# Orbital mechanics question... **URL:** \ **Category:** Factual Questions\ **Created:** [September 1, 2006, 10:22pm UTC](https://boards.straightdope.com/t/orbital-mechanics-question/371130 "2006-09-01T22:22:16Z")\ **Posts on this page:** 5\ **Page:** 1
**Author:** ![Subway\_Prophet](https://avatars.discourse-cdn.com/v4/letter/s/ecb155/32.png) [@Subway\_Prophet](https://boards.straightdope.com/u/Subway_Prophet)\ **Post date:** [September 1, 2006, 10:22pm UTC](https://boards.straightdope.com/t/orbital-mechanics-question/371130/1 "2006-09-01T22:22:16Z")
I imagine this is a fairly common physics two-body problem, if two-body problems are indeed common in physics, but I’m having some difficulty with Kepler’s Third Law. Pretend you have two bodies, one of which is so much more massive than the other, that the barycenter is practically the larger object’s center of mass. The other body orbits the first one in a perfectly circular orbit, in order to keep the math simple. Also pretend that you can move the smaller body at will, and insert it into other arbitrary orbits. So: Is there a simple relationship between the orbital radius and the orbital speed? Kepler’s Third suggests that if you multiply the radius by n, the orbital period is multiplied by n to the power of 2/3. (At least, how I understand it.) But according to the equation I found at [Wikipedia’s article on orbital speed](http://en.wikipedia.org/wiki/Orbital_speed), this does not seem to be so simple a relationship - the smaller body’s specific orbital energy puts a kink in the “directly proportional” part of Kepler’s Third. If the difference in mass between the two bodies is extreme (think Sun and Earth, for example), is the smaller body’s specific orbital energy significant? Or have I completely misunderstood the math involved? 🙂 FWIW, this is for a computer simulation I’m writing - not homework, and I’m not planning on adjusting any planetary orbits any time soon. 🙂 ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [September 1, 2006, 10:42pm UTC](https://boards.straightdope.com/t/orbital-mechanics-question/371130/2 "2006-09-01T22:42:30Z")
Kepler’s Third Law is an exact relationship (in two-body Newtonian mechanics) between the orbital _period_ and the orbit’s semimajor axis. It’s not quite the relationship you give, though. The period scales as the 3/2 power of the semimajor axis, as in the formulas in [Orbital period](http://en.wikipedia.org/wiki/Orbital_period). The orbital speed, however, varies along the orbit, as the equations in that Wikipedia link show. The speed variation depends on the shape of the orbit; a nearly-circular orbit will show only very small speed variations while a very eccentric orbit will have large speed variations. (The “r” in those formulas is the time-varying distance between the two objects; when r is small v is large, and vice versa.) This is just due to conservation of energy: as the planet drops farther down the star’s potential well, it must gain kinetic energy: i.e., speed up. (It is also a consequence of Kepler’s Second Law, if you like.) ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [September 1, 2006, 10:57pm UTC](https://boards.straightdope.com/t/orbital-mechanics-question/371130/3 "2006-09-01T22:57:56Z")
Without getting into the math… For a circular orbit, the speed it fixed at a given radius. Think of it this way: Instead of a planet, twirl a ball around on a string. You have to get it up to a certain speed for the string to rotate parallel to the ground. Any slower, and the ball can sill orbit, but it’s radius is smaller (the string is no longer parallel to the ground). If you speed it up (and don’t hold the string too tightly), the radius will grow (ie, the ball will move out to a larger orbit) and you’ll have to swing it faster to keep the string parallel to the ground. ---
**Author:** ![Subway\_Prophet](https://avatars.discourse-cdn.com/v4/letter/s/ecb155/32.png) [@Subway\_Prophet](https://boards.straightdope.com/u/Subway_Prophet)\ **Post date:** [September 1, 2006, 11:43pm UTC](https://boards.straightdope.com/t/orbital-mechanics-question/371130/4 "2006-09-01T23:43:22Z")
> [@Subway Prophet](#): > > Kepler’s Third suggests that if you multiply the radius by n, the orbital period is multiplied by n to the power of 2/3. > [@Omphaloskeptic](#): > > It’s not quite the relationship you give, though. The period scales as the **3/2** power of the semimajor axis :smack: :o :smack: :o That’s it. I knew my math was bad. Thanks! 🙂 ---
**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:** [September 2, 2006, 7:06pm UTC](https://boards.straightdope.com/t/orbital-mechanics-question/371130/5 "2006-09-02T19:06:59Z")
> [@](#): > > You have to get it up to a certain speed for the string to rotate parallel to the ground. You might want to get into the math enough to calculate that speed… 🙂