# Can a baseball be a shooting star? **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [June 20, 2003, 3:18pm UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179 "2003-06-20T15:18:37Z")\ **Posts on this page:** 7\ **Page:** 2
**Author:** ![The\_Ryan](https://avatars.discourse-cdn.com/v4/letter/t/7feea3/32.png) [@The\_Ryan](https://boards.straightdope.com/u/The_Ryan)\ **Post date:** [June 28, 2003, 2:40am UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179/21 "2003-06-28T02:40:27Z")
45 minutes seems like an awfully long time for a ballistic missile aimed at a mobile target. Chronos: > [@](#): > > So it’ll pull apart into chunks small enough that they can hold together. But even then, those chunks will still be in orbit. There’s a minimum size at which the Roche limit applies? Also, are the rings of Saturn et al. former satellites that were below the Roche limit? > [@](#): > > When you’ve got nothing but Newtonian gravity in a system, with one body orbiting another, there is no such thing as an unstable orbit. Are you considering tidal forces to not be Newtonian? ---
**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:** [June 28, 2003, 5:15am UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179/22 "2003-06-28T05:15:24Z")
Tidal forces are Newtonian, but they can’t destabilize an orbit. And the Roche limit only applies for things which are held together primarily by gravity: If I have a baseball, say, or the Space Shuttle, in low Earth orbit, then the tidal forces pulling the baseball apart will be stronger than the gravitational force holding the ball together… But so what? The gravitational force holding a baseball together is pitifully small anyway, compared with the electromagnetic forces holding it together. I’m pretty sure that the rings of the gas giant planets are below the Roche limits, but it’s hard to say whether they were once a larger body which got pulled apart, or if they just never formed a moon to begin with. ---
**Author:** ![The\_Ryan](https://avatars.discourse-cdn.com/v4/letter/t/7feea3/32.png) [@The\_Ryan](https://boards.straightdope.com/u/The_Ryan)\ **Post date:** [June 28, 2003, 5:29am UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179/23 "2003-06-28T05:29:46Z")
> [@](#): > > Tidal forces are Newtonian, but they can’t destabilize an orbit. But don’t they decrease the momentum of the orbiting body? And from the article, I don’t understand this: > [@](#): > > Thrusting down, for instance, will lower you temporarily, but now you’re going too fast to stay in that lower orbit, and you’ll end up oscillating back above your original orbit. Are you saying that the closer you are to Earth, the slower the orbiting speed is? ---
**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:** [June 28, 2003, 9:31pm UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179/24 "2003-06-28T21:31:42Z")
Tidal forces don’t do anything to the momentum of an orbiting body. All they do is try to pull the body apart. Where did you hear otherwise? And a closer circular orbit is faster than a further circular orbit, but when you go down, you gain speed (from potential energy being converted to kinetic). The net effect is that, if you try to thrust yourself down, the circular orbit speed for your height will increase, but your speed will increase even more. So you end up too fast to stay at that height. ---
**Author:** ![The\_Ryan](https://avatars.discourse-cdn.com/v4/letter/t/7feea3/32.png) [@The\_Ryan](https://boards.straightdope.com/u/The_Ryan)\ **Post date:** [June 29, 2003, 5:38am UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179/25 "2003-06-29T05:38:25Z")
> [@](#): > > \*Originally posted by Chronos \* > Tidal forces don’t do anything to the momentum of an orbiting body. Perhaps we’re talking about different things. In physics, “tidal forces” usually refers to the differential in the gravitational force, but in this case I’m using the term to refer to the forces regarding tides (as in changes in sea level). The Earth is losing angular momentum because of the tides, right? So isn’t that momentum being transfered to the moon? > [@](#): > > And a closer circular orbit is faster than a further circular orbit, but when you go down, you gain speed (from potential energy being converted to kinetic). Hmm, I think I’m starting to see what you’re saying. As the ball “falls”, it starts to move towards the Earth faster and faster. But unless it hits the Earth on the first pass, all that velocity will be turned into velocity away from the Earth, like a bull whose velocity towards the matador is transformed into velocity away from him by the simple act of the matador stepping aside. ---
**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:** [June 29, 2003, 8:02pm UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179/26 "2003-06-29T20:02:24Z")
OK, I think I see what you’re getting at. While in principle, dragging the oceans around the planet will change the momentum of an orbiting body, the effect will be negligible for any artificial satellite, since they have so little effect on the oceans. Even for the Moon, which is easily the most tidally-significant object for the Earth, the effect is miniscule. I can’t imagine that this would be enough to bring down a satellite before the Sun engulfed the Earth, regardless of whether the satellite is within the Roche limit. > [@](#): > > But unless it hits the Earth on the first pass, all that velocity will be turned into velocity away from the Earth, like a bull whose velocity towards the matador is transformed into velocity away from him by the simple act of the matador stepping aside. Essentially correct, if you allow for the fact that this isn’t all happening in a straight line, like with the bull. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [July 4, 2003, 2:10pm UTC](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179/27 "2003-07-04T14:10:59Z")
Description of gravity assist maneuvers. [http://www.jpl.nasa.gov/basics/bsf4-1.html](http://www.jpl.nasa.gov/basics/bsf4-1.html) Though I think the term is really not accurate. It is not the gravity that is assisting, but the angular momentum. The gravity slows down as much as it speeds up, it’s the orbit of the sun that gives the extra speed. Picture of Phobos. It mentions the decaying orbit, but doesn’t explain. [http://antwrp.gsfc.nasa.gov/apod/ap030701.html](http://antwrp.gsfc.nasa.gov/apod/ap030701.html) [Previous page](https://boards.straightdope.com/t/can-a-baseball-be-a-shooting-star/183179.md?page=1)