exerted by the third body on the first.

So, straight off the bat we see the Earth-Moon example. Keep in mind that the Earth is many times more massive than the Moon, and the Moon is many times more massive than Iapetus.

So let's look at some basic info about Iapetus that is relevant to your question using info from here. The surface-level gravity on Iapetus is 0.223 m/s2, or roughly 0.2 Earth Gs. Escape velocity is also pretty low - 0.573 km/s, or 573 m/s (Earth's escape velocity is 11.186 km/s). This means that although you can't just jump and fly off into space, it won't take much. In fact, if you're in a car-shaped rocket on a flat, race-track like patch of land (approximately tangent to the overall curvature of Iapetus), you could very likely accelerate right off the moon on the end of the track. The fastest car on Earth is the Hennessey Venom F5, at 301 miles per hour (or 484.413 km/h). But this is on Earth, where air-friction is a thing and let's not forget the mass of the car, which is a large part of why a car travels fast at all (and I'm not even going to talk about downforce). So, if we instead attach a rocket at the end of this vehicle (not even a massive one, just like the kind used in missiles), we'll be off-world in no time.

TLDR: Iapetus is too small to affect anything even the size of a car with tidal forces. Iapetus does, however experience tidal forces relative to Saturn. If you want to get needlessly technical about definitions, everything has a tidal force on everything because gravity affects objects even at infinity, hence there must be some perturbations - just not any noticeable ones. I know most of my answer didn't even deal with tidal forces, but perhaps the info I gave you will spark other ideas. I hope this helps!

Edit: @Hypnosifl pointed out that you were looking for info on a moon LIKE Iapetus, but not actually Iapetus, specifically if it has a greater mass. More specifically, we're talking about a planet like Iapetus that has the gravity of Earth somehow. This is given by the equation:

g = GM/r2

Now g here is equal to one of our Gs (9.8 m/s^2), but r=735 km. With some quick maths, M must be equal to: 7.9373388306×1022 kg. This is no moon at all, and must be made completely of some unrealistically dense material! This object must be spinning really fast, so fast that it's tearing itself apart, and nothing could possibly last long on its surface, regardless of the 1G it has. That kind of eliminates the question of tidal forces or anything. An Iapetus-sized object that has such a high value of G is pretty unrealistic, even in scifi. Let's assume it's ridiculously dense as described in the comment. At this point I'd like someone who knows better to please chip in as to my knowledge, I know that tidal forces won't be what you need to worry about, but I know what it would be.

arpanet101
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  • The question is about what Iapetus would be like if it hypothetically had a much larger mass so that it had 1 G gravity at the surface, not about what the real Iapetus is like. – Hypnosifl Aug 03 '19 at 19:59
  • Is that so? The question doesn't seem to overtly state anything about modifying Iapetus's characteristics. I'll change my answer in accordance to your comment - thanks for letting me know! – arpanet101 Aug 03 '19 at 20:01
  • You may have missed the bolded part of the first line: "I been trying to imagine what it would be like to move around a small moon, around the size of Saturn's moon Iapetus, which possess Earth-like gravity on the surface." – Hypnosifl Aug 03 '19 at 20:05
  • Sorry my bad - will fix. Since this is a moon, can you tell me about its planet? – arpanet101 Aug 03 '19 at 20:10
  • The exponent in your mass number seems to be off by 1, if g = GM/r^2 then M = gr^2/G so for g=9.8 m/s^2, r=735000 m, and G=6.67408 * 10^-11 m^3 / (kg * s^2) we should have M = 7.93 * 10^22 kg. And that isn't anywhere near the mass or density of a neutron star, it says here a neutron star has a mass at least 1.1 times the mass of the Sun, 1.989 * 10^30 kg. The volume of Iapetus is (4/3)pi(735000^3) = 1.66 * 10^18 m^3, so density would be about 48000 kg/m^3 = 48 g/cm^3, whereas neutron stars are about 10^17 kg/m^3. – Hypnosifl Aug 03 '19 at 20:55