# How does escape velocity work with rotational gravity? [closed]

Every planetoid who gains gravity "normally" like Earth and Io and every other piece of matter in the universe gets it from their mass and radius. Escape velocity is also from mass and radius. My question is, how does escape velocity work if matter gains rotational gravity from its inertia, with negligible true gravity?

• I don't get what you are asking. Inertial mass and gravitational mass are the same, according to our understanding of physics. What is rotational gravity?
– L.Dutch
Sep 11, 2019 at 19:01
• It seems you're describing the classic sci-fi ring-shaped space station that generates artificial gravity on the inside surface of the ring through rotation. In that case, the analogue of escape velocity might be the velocity you need to jump with to not return to the surface of the ring (while staying inside the ring). But that will be highly dependent on the direction you jump, whereas moving at escape velocity in any direction will allow you to escape a planet. Sep 11, 2019 at 19:08
• @NuclearWang, if that is the question, than it is a duplicate of this other question of mine worldbuilding.stackexchange.com/q/150259/30492
– L.Dutch
Sep 11, 2019 at 19:09
• And please add some more tags. Hard science cannot be the only tag of a question
– L.Dutch
Sep 11, 2019 at 19:12
• You ask for hard science but do not apply it to your question. There is no such thing in hard science as "rotational gravity" only gravity. I believe what you are referring to is faux or emulated gravity that one can generate through centrifugal force (which isn't real gravity because there are some problematic factors in how centrifugal force behaves as opposed to gravity). You also include no numbers for speed, mass, etc. Sep 11, 2019 at 19:24

## 2 Answers

"Rotational gravity" is just false gravity emulated by centrifugal force. So, escape velocity would be either a) the inverse of the centrifugal force being generated or, (and much simpler) a corridor or tunnel to the outside of your ring/cylinder/sphere, where a craft would literally "fall" away at your rotational gravity pace (though it would stop accelerating the instant it left contact with the rotating body).

IOW: Escape velocity = "let go"

https://en.wikipedia.org/wiki/Centrifugal_force

In the article, look at the basic diagram for rotational inertia, that would become your departing velocity the instant you were no longer bound to the rotating body.

• A velocity cannot be a force nor a tunnel...
– L.Dutch
Sep 11, 2019 at 19:10
• That apart, the question is tagged hard science. This answer doesn't meet the hard science requirements.
– L.Dutch
Sep 11, 2019 at 19:13
• @L.Dutch I swear it wasn't tagged as hs when I looked before. And the question itself is not using hard-science as there is no such thing as "rotational gravity" I believe he was referring to the principle of using centrifugal force to emulate gravity on a space station or other construct, thus my answer to "how escape velocity would work with rotational gravity" is "let go" as you would not have to attain a velocity to escape the rotational gravity, simply bypass the barrier holding you in the rotation and you have "escaped" Sep 11, 2019 at 19:19
• Then if the question is poorly asked don't answer it and ask for clarification. Answering poorly asked questions simply encourages not putting effort into the question.
– L.Dutch
Sep 11, 2019 at 19:25
• @L.Dutch added link to a diagram of how centrifugal force works. It's as hard science as the matter requires. Sep 11, 2019 at 19:32

It depends a lot on where you are going and the size of your ring

Let's say you are in a Nivan style ring world and you are trying to go out deeper into space. The most practical approach to escape velocity may be to just open a hole in the floor. Then you just need to time your escape such that you go flying off in the right direction effectively "falling" to your destination.

Now, let's say there is somewhere you want to go inside of the ring and you want to know how fast you need to go to reach zero apparent Gs, the answer lies in matching the Tangent Velocity of the ring by moving against its rotation. Once you reach the Tangent Velocity relative to the ring's surface, you will become stationary in space and no longer under the influence of any artificial gravity. At that point you can fly off at whatever speed you desire unaffected by the ring's rotation.

Larger rings require a higher Tangent Velocity than smaller rings to achieve the same apparent gravity. As you can see demonstrated in this trajectory calculator I wrote for L. Dutch's previous previous question about rotational gravity physics, you see that a Nivian Ring world rotates at over 1 million m/s whereas a Stanford torus only rotates at 94 m/s despite both having the same apparent gravity.

Just use the attached calculator and you can find the "escape velocity" of any ring by calculating for the tangent velocity.

Side-note: don't accelerate with the direction of rotation, that will actually increase your apparent gravity.