###### By Joshua Winn, Princeton University

## We can explain Kepler’s laws if we assume the Sun is pulling on planets with a force whose strength varies as one over *r*-squared. We can discover the law of gravity, by following Kepler’s laws. Let’s think about the two big conservation laws: the conservation of angular momentum, *L*, and energy, *E*.

### Formulating a New Equation

Both *L* and *E* remain constant throughout a planet’s elliptical orbit, even while the planet is moving and changing speed. So, we should be able to derive expressions for *L* and *E* purely in terms of constants: *G*, big-*M*, little-*m*, *a*, and *e*. First, let’s do it for angular momentum. In general, *L* equals *m *times* r *times* v_theta _{.}* Remember, only the angular component of the velocity matters. Since

*v_theta*equals

*r*

*d-theta*/

*dt*, we can also write

*L*as

*mr*-squared

*d-theta*/

*dt*.

If we consolidate Kepler’s second and third laws into one equation, *r*-squared *d-theta/dt* equals the square root of *Ka* times one minus *e*-squared. If we multiply this by little-*m*, and substitute *GM* for *K*, we arrive at a new formula for angular momentum: little-*m* times the square root of *G* times big-*M *times *a *times one minus *e*-squared.

As an immediate application of this formula, we can prove Kepler’s third law for the general case of an elliptical orbit. We start with Kepler’s second law: 1/2 *r-*squared *d-theta*/*dt* equals *pi* *a*-squared times the square root of one minus *e*-squared (that’s the area of the ellipse), divided by *P*, the orbital period. Notice the left side of this equation is the angular momentum, *L*, divided by 2*m*.

Now, let’s use our nifty new formula for *L*. When we insert that, the little-*m*’s cancel out, as do the one minus *e*-squared’s, and if we solve for *P*, we find that it’s 2*pi* over the square root of *G* time big-*M* times *a *to the 3/2, which is, of course, Kepler’s third law. Thus, endeth the proof.

This article comes directly from content in the video seriesIntroduction to Astrophysics. Watch it now, on Wondrium.

### Energy Has Two Parts

That’s enough playing around with angular momentum. How about energy, the other conserved quantity? Energy has two parts: kinetic, 1/2 *mv*-squared, and potential, minus-*G *times big-*M *times little-*m* over *r*. Their sum must be equal to some combination of the constants *G*, big-*M*, little-*m*, *a*, and *e*, and let’s try to figure out what it is.

Since energy is constant, we can calculate it at any point we want in the planet’s orbit, and we’ll get the same answer, so let’s make life simple by choosing *theta *equals zero. That’s when the planet makes its closest approach to the Sun, and *r* equals *a* times one minus-*e*. What about the velocity? Well, we can figure that out with another application of our new angular momentum formula.

In general, *L* is equal to *m* times *r* times *v*_*theta*. Here, at *theta *equals zero, *v*_*theta* is simply *v*, because at that point the velocity vector is totally perpendicular to the radius vector: *r* is in the *x*-direction and *v* is in the *y*-direction. So, at *theta *equals zero, *L* equals *m *times *a* times one minus-e times *v.* We solve for *v*, and plug in our new expression for *L*. Then, we insert that expression for *v* into the energy equation, and we simplify. The algebra leads to a cascade of cancellations, and a result that’s refreshingly simple. The energy is minus-*G *times the product of the masses divided by 2*a*.

### What Does the Energy of a Planet’s Orbit Depend On?

All the terms related to eccentricity ended up canceling out. It turns out that energy depends only on the semimajor axis of the ellipse, not its eccentricity. If you have a nearly circular orbit with radius 1 AU, like the Earth, and you compare it to a planet on a highly elliptical orbit, with *a *equals 1 AU and an eccentricity of 0.9, they both have the same energy.

They’ll also have the same orbital period, one year, because Kepler’s third law says *P* depends on *a*, but not on *e*. The planet in the elliptical orbit whips around the Sun near its closest approach and moves more slowly when it’s far away, and the 2 effects cancel each other exactly to give the same period as the Earth. It’s an interesting coincidence.

### Common Questions about How to Calculate the Energy of a Planet’s Orbit

**Q: How can we derive expressions for L and E purely in terms of constants to calculate a planetary orbit’s energy?**Both the L and E of a planet’s orbit stay the same through the planet’s elliptical orbit. Even when the planet is moving and changing speed.

**Q: How can we find out a planetary orbit’s energy simply?**Through a planet’s orbit, the sum of its kinetic and potential energy must stay the same. So we can pick any position of the planet’s orbit we want and when we calculate it’s energy we’ll know its the same in all parts of the orbit. To make things simpler its best to choose theta equals zero. That’s when the planet makes its closest approach to the Sun, and r equals a times one minus-e.

**Q: What does the energy of a planet’s orbit depend on?**The energy of a planet’s orbit depends solely on the semimajor axis of the ellipse, not its eccentricity. So a nearly circular orbit and a highly elliptical orbit may have the same energy. If they have the same energy, they can also have the same orbital period.