# Deriving energy for elliptical orbit

So I wanted to derive the total energy for an elliptical orbit, $$E = -GmM/2a,$$ and while I was doing it, I ran into this hurdle. So at the closest point to the focus, the orbiting object is at a distance of $$a(1-e)$$ from the focus, where a is the semi-major axis and e is the eccentricity. So if we were to take the centripetal force at this point we should get $$\frac{mv^2}{r} = \frac{GmM}{r^2}$$ and $$r = a(1-e)$$ so then we would get $$\frac{1}{2}mv^2 = \frac{GmM}{2a(1-e)}$$ which is the kinetic energy. If we were to add this to the potential energy $$U = -\frac{GmM}{a(1-e)},$$ we get the total energy as $$E = -\frac{GmM}{2a(1-e)}.$$ Isn't this wrong because the total energy of an ellipse is $$E = -GmM/2a$$? Why did I end up with a total energy not equal to that?

I asked my teacher about this and she said that we can only use mv^2/r for circular orbit but at the closest point in the ellipse, isnt the force perpendicular to the velocity so there shouldn't be any tangential acceleration, and therefore we can use the centripetal force equation?

The expression you use for centripetal force, $$F = mv^2/r,$$ is only valid for circular orbits. For a planet in an elliptical orbit at its closest approach to the star it is orbiting, it will have a higher speed than that given by equating the centripetal and gravitational force. The planet will start moving farther away from the star after its closest approach, which means the star's gravity is not strong enough to hold the planet to a constant distance--that is, a circular orbit. So, the gravitational force must be weaker than $$mv^2/r.$$
• Thank you very much. But what I still don't fully understand is what exactly the radial component of the force would be in this equation? I always thought that the radial component was mv^2/r and the tangential component of the acceleration was m$\alpha$r. Dec 2, 2022 at 4:00
• @AnandatheerthaBapu Since the planet is at its closest distance from the star, it is moving perpendicular to the line joining the planet and star. So, there is zero tangential force; gravity pulls purely radially. The only force is gravity, so that radial component must be $GMm/r^2.$ Because this force is less than $mv^2/r,$ the planet will move in a wider arc than a circle with a radius of $r.$ That's why it will start to move farther from the star. Dec 2, 2022 at 6:36
• @AnandatheerthaBapu At the closest distance, $GMm/r^2 < mv^2/r,$ so the planet starts moving away from the star. At the farthest distance, $GMm/r^2 > mv^2/r,$ so the planet starts falling back towards the star. Dec 2, 2022 at 6:39
It is not the expression for acceleration which is not valid. The expression for radial acceleration $$a_{r}= \frac{v^2}{r}$$ is always valid but the radius in the denominator is the radius of curvature of the trajectory and not the distance between the moving object and the center of force. For a circular trajectory, the radius of curvature is the same at any point and equal to the radius of the circle. But this is not so for other curves. For an ellipse, the radius of curvature changes from a minimum value at the two ends of the major axis to the maximum value at the two ends of the minor axis. You can tell even by looking at the curve that is "less curved" at the points intersecting the minor axis so the radius is larger. The radius of curvature in the centripetal acceleration and the radius "r" in the expression of the force are not the same thing. For the point where you do the calculation, the radius of curvature (the minimum value) is given by $$R=\frac{b^2}{a}$$ or in terms of eccentricity $$R=a(1-e^2)$$. The distance between the planet and the star is $$r= (1-e)a$$. So, Newton's law will provide: $$\frac{mv^2}{R}=\frac{GmM}{r^2}$$ The kinetic energy is then $$KE=\frac{GmM(1+e)}{2(1-e)a},$$ the potential energy is $$PE=-\frac{GmM}{r}= -\frac{GmM}{(1-e)a}$$ and the total is what you expected: $$-\frac{GmM}{2a}$$