Planet moving in elliptical orbit around a star

[songoku]

Homework Statement

Please see below

Relevant Equations

Not sure

(a) I use centripetal force to solve this question, even though the question states elliptical orbit

$$\frac{GMm}{r^2}=m\omega^2 r$$
$$\omega^2 r^3=GM$$

So,
$$\omega_{A}^{2} r_{A}^{3}=\omega_{B}^{2} r_{B}^{3}$$
$$\frac{r_A}{r_B}=\left(\frac{\omega_{B}}{\omega_{A}}\right)^{\frac{2}{3}}$$

Is this correct?

Thanks

 

[haruspex]

I don't think you can use that formula in an elliptical orbit. There can be a radial component to the acceleration.
What do you know is conserved?

 

[Orodruin]

I don't think you can use that formula in an elliptical orbit.

Just let me emphasise that you most definitely cannot use that formula. It assumes that the radius is constant, which directly contradicts the assumptions of the problem.

As an additional hint, all you need to know here is that the potential is central. The actual functional behaviour of the force magnitude is irrelevant.

 

[songoku]

I don't think you can use that formula in an elliptical orbit. There can be a radial component to the acceleration.

Is the radial component of acceleration the centripetal acceleration? Or is it something else?

What do you know is conserved?

I think it should be angular momentum, so:
$$m v_A r_A = m v_B r_B$$

I don't think I can use $v=\omega r$ because it is not circular motion so how to relate the equation to $\omega$?

As an additional hint, all you need to know here is that the potential is central. The actual functional behaviour of the force magnitude is irrelevant.

Sorry, what do you mean by "potential is central"?

Thanks

 

[Orodruin]

I think it should be angular momentum, so:
$$m v_A r_A = m v_B r_B$$

Not exactly. The angular momentum depends only on the tangential component of the velocity.

I don't think I can use $v=\omega r$ because it is not circular motion so how to relate the equation to $\omega$?

As such, $\omega r$ is definitely equal to the tangential component of the velocity by definition.

Sorry, what do you mean by "potential is central"?

A central potential depends only on $r$, meaning the force is always pointing to the center.

 

[kuruman]

Just let me emphasise that you most definitely cannot use that formula. It assumes that the radius is constant, which directly contradicts the assumptions of the problem.

An apparent paradox is that, if one assumes that the planet moves in a straight line at constant velocity, one gets the correct answer for part (a).

 

[Super Man]

Seeing the question, the first part can be done using Kepler law which you have done correctly. For second part basic conservation of energy can be utilised to solve it.

 

[haruspex]

Is the radial component of acceleration the centripetal acceleration? Or is it something else?

Sorry, I should have been clearer. A radial component other than centripetal acceleration, i.e. $\ddot r$

 

[haruspex]

the first part can be done using Kepler law which you have done correctly

If you mean Kepler III, the radius in that law refers to the semi-major axis. It is a law relating all planets orbiting the same star, not all positions of a given planet. It does not apply to the first part of this question.

 

[songoku]

Not exactly. The angular momentum depends only on the tangential component of the velocity.

Maybe you mean something like this?
$$mv_Ar_A\sin \theta=mv_Br_B\sin \beta$$

where $\theta$ and $\beta$ is the angle between $v$ and $r$ at point A and B. Using $v=\omega r$:
$$\omega_A r_{A}^{2}\sin \theta=\omega_B r_{B}^{2}\sin \beta$$

The question does not give information about $\theta$ and $\beta$. How to continue?

Sorry, I should have been clearer. A radial component other than centripetal acceleration, i.e. $\ddot r$

Is $\ddot r$ the angular acceleration?

Thanks

 

[haruspex]

Using $v=\omega r$:

No, $v_B\sin(\beta)=\omega r_B$, etc.

Is $\ddot r$ the angular acceleration?

No, it is the rate of increase of the rate of increase of the radius. In a circular orbit it is zero.
The component of the acceleration vector in the radial direction is $\ddot r-r\omega^2$. See e.g. https://proofwiki.org/wiki/Acceleration_Vector_in_Polar_Coordinates

Note also that the centripetal acceleration is not $\omega^2r$. Centripetal acceleration is that component of the total acceleration which is normal to the velocity. If the velocity has a radial component then the centripetal acceleration is not entirely radial. The full expression for it in polar coordinates is horrendous.

 

[kuruman]

Maybe you mean something like this?
$$mv_Ar_A\sin \theta=mv_Br_B\sin \beta$$

where $\theta$ and $\beta$ is the angle between $v$ and $r$ at point A and B. Using $v=\omega r$:
$$\omega_A r_{A}^{2}\sin \theta=\omega_B r_{B}^{2}\sin \beta$$

For angular momentum use $~~\mathbf L =\mathbf r\times\mathbf p$.

Remember that $~~\mathbf p=m\mathbf v$ and $\mathbf v=\mathbf \omega \times \mathbf r$. Put it together using the vector triple product rule.

 

[Orodruin]

Seeing the question, the first part can be done using Kepler law which you have done correctly.

This is incorrect.

Please refrain from confusing the OP by providing wrong answers.

 

[songoku]

I understand.

Thank you very much for all the help and explanation haruspex, Orodruin, kuruman, Super Man