Velocity ratio - bodies in orbit

Main Question or Discussion Point

Hi, I was just going over some equations for velocity with respect to bodies revolving around the sun. I wanted to figure out the ratio of tangential linear velocity (i.e. speed) of a body when it is at its perihelion to its velocity at aphelion.

In other words, I wanted to solve for [tex]\frac{v_a}{v_p}[/tex] (velocity at aphelion/velocity at perihelion) in terms of (Ra and Rp, distance at aphelion and perihelion, respectively).

I figured there are two ways of doing this.
One, we can say that the centripetal force keeping the body in orbit is entirely due to the gravitational force of (sun on body).

So setting up a force equality:
[tex]\frac{mv^{2}}{R}=\frac{GMm}{R^2}[/tex]

where m is the mass of the body, M the mass of the sun, G the gravitational constant, and R, the distance between the centers of mass of the sun and body.

Canceling out like terms, we now get:
[tex]v^2=\frac{GM}{R}[/tex]

Indicating that the linear velocity is inversely proportional to the square root of the distance between centers of mass.
So the ratio va/vp is:
[tex]\frac{v_a}{v_p}=\sqrt{\frac{R_p}{R_a}}[/tex]

Now, solving it another way, if we see that there are no external torques acting on the system, such that it is in angular equilibrium, then the angular momentum at each point in the orbit should be equal, in other words:

[tex]L_a = L_p[/tex]
[tex]mv_aR_a = mv_pR_p[/tex]

Rearranging this then, we see that the velocities are inversely proportional to just the distances, i.e.

[tex]\frac{v_a}{v_p}=\frac{R_p}{R_a}[/tex]

So, which is right? Velocity inversely proportional to the square root or just the straight distance?
I'm assuming that they both are right, and I forgot to integrate something, or I assumed too much in setting this up. Can anyone shed light on this?
Thanks!
 
Last edited:

Answers and Replies

clem
Science Advisor
1,308
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The R in mv^2/R is NOT the distance from the sun.
It is the radius of curvature of the orbit.
You result only holds for a circular orbit.
 
Aha, then the equation [tex]F=\frac{mv^2}{R}[/tex] only holds true for circular motion then, and thus R can never be varied, correct?

Or, R can be varied, but it must be computed as the radius of curvature of motion rather than the distance of the object from the center.
 
D H
Staff Emeritus
Science Advisor
Insights Author
15,329
681
So setting up a force equality:
[tex]\frac{mv^{2}}{R}=\frac{GMm}{R^2}[/tex]
That isn't valid. You are implicitly assuming a circular orbit here.

Now, solving it another way ...

[tex]\frac{v_a}{v_p}=\frac{R_p}{R_a}[/tex]
You will get the same result if you look at the problem from the perspective of conservation of energy. Conservation of energy dictates that

[tex]v^2 = GM\left(\frac 2 r - \frac 1 a\right)[/tex]

This is the vis-viva equation. The semi-major axis is related to the apofocus and perifocus via [itex]2a = r_a + r_p[/itex]. With this, the same relationship as you found with conservation of angular momentum arises from the vis-viva equation.
 
Thanks a lot, that was a big help. I hadn't caught that the centripetal force equation was only for circular orbits.
 

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