# Kepler's laws to determine radius of circular orbit

"In nonrelativistic limit of general relativity there is a correction to the Newtonian gravitational potential energy $−h/r^3$ with $h = αL^2/(mc)^2$, where $c$ is the speed of light, $α = GMm$ and $L$ is the angular momentum"

Using this knowledge, I'm supposed to find the radius of circular orbits for a given $m$ and $L$ and decide which of them is stable.

My question has to do with how I can actually determine the radii, but bear with me as I show my process thus far:

I e-mailed my professor and was told that I must SUBTRACT this correction factor from gravitational potential energy, which gives me:

$$V(r) = \frac{-GMm}{r} - \frac{GMmL^2}{r^3(mc)^2}$$

I can find the effective potential to be:

$$V_\text{eff}(r) = \frac{L^2}{2mr^2}-\frac{GMm}{r} - \frac{GMmL^2}{r^3(mc)^2}$$

Based on information in my textbook, I'd imagine that I must graph the effective potential and the straight line of my constant energy $E$, and the two radii will be the points where the energy line intersects the $V_\text{eff}(r)$ curve. My problem arises when I try to graph it.

If I draw a qualitative graph simply by using $\frac{1}{r^2} - \frac{1}{r} - \frac{1}{r^3}$ I get a curve with no apparent extrema that approaches 0 from the $-y$ axis as $r$ goes to infinity. With other graphs I've drawn using $V_\text{eff}$, I've gotten curves that make sense - I see a local maximum or minimum, and I assume that the planet could stay "trapped" between the two "walls" of the minimum. In this case, of course, I see none of that.

My question is (hopefully) a lot more general than just this example, "How do I find the potential radii of the orbiting planet using Kepler's Laws?" Unless I've made a mistake in my process, I don't believe I can find them using this method. I'd imagine I could find them with a lot of calculus and rearranging, but I'm sure there must be a simpler way.

• What's the $1/r^2$ term for? Commented Mar 23, 2015 at 22:13
• $V_{eff}(r) = \frac{L^2}{2mr^2} + V(r)$, so the $1/r^2$ term comes from $\frac{L^2}{2mr^2}$!
– Alex
Commented Mar 23, 2015 at 22:16
• when we say the "non relativistic limit" of general relativity, this implies that the $1/r^3$ is much smaller than the other two terms for this approximation to be valid Commented Mar 23, 2015 at 22:55
• Try plotting $r^{-2}-r^{-1}-0.01\cdot r^{-3}$ instead. The correction is small. $c^{-2}$ is small.
– user12029
Commented Mar 23, 2015 at 23:15
• $1/r^2$: Aha. Your OP had a typo ... Commented Mar 23, 2015 at 23:47

Your equation has the form $$V_\text{eff}(r) = \frac{\alpha}{r^2} - \frac{\beta}{r} - \frac{\gamma}{r^3}$$ If you set $\alpha=\beta=\gamma=1$, then you're overestimating the $r^{-3}$ term, which is supposed to be a small correction. You will only find two extrema if the derivative has two roots: $$V_\text{eff}'(r) = -\frac{2\alpha}{r^3} + \frac{\beta}{r^2} + \frac{3\gamma}{r^4} = 0$$ which implies $$\beta r^2 - 2\alpha r + 3\gamma = 0$$ This equation has the discriminant $$\Delta^2 = 4(\alpha^2 - 3\beta\gamma)$$ So $V(r)$ has two extrema if $$\alpha^2\geqslant 3\beta\gamma$$ which is true if $\gamma$ is sufficiently small.

• Thanks a lot for helping me out here! That brings me to my second question, how to find the radii. Since the orbits are circular, $r_{min} = r_{max}$ so they would just be at the extrema, right? Is there a simpler way to solve this? I've had no luck graphically since the values are so big and the formula is too complicated to yield a nice derivative equal to zero.
– Alex
Commented Mar 24, 2015 at 0:18
• @Alex Yes, the radii are at the extrema. Just solve $V_\text{eff}'(r)=0$. Commented Mar 24, 2015 at 0:38

Finding the potential radii is actually quite simple. I already have: $$V_{eff}(r) = \frac{L^2}{2mr^2}-\frac{GMm}{r} - \frac{GMmL^2}{r^3(mc)^2}$$

I was mistakenly graphing $1/r^2 - 1/r - 1/r^3$, when in actuality it makes more sense to take $1/r^2 - 1/r - 0.1/r^3$, since $1/r^3$ is sufficiently small.

Since the orbits are circular, the potential radii will be at the extrema of the function $V_{eff}(r)$, so I simply need to take:

$$V'_{eff}(r) = 0$$ $$\frac{-L^2}{mr^3}+\frac{GMm}{r^2} + \frac{3GMmL^2}{r^4(mc)^2} = 0$$ Multiplying each side by $r^4$, I can see that I have an equation similar to: $$ar^2 + br + c = 0$$ So I can use the quadratic formula to get: $$r = |\frac{-L^2/m +- \sqrt{L^4c^2 - 12G^2M^2m^2L^2}/mc}{2GMm}|$$ Referring to my graph, I can see that the first of these radii will be unstable because the energy is greater than it so it will go to infinity, whereas for the second (larger) radii, the total energy is greater than the local minimum so the planet/particle will be trapped in the potential well, so it is stable.

Substituting accepted values for G, M, m, c and L into the equation which yields a larger r (where I subtract at the +- sign), I get the actual radius from the earth to the sun, ~$1.5*10^8 km$!