1

For a planet under the effect of gravitational force the mechanical energy can be written as

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

Where $\mu$ is the reduced mass of the system planet-star.

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

This is a kinetic term in the mechanical energy expression (actually the one regarding $v_{\theta}$), but, since $L$ is constant, it has been rewritten as a function of the $r$ coordinate only. I read that this "makes" it a sort of potential term. That's clear, since it does not involve the velocity, but what is its real meaning?

The name "centrifugal" is because it sort of makes a "centrifugal barrier" which is the reason why the planets with non zero angular momentum stay in orbit instead of "fall" on the star.

I can see this from this graph if I think about the fact that kinetic energy cannot be negative. enter image description here

But I still do not get the physical meaning of $U_{centrifugal}$. To what extent is $U_{centrifugal}$ to be considered a potential term and how does it prevent the orbiting planet to meet the star, mathematically, from $(1)$?

Qmechanic
  • 201,751
Sørën
  • 2,577
  • 4
  • 39
  • 95
  • 1
    This question (v2) is essentially a duplicate of http://physics.stackexchange.com/q/83190/2451 and http://physics.stackexchange.com/q/37968/2451 . The centrifugal potential $V_{\rm cf}=\frac{L^2}{2\mu r^2}>0$ is positive, and it is minimized for large radial position $r\to \infty$, as one would expect. – Qmechanic Apr 09 '16 at 15:17
  • 2
    If you are in a rotating frame then you "feel" a centrifugal force $F= \dfrac {mv^2}{r}$ which in terms of angular momentum $L$ which is conserved is $F= \dfrac{L^2}{mr^3}$. Now doing the usual potential energy calculation for the centrifugal force $\int_\infty^r Fdr = +\dfrac {L^2}{2mr^2}$ which says that work has to be done against the centrifugal force to bring the mass in from infinity. Adding this to the negative gravitational potential energy gives you a minimum of potential energy and hence the possibility of a stable orbit. – Farcher Apr 09 '16 at 15:36
  • 1
    The "duplicate" link is unnecessarily complicate to answer this simple question. There is a nice video about this on YouTube. https://www.youtube.com/watch?v=koXXTu39Qpc – Farcher Apr 09 '16 at 15:39

0 Answers0