# Ratio of speeds in an elliptic orbit

1. Dec 31, 2007

### mike115

[SOLVED] Ratio of speeds in an elliptic orbit

Hi, I need some clarification on a problem.

The problem is: The diagram shows a view of the earth's elliptic orbit about the sun. (This was the closest picture I could find.) In terms of Ra (the distance between the sun and point A) and Rb (the distance between the sun and point B), what is the ratio Vb/Va?

Using conservation of angular momentum, L = I*w = m*r^2*v/r = m*r*v,
m*rA*vA = m*rB*vB
vB/vA = rA/rB
So the velocity is inversely related to the distance from the sun.

However, the answer given is the square root of rA/rB. I can get this answer by equating Newton's gravitational law with the centripetal force, but doesn't centripetal acceleration only work for circular orbits?

Any help would be appreciated. Thanks!

#### Attached Files:

• ###### meug1.gif
File size:
1.5 KB
Views:
90
Last edited: Dec 31, 2007
2. Dec 31, 2007

### D H

Staff Emeritus
You can't use an inertia tensor here. Inertia tensors are used to describe the behavior of rigid bodies. The Sun-Earth system is not a rigid body.

3. Dec 31, 2007

### Rainbow Child

The angular momentum is conserved, so this answer is correct.

You can' t use centripetal acceleration in the naive form

$$\vec{a}_r=-r\,\dot{\theta}^2\,\vec{e}_r$$

in a non-circular orbit. The full form of centripetal acceleration is

$$\vec{a}_r=(\ddot{r}-r\,\dot{\theta}^2)\,\vec{e}_r$$

The distance $$r$$ varies with time so the first term of the left hand side is not zero.

By the way, you can use for the inertia momentum the formula

$$I=m\,r^2 \quad (1)$$

since it's definition (for a ridig body) is

$$I=m_1\,r_1^2+\dots +m_\nu\,r_\nu^2\Rightarrow I=\int r^2\,d\,m$$

and for a point particle reproduces (1).

4. Dec 31, 2007

### D H

Staff Emeritus
The Sun-Earth velocity vector is normal to the Sun-Earth position vector at perihelion and apohelion. Thus the magnitude of the angular momentum vector at these points is simply the product of the magnitudes of the Sun-Earth distance and the magnitude of the Sun-Earth velocity vector. Since angular momentum is a conserved quantity,

$$v_A r_A = v_B r_B$$

or

$$\frac{v_A}{v_B} = \frac{r_B}{r_A}$$

The same result can be derived from the vis-viva equation,

$$v^2=\mu\left(\frac 2 r - \frac 1 a\right)$$

At apohelion, $r_A=(1+e)a$ and at perihelion, $r_B=(1-e)a$. Thus

$$v_A^2=\mu\left(\frac 2 {r_A} - \frac 1 a\right) = \frac {\mu}{r_A}(2 - (1+e)) = \frac {\mu}{r_A}(1-e) = \frac {\mu} a \frac {r_B} {r_A}$$

Similarly,

$$v_B^2=\mu\left(\frac 2 {r_B} - \frac 1 a\right) = \frac {\mu}{r_B}(2 - (1-e)) = \frac {\mu}{r_B}(1+e) = \frac {\mu} a \frac {r_A} {r_B}$$

Thus

$$\frac {v_A} {v_B} = \frac {r_B} {r_A}$$

Conservation of energy and conservation of angular momentum yield the same result.

Where did you get the "answer" $v_B/v_a = \sqrt{r_A/r_B}$?

Last edited: Dec 31, 2007