# Orbital mechanics and sideways forces

**URL:** <https://boards.straightdope.com/t/orbital-mechanics-and-sideways-forces/550589>\
**Category:** Factual Questions\
**Created:** [August 18, 2010, 12:20am UTC](https://boards.straightdope.com/t/orbital-mechanics-and-sideways-forces/550589 "2010-08-18T00:20:08Z")\
**Posts on this page:** 3\
**Page:** 2

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**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\
**Post date:** [August 18, 2010, 11:57pm UTC](https://boards.straightdope.com/t/orbital-mechanics-and-sideways-forces/550589/21 "2010-08-18T23:57:57Z")

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> [@MikeS](#):
>
> Nitpick: I’m pretty sure that two ellipses with one shared focus will intersect in two points at most.

And you’d be right. I’m not sure where that came from.

Stranger

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**Author:** ![Cheshire\_Human](https://avatars.discourse-cdn.com/v4/letter/c/5f8ce5/32.png) [@Cheshire\_Human](https://boards.straightdope.com/u/Cheshire_Human)\
**Post date:** [August 19, 2010, 1:37am UTC](https://boards.straightdope.com/t/orbital-mechanics-and-sideways-forces/550589/22 "2010-08-19T01:37:00Z")

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> [@Stranger\_On\_A\_Train](#):
>
> > [@MikeS](#):
> >
> > Nitpick: I’m pretty sure that two ellipses with one shared focus will intersect in two points at most.
> 
> And you’d be right. I’m not sure where that came from.
> 
> Stranger

Maybe because 2 ellipses with _ **random orientation** _ to each other can only intersect at at most 4 points. 😉

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**Author:** ![dracoi](https://avatars.discourse-cdn.com/v4/letter/d/90db22/32.png) [@dracoi](https://boards.straightdope.com/u/dracoi)\
**Post date:** [August 19, 2010, 5:09pm UTC](https://boards.straightdope.com/t/orbital-mechanics-and-sideways-forces/550589/23 "2010-08-19T17:09:30Z")

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> [@flight](#):
>
> I solved a closed-form solution to this. It holds for velocity changes in or opposite the direction of motion when you start with a circular orbit. It gets a bit more complex when you start with an elliptical orbit and depends on where in the orbit you make the change. Anyway, it looks like this:
> 
> %r = (%V)^2/(2 - (%V)^2)
> 
> where %r is the percent change (divided by 100) in the distance to the Earth at the opposite end of the orbit and %V is the percent change (divided by 100) in velocity. So, if I throw the wrench forward at 1% of my velocity (%V = 1.01), the distance from the Earth at the opposite side of its orbit (the most extreme change) will be 4.1%.
> 
> For small changes (which is almost always going to be the case, the percentage change in radius is 4 times the percentage change in velocity. I think this is what you were looking for.

That’s awesome! There are definitely enough near-circular orbits to make this a useful shortcut sometimes.

And thanks to everyone for the additional responses. I’m bookmarking things to come back and read in more detail when I have time to really digest some of the resources posted.

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