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First of all, I want to make clear that I don´t know how to calculate the gravity of a planet in m / s². Starting from there, I don´t know any formula referring to this knowledge and that is why I am doing this topic.

My question is, what would be the size of an earth-like planet to have a fifth of the Earth's gravity? I know that aspects such as density and size influence, but my knowledge of the subject is so poor that i couldnt calculate something like this by myself. I would be grateful to anyone who could give me an answer on this topic.

JAMS
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    Yes, I am asking a question for Worldbuilding purpose of my story. That you dont see the utility, it's your business, dude. – JAMS Mar 25 '18 at 00:16
  • Acting in a self-entitled and rude way is not going to get you any friends around here. You're expected to justify your use of the site to the members, not make demands. Park your ego and the door and read the site policies. Also note that giving members a more useful context for your request (that's request !) helps them help you. – StephenG - Help Ukraine Mar 25 '18 at 00:28
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    I asked a simple question, that two people had no problem answering me in a good way. Instead, your only contribution is to say that I am trying to undermine the purpose of this forum. I just dont see your good will. – JAMS Mar 25 '18 at 00:33
  • I'm trying to encourage you and other new members to conform to the purpose and intent of the site. Most of us do not want it to become a dumping ground for any old "how/what/why" question in science - that's not what the site is for. And that is what your question reads like. Until you explain why your question is worldbuilding material it's actually your lack of good will that's in question. – StephenG - Help Ukraine Mar 25 '18 at 00:36
  • I just wanted to know how big a body with these characteristics should be, and how would be its gravitational interaction with other bodies, or how it would be seen from another planet. – JAMS Mar 25 '18 at 00:42
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    @StephenG The OP's response to your initial comment was no more high-handed than your own. It isn't easy dealing with questions where the OP's starting position is less informed than usual. Frankly you could have handled the situation with more tact. – a4android Mar 25 '18 at 03:11
  • @a4android I am not here to hand-hold self entitled people too lazy to look up "gravity" on line - just how long would it have taken to find Newton's formula ? There must be a hundred videos on YouTube explaining it and that's ignoring answers to questions on SE. I don't bother with tact for people who make no effort to do prior research and don't even appear to be here for worldbuilding. On Physics SE and Astronomy SE I tend to make more allowance for people who make even a minimal effort and view them partly as educational sites. This is not an educational site in that sense. – StephenG - Help Ukraine Mar 25 '18 at 03:22
  • Good thing that you're "not here to hand-hold self-entitled people too lazy to look up "gravity" on line". There is always the option to ignore and let other people more willing than you offer their help. – JAMS Mar 25 '18 at 03:52
  • @StephenG I'm sure you don't reminding about WB's be nice policy. However, you may need to re-read it at https://worldbuilding.stackexchange.com/help/be-nice. While you can choose whether you want to be helpful or not, however if you participate, then tact and good manners come with the territory. There are questions that leave me aghast, I find it's best to walk away. – a4android Mar 25 '18 at 04:03
  • @a4android It's every member's duty to help police the site. Walking away and ignoring off-topic Qs is not the way we're supposed to behave. Most people, when asked to justify apparently off-topic Qs do make an effort to do so, in my experience, and the extra context is often useful and can generate useful answers and comments. Ignoring things does no good. – StephenG - Help Ukraine Mar 25 '18 at 10:53

2 Answers2

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You just need to know the right term to Google =)

Wikpedia's page on Surface Gravity includes:

$$g=\frac{4\pi}{3}G\rho r$$

The parts that matter to you are density ($\rho$) and the radius of the planet ($r$). If you assume that the planet is made out of the same stuff as Earth, then gravity scales linearly with radius. 1/5th gravity = 1/5th radius, or 1,274.2km. That's a little smaller than the radius of the moon. If you change the density, you could tweak this gravity this way and that.

It's actually a neat little problem, and worth reading Wikipedia on how it works out. Basically, as you increase the radius, the mass increases by $r^3$, but you also move the surface further away from the center of the planet, reducing it by $r^2$. Those two effects partially cancel to give you the $r$ that you see in the equation above.

Cort Ammon
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Based on this information, 1/5 of Earth's gravity would not be much more than is seen on the moon (i.e. Earth's moon), so a little larger than that seems reasonable:

Gravity on the Moon: This is one astronomical body where human beings have been able to test out the affects of diminished gravity in person. Calculations based on its mean radius (1737 km), mass (7.3477 x 10²² kg), and density (3.3464 g/cm³), and the missions conducted by the Apollo astronauts, the surface gravity on the Moon has been measured to be 1.62 m/s2 , or 0.1654 g.

i.e. 17% of earth's gravity, and you're aiming for 20%.

edit: see also this question

Nick
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