# Centripetal Acceleration of an Ellipse

1. Sep 8, 2013

### Philosophaie

The centripetal acceleration of a circle is: $$a_c = \frac{v^2}{r} * u_n.$$ The acceleration of an ellipse is different. It increases from from apoapsis to periapsis as the position changes from furthest point in the orbit to the closest. Then decreases from from periapsis to apoapsis as the position changes from closest point to the furthest. Is there a way to calculate this centripetal acceleration of an ellipse due to the change in the normal and the varying of position due to the gravitating body?

2. Sep 8, 2013

### Andrew Mason

The only way to make a body prescribe an elliptical path due to a centripetal force (ie. a force that is always directed toward a central point whose location does not change) is to have that central point located at a focus of the ellipse and have the central force vary as 1/r^2 where r is the distance from the body to that central point (focus).

So if you know its tangential speed at the perigee or apogee (radial component is 0 so a = v^2/r) you can work out the acceleration at any other point eg.:

$$\frac{a_r}{a_{apogee}} = \frac{r_{apogee}^2}{r^2}$$

AM

Last edited: Sep 8, 2013
3. Sep 8, 2013

### eigenperson

Really? That may be the only sensible way to do it, but I'm pretty sure that given any point whatsoever inside the ellipse, you can produce an elliptical trajectory by a centripetal force that varies appropriately.

4. Sep 8, 2013

### rcgldr

Doesn't this redefine "centripetal force"? Using the normal definition, if the only force on an object is centripetal (perpendicular to velocity), then the speed is constant, and only the radius of curvature changes over time. There was a recent thread that worked out the math for an objects traveling at constant speed in an elliptical path.

5. Sep 8, 2013

### Andrew Mason

Not if you want to produce an elliptical path that keeps repeating itself (ie. an ellipse that does not precess).

You will find that the appropriate variation of the central force to produce a constant elliptical path is a variation of 1/r^2 where r is the distance from the body to the central point. The special case is the circle where there is no variation in r (ie the magnitude of the central force is constant).

The mathematical proof involves finding a very messy solution to a differential equation. Newton figured it out using a geometrical method which is very difficult to follow (See: Feynman's attempt to explain it in "Feynman's Lost Lecture").

AM

6. Sep 9, 2013

### Andrew Mason

I was assuming that the OP was referring to an object prescribing an elliptical orbit due to a central force - a force that is always pointing toward the same central point. The OP referred to a gravitating body.

If a body's motion is being affected only by a central force, Fc, the acceleration toward that central point (ie. the centripetal acceleration) is simply ac = Fc/m. The centripetal acceleration is perpendicular to velocity only for circular motion or at only two points if the motion is elliptical.

AM

7. Sep 9, 2013

### vela

Staff Emeritus
Centripetal means "toward the center," not "perpendicular to velocity."

8. Sep 9, 2013

### A.T.

"Centripetal" means "toward the instantaneous center of the path curvature", which is the same as "perpendicular to velocity". See:
http://en.wikipedia.org/wiki/Centripetal_force

Last edited: Sep 9, 2013
9. Sep 9, 2013

### vela

Staff Emeritus
I stand corrected.

10. Sep 9, 2013

### voko

No, this is not the only way. A radial harmonic oscillator, whose force is proportional to r, will also have an elliptic orbit, with the center of revolution in the center of the ellipse, not at a focus. According to Bertrand's theorem, this and the inverse-square law are the only two systems that admit closed orbits, which then happen to be elliptic.

Note that any central attractive force admits a circular orbit.

However, since the original question explicitly involved a "gravitating body", we probably needn't consider anything except the inverse-square law to answer it.

11. Sep 9, 2013

### harrylin

Interesting! I see that it's also how Newton used it:
http://gravitee.tripod.com/booki3.htm (simply press "cancel").
The Wikipedia article also cites his definition:
In fact your answer was correct! Centripetal simply means "seeking the center". :tongue2:
It appears that there are two slightly different definitions of "centripetal force"...

12. Sep 9, 2013

### A.T.

Yes, but "center" is ambiguous. The most common and general definition that works for every curved path references the instantaneous center of curvature of the path.

13. Sep 9, 2013

### Staff: Mentor

To the OP: are you just interested in the force vector on an elliptical orbit, or are you interested in breaking it into components which are parallel and perpendicular to the velocity?

14. Sep 9, 2013

### AlephZero

AFAIK the usual description of the OP's scenario is a central force (i.e. a force that is always directed towards a fixed point). not "centripetal".

In general, the central force has two components, normal and tangential to the path of the object around the orbit. Since "normal" is a well defined geometrical notion for any curve in space (see http://en.wikipedia.org/wiki/Frenet–Serret_formulas) calling the normal direction "centripetal" doesn't seem to add any value IMO - the central force only acts along the normal direction at two points around the orbit, except in the special case where the orbit is a circle.

15. Sep 10, 2013

### D H

Staff Emeritus
Yet that description of gravitation as a central force is not what the OP wants. Just take a look at other threads recently started by the OP. He wants centripetal force, that is, the component of force normal to the velocity vector.

I'll make one last vain attempt to steer Philosophaie away from this concept.

Philosophaie: Centripetal force, the component of force normal to the velocity vector, is not going to help you in your quest to understand orbits. Gravitation is a central force rather than a centripetal force. Learn how to use polar coordinates. Learn about true anomaly, eccentric anomaly, mean anomaly, and the relationships between them.