Approximations in elliptical orbits

I am learning about orbits and getting very confused as to what is exactly true and what is an approximation.

Namely the following points:

• It is often said that the planet has an elliptical orbit with the sun at one focus. I think this is an approximation in assuming that the mass of the sun is so much greater than that of the planet that its motion is negligible compared with the planet, but in fact both have an elliptical orbit about their common center of mass (as in this SE post).

But the derivation we used in our classes for the elliptical motion of a planet is about the sun, and not for both bodies about a common center of mass. The other derivations I have seen also show this orbit about the sun. The derivations all initially assume the sun as a center, and use the gravitational potential around the sun. I am guessing that the approximation comes in because, if you assume the sun at one focus, then the kinetic energy is not simply $0.5mv^2$ because the sun is actually a non inertial reference frame. Is this correct?

Finally, is angular momentum really conserved if we place the sun at a new focus? Why/why not? Linear momentum is clearly not conserved. It would be if we considered the whole system (the planet and the sun). So is linear momentum conserved?

Ignoring that the derivation assumes a two body problem and Newtonian gravity, there's no approximation here. With these assumptions, a bound orbit of one body about another is an ellipse, with either body viewed as being fixed, and the fixed body being one of the foci of the ellipse.

This is of course a non-inertial perspective. The fixed body is accelerating toward the orbiting body. From the perspective of an inertial frame in which the center of mass is fixed rather than one of the bodies, both bodies are in elliptical orbits about the center of mass, with the center of mass being the common focus of the two ellipses.

To show that these are indeed equivalent, suppose we know that object B is orbiting the center of mass of a two-body system in an ellipse, with the center of mass at one of the two foci of the ellipse. This means that the polar coordinates of object B with the origin at the center of mass is $$\vec r_B = \frac {a_B (1-e^2)}{1+e \cos\theta} \,\,\hat r$$ where $a_B$ is the semi-major axis length of the orbit, $e$ is the eccentricity of the orbit, and $\theta$ is the angle on the orbital plane subtended by B's closest approach to the center of mass, the center of mass, and object B itself.

What about the other object? It's position in this center of mass system is constrained by $m_A \vec r_A = -m_B \vec r_B$. Thus it too moves in an ellipse with the center of mass as one of the foci: $$\vec r_A = \frac {a_A (1-e^2)}{1+e \cos\theta}(-\hat r)$$ where $a_A = \frac{m_B}{m_A} a_B$.

Finally, what about the displacement vector between the two? This is $$\vec r_B - \vec r_A = \left(1 + \frac {m_B}{m_A}\right) \frac {a_B(1-e^2)}{1+e\cos\theta} \,\, \hat r \equiv \frac {a(1-e^2)}{1+e\cos\theta}\,\,\hat r$$ where $a \equiv \frac{m_A+m_B}{m_A}a_B = \frac{m_A+m_B}{m_B}a_A$.

This is of course yet another ellipse. It can be looked at in two different ways: From the perspective of object $B$ orbiting a fixed object $A$, in which case object $A$ is at one of the foci of this ellipse, or from the perspective of object $A$ orbiting a fixed object $B$, in which case object $B$ is at one of the foci of this ellipse.

• Many apologies for the delayed response, and also I do not seem to be able to change my accepted answer. I did up vote your answer however. Thank you for your explanation- it was very clear and explained a lot. – 21joanna12 May 10 '17 at 13:42

First, the fact that the planet has an elliptical orbit with the sun at one focus can be derived with no approximation. This is very well explained in David Hammen's answer. The reference frame which corresponds to this view is not inertial.

In the chapter 8 of Marion all dynamics are derived always from the center of mass frame. As you may already know, the problem of two bodies can be solved analytically, and this can be done by placing our inertial frame on the center of mass and working with the reduced mass $\mu$.

An approximation that is sometimes done is the following: as the sun is usually much more massive than the planet, you can take the approximation of placing the center of mass of the whole system on the center of the sun. As the mass of the sun $m_S$ is much bigger than the planet mass $m_P$ ($m_S >> m_P$), you can also take the approximation $\mu \approx m_P$. This is equivalent to consider the sun as an inertial frame. This means that usual mechanics ("inertial frame" mechanics) cannot be applied directly, but are a good approximation. If you want to solve exactly the dynamics of the system you must move to the center of mass and work from there with the reduced mass.

The total angular momentum is a conserved quantity of the total system because there's never an external torque $\tau=r \wedge F$ applied on the system. It doesn't matter which observer you are, the direction of force $F$ is always parallel to the position vector $r$, so $\tau=0$.

On the center of mass frame, the total linear momentum is conserved for the whole system, as there's no external force acting on it; but not for each of the bodies separately because gravity acts on each of them.

If you put your frame on the sun or on the planet (non inertial frames), then the linear momentum is no longer conserved, even for the whole system, because the only contribution comes from the planet, and that's not a conserved quantity.

Note: it is the center of mass of the sun which is moving in an elliptical orbit around the center of mass of the two bodies, as both are into the sun.

• Thank you for your reply. However it is still not clear to me as to how angular momentum can be conserved in this approximation. As you said, clearly linear momentum cannot be. But how is angular momentum conserved? – 21joanna12 May 8 '17 at 16:30
• There's not a net external torque acting on the system, so angular momentum is conserved. As $\tau=r \wedge F$, we know that $r$ and $F$ are parallel because of being a central force, so $\tau=0$ and as a consequence angular momentum is constant. Even with the approximation you would have those two vectors parallel. – falgenint May 8 '17 at 16:39
• Ah I see! The gravitational force acts along the line through each mass and the centre of mass, so whether I consider the situation from either mass' point of view or from the centre of mass, there is no torque acting so angular momentum is conserved (even though each mass itself is not an nertial reference frame) – 21joanna12 May 8 '17 at 16:48
• Yes, that's it! – falgenint May 8 '17 at 16:55
• This answer is very wrong. There is no approximation here. Given a two body system, either of the two bodies can be viewed as fixed and the other orbiting about it in an ellipse. This is the standard reduced mass derivation. Given this, it's a mathematical necessity that from the perspective of a center of mass frame, the two objects will each orbit the center of mass in an ellipse. – David Hammen May 8 '17 at 18:40