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$$E = \frac { 1 } { 2 } m v _ { r } ^ { 2 } + \frac { L ^ { 2 } } { 2 m r ^ { 2 } } + U ( r )$$

This is the formula of total energy of an object with mass $m$ orbiting an assumed stationary body of mass $M$. The velocity is broken down into 2 components: tangential and radial velocity ($v(\phi)$ and $v(r)$ respectively). Plug in the value of $v(\phi)$ in term of the angular momentum and you will arrive at the formula. $U(r)$ is the potential energy of the object based on the gravity force exerted on it. When you integrate $U_{eff}$ which is the effective potential (sum of the last 2 terms), you will have gravity force and centripetal force. I dont understand this because centripetal force is supposed to be a net force. And since there is only gravitational force in play, wouldn't $Fg = Fc$? That would mean $$U_{eff} = 2Fg = 2Fc$$ The integrated form of the first term should be centrifugal force which has the same magnitude of the centripetal force but of different direction. So $E= 3Fc$ and the only force acts upon the orbiting object is the gravity force between the 2 masses since the centrifugal and centripetal force cancels one another out? But wouldnt this make one mass spiral into the other?

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  • $\begingroup$ @AaronStevens That's correct, because there is no work done in those instances. $\endgroup$ – Zack Hutchens Jan 4 at 15:12
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When you integrate $U_{eff}$ which is the effective potential (sum of the last 2 terms), you will have gravity force and centripetal force.

First, $\mathbf F=-\nabla U$, so you would not integrate the potential energy to get the force. The effective force is given by $$\mathbf F_{eff}=\left(\frac{L^2}{mr^3}-\frac{GmM}{r^2}\right)\mathbf{\hat r}$$ Second, the term $\frac{L^2}{2mr^2}$ represents the energy from a centrifugal force, not a centripetal force. Of course in an inertial reference frame there is only the force of gravity, but the problem becomes easier to work with using an effective potential. This is because we are working in polar coordinates. Or you can view it as working in a rotating frame rotating at a rate of $\dot\phi$.

And since there is only gravitational force in play, wouldn't $F_g = F_c$?

This would only be true in a circular orbit, or at least where the position and velocity vectors are orthogonal. Or using the effective force, this is when $\mathbf F_{eff}=0$. However this is not always true for general elliptical orbits. In general orbits these two terms alternate between which one is larger than the other.

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  • $\begingroup$ Yes it should have been differentiation not integration. What type of force would I receive when I differentiate the first term in the formula? $\endgroup$ – Jung Jan 4 at 17:35
  • $\begingroup$ @Jung It's the centrifugal force $\endgroup$ – Aaron Stevens Jan 4 at 19:26
  • $\begingroup$ so that means both term $\frac{{L}^{2}}{2m{r}^{2}}$ and $\frac { 1 } { 2 } m v _ { r } ^ { 2 }$ originate from centrifugal force? $\endgroup$ – Jung Jan 4 at 22:22
  • $\begingroup$ But $\frac { 1 } { 2 } m v _ { r } ^ { 2 }dr$ would be equal to 0 $\endgroup$ – Jung Jan 4 at 22:28
  • $\begingroup$ @Jung Oh sorry I thought you meant the first part of $U_{eff}$. The $\frac12mv_r^2$ is a kinetic energy term. The $\frac{L^2}{2mr^2}$ is centrifugal. $\endgroup$ – Aaron Stevens Jan 5 at 7:37

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