- #1

Soren4

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$$E=\frac{1}{2}\mu {\dot{r}}^2+\frac{L^2}{2\mu r^2}-\gamma \frac{m M}{r^2} \tag{1}$$

Where [itex]\mu[/itex] is the reduced mass of the system planet-star.

Consider now the term $$U_{centrifugal}=\frac{L^2}{2\mu r^2}$$

I don't understand this explanation found on

*Morin.*

The [itex]L^2/2mr^2[/itex] term in the effective potential is sometimes called the angular momentum barrier.It has the effect of keeping the particle from getting too close to the origin.Basically, the point is that [itex]L ≡ mr^2\dot{ θ}[/itex] is constant, so as r gets smaller, [itex]\dot{θ}[/itex] gets bigger. But [itex]\dot{θ}[/itex] increases at a greater rate than r decreases, due to the square of the [itex]r[/itex] in [itex]L=mr^2\dot{ θ}[/itex]. So eventually we end up with a tangential kinetic energy, [itex]mr^2\dot{ θ}^2/2[/itex],that is greater than what is allowed by conservation of energy.

Why [itex]mr^2\dot{ θ}^2/2[/itex] is greater than maximum KE allowed by conservation of energy in this case?

In general how does [itex]U_{centrifugal}[/itex] prevent the planet to collide with the star (provided the planet has non zero angular momentum)?