How Do Velocity and Angular Momentum Change in Elliptic Orbits?

[baron.cecil]

When deriving the motions of the planets in our solar system, do we typically ignore Newton's law of gravitation between the planets themselves, since the force is negligible?

 

[mathman]

Accurate descriptions of planetary orbits do include the effects of other planets. The orbit without other planets can be computed analytically, but the perturbations due to the other planets then must be done numerically.

Further complication - the orbit of Mercury requires general relativity. It was one of the first tests of the theory.

 

[baron.cecil]

New Question

So in planetary motion, if satellites travel faster at the perigee than the apogee, that implies some change in velocity. Is it the tangential velocity that is changing, or the angular velocity? If universal gravitation is the only force acting between the two objects, how can the velocity change?

 

[D H]

So in planetary motion, if satellites travel faster at the perigee than the apogee, that implies some change in velocity.

Yes.

Is it the tangential velocity that is changing, or the angular velocity?

Both.

If universal gravitation is the only force acting between the two objects, how can the velocity change?

The velocity must change in order for angular momentum and mechanical energy to be conserved. The specific angular momentum¹ of the orbiting body² is

$$\aligned \vec L &= \vec r \times \vec v \\ L &\equiv ||\vec L|| = r v_{\perp} \endaligned$$

Angular momentum is a conserved quantity, and thus the tangential component of velocity $v_{\perp}$ is inversely proportional to the distance from the central body: $v_{\perp} = L/r$. Angular velocity is $v_{\perp}/r = L/r^2$, so this is also a time-varying quantity in an elliptic orbit.

Mechanical energy is also a conserved quantity. The total specific mechanical energy for a given orbit is

$$E = -\frac 1 2 \,\frac {GM}a$$

Specific mechanical energy is simply the sum of specific kinetic energy $v^2/2$ and gravitational potential $-GM/r$. Thus

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

This is the vis-viva equation.



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¹ The mass of the orbiting body m will simply drop out of the equations of motion assuming m is very, very small compared to M, the mass of the central body. The math becomes a bit easier if that little m mass is eliminated as soon as possible. The prefix specific means "divided by mass". Think specific gravity, specific impulse, ...

² I am assuming the mass of the orbiting body is much smaller than the mass of the central body. Things get a bit messier (but not too much) if this simplifying assumption is not valid. Note well: This assumption fails for Jupiter's orbit about the Sun, and for the Moon's about the Earth.