# Space elevator question: gravity conditions in space station **URL:** \ **Category:** Factual Questions\ **Created:** [January 16, 2011, 9:34pm UTC](https://boards.straightdope.com/t/space-elevator-question-gravity-conditions-in-space-station/567865 "2011-01-16T21:34:40Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [January 18, 2011, 12:10am UTC](https://boards.straightdope.com/t/space-elevator-question-gravity-conditions-in-space-station/567865/21 "2011-01-18T00:10:00Z")
The equation for centrifugal force (or alternately for centripetal force; confusingly it’s the same equation) is F = m_v^2/r, or F = m_omega^2_r (where omega is the “angular speed”, or v/r). For something like this, the omega version of the equation is probably more useful, since a rigid body like a space station will have a single value of omega, but a different v at every point. The force of gravity is F = m_g, where g = 9.8 m/s^2. So the rotation of your station is given by omega^2\*r = g, or omega = sqrt(g/r). ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [January 18, 2011, 12:26am UTC](https://boards.straightdope.com/t/space-elevator-question-gravity-conditions-in-space-station/567865/22 "2011-01-18T00:26:45Z")
> [@Chronos](#): > > … find something 3 feet across, and stand 10 feet away from it. That’s about how big Earth would look. This is still well larger than the apparent size of the Sun Indeed - around 40 times the angular size of the Sun. ---
**Author:** ![Mangetout](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mangetout/32/19_2.png) [@Mangetout](https://boards.straightdope.com/u/Mangetout)\ **Post date:** [January 18, 2011, 8:07am UTC](https://boards.straightdope.com/t/space-elevator-question-gravity-conditions-in-space-station/567865/23 "2011-01-18T08:07:58Z")
> [@TriPolar](#): > > Yes, but there will be plenty of shear forces that have to be accounted for, at least within the atmosphere. Conservation of angular momentum is going to be a big problem isn’t it? ---
**Author:** ![knockout](https://avatars.discourse-cdn.com/v4/letter/k/ebca7d/32.png) [@knockout](https://boards.straightdope.com/u/knockout)\ **Post date:** [December 2, 2015, 6:32am UTC](https://boards.straightdope.com/t/space-elevator-question-gravity-conditions-in-space-station/567865/24 "2015-12-02T06:32:31Z")
Reviving an old thread, yes I know… To follow up in this discussion, how would you calculate the gravity that would be felt by people on different locations along the tether line? Let’s say for example, there’s some kind of station located along the path in the mesosphere, how would you calculate the gravity of this location? ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [December 2, 2015, 2:04pm UTC](https://boards.straightdope.com/t/space-elevator-question-gravity-conditions-in-space-station/567865/25 "2015-12-02T14:04:32Z")
The very short answer is there are two force components to deal with. 1) What is the gravity at that altitude? This force pulls downwards. 2) What is the centrifugal force generated by the orbiting motion? This force pulls upwards. The sum of the two (AKA net of the two) is the answer. For any altitude the first force is given by F=GM[sub]E[/sub]M[sub]S[/sub]/R^2 where G is the gravitational constant, M[sub]E[/sub] is the mass of the Earth, M[sub]S[/sub] is the mass of your station or person or instrument or …, and R is the distance from the center of the Earth to your altitude on the cable. The latter force is (ref post #21) given by F = M[sub]S[/sub]\*v^2/R where M[sub]S[/sub] and R are the same as in the first equation, and v is the tangential speed. Tangential speed is distance / time which is 2_pi_R / 24 hours. Mix all the math together in the units of your choice and … _Viola_! (sic :)). If you want to restate the force as its ratio to conventional ground level gravity, you can eventually simplify the equation into something that amounts to RatioOfLocalvsNormalGroundLevelGravity = SomeSimpleFunction®. All the other variables drop out and all the constants meld into one or two. Somebody more ambitious than I can pick a unit convention (SI or customary) and do the scutwork algebra to simplify. [Previous page](https://boards.straightdope.com/t/space-elevator-question-gravity-conditions-in-space-station/567865.md?page=1)