Ratio of speeds in an elliptic orbit

[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?

Which answer is correct?

Any help would be appreciated. Thanks!

 

[D H]

Using conservation of angular momentum, L = I*w

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.

 

[Rainbow Child]

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.

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

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?

Which answer is correct?

Any help would be appreciated. Thanks!

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).

 

[D H]

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" $v_B/v_a = \sqrt{r_A/r_B}$?