# Terminal Velocity

**URL:** <https://boards.straightdope.com/t/terminal-velocity/217033>\
**Category:** Factual Questions\
**Created:** [December 4, 2003, 9:56pm UTC](https://boards.straightdope.com/t/terminal-velocity/217033 "2003-12-04T21:56:07Z")\
**Posts on this page:** 5\
**Page:** 2

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**Author:** ![Musicat](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/musicat/32/20189_2.png) [@Musicat](https://boards.straightdope.com/u/Musicat)\
**Post date:** [November 13, 2013, 8:28pm UTC](https://boards.straightdope.com/t/terminal-velocity/217033/21 "2013-11-13T20:28:56Z")

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> [@Musicat](#):
>
> Might be hard to get. Q.E.D. is no longer with us. _Really_ no longer with us, RIP.

[http://boards.straightdope.com/sdmb/showpost.php?p=16390540&postcount=26](http://boards.straightdope.com/sdmb/showpost.php?p=16390540&postcount=26)

☹

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**Author:** ![Stealth\_Potato](https://avatars.discourse-cdn.com/v4/letter/s/d78d45/32.png) [@Stealth\_Potato](https://boards.straightdope.com/u/Stealth_Potato)\
**Post date:** [November 13, 2013, 11:11pm UTC](https://boards.straightdope.com/t/terminal-velocity/217033/22 "2013-11-13T23:11:11Z")

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> [@Learjeff](#):
>
> Well, summagun. Really? Hmmm. I believe you, but I’d appreciate an intuitive explanation.
> 
> Not that orbital stuff is ever very intuitive, in general!

Intuitively, you can think about the definition of escape speed: it’s the speed at a given distance from some object (say, a planet) for which the escaping object will come to rest “at infinity.” So, if you start infinitely far away from that object and begin (infinitesimally) accelerating toward it, you’ll have recovered all that gravitational potential energy by the time you reach it – i.e., you’ll be at escape speed. Thus, no matter where you start in relation to a body, you can never exceed its surface escape speed using its gravity alone.

And hearty salute to our **Q.E.D.** He was a good man, and a good doper, what’s better!

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**Author:** ![Canadjun](https://avatars.discourse-cdn.com/v4/letter/c/76d3ee/32.png) [@Canadjun](https://boards.straightdope.com/u/Canadjun)\
**Post date:** [November 13, 2013, 11:17pm UTC](https://boards.straightdope.com/t/terminal-velocity/217033/23 "2013-11-13T23:17:00Z")

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Glad **Learjeff** made that necropost - I was “this” close to making a similar one; now I won’t embarrass myself :D. Thanks **Stealth Potato** for the excellent explanation!

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**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:** [November 14, 2013, 1:47am UTC](https://boards.straightdope.com/t/terminal-velocity/217033/24 "2013-11-14T01:47:18Z")

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> [@](#):
>
> Quoth **Learjeff** :
> 
> Cube, not square.

No, the force is proportional to the square of the speed. You might be thinking of the power lost to the drag.

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**Author:** ![Learjeff](https://avatars.discourse-cdn.com/v4/letter/l/94ad74/32.png) [@Learjeff](https://boards.straightdope.com/u/Learjeff)\
**Post date:** [November 14, 2013, 2:42pm UTC](https://boards.straightdope.com/t/terminal-velocity/217033/25 "2013-11-14T14:42:58Z")

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> [@Stealth\_Potato](#):
>
> Intuitively, you can think about the definition of escape speed …

Thanks, that does make sense, and frankly, I should have been able to come up with that myself!

> [@Chronos](#):
>
> No, the force is proportional to the square of the speed. You might be thinking of the power lost to the drag.

Thanks for the correction. Ignorance fought!

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