# Kepler's third law for elliptical orbits

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1. Feb 21, 2016

### DanHelzer

Kepler's third law states T^2=(4pi^2/GM) x r^3 for CIRCULAR orbits. My question is, in the derivation for this equation ma=GMm/r^2 why can centripetal acceleration be used to replace a at m(v^2/r)=GMm/r^2 yielding v^2/r=GM/r^2 when the orbit is not circular. Planets have elliptical orbits so why can centripetal acceleration be used so that for elliptical orbits the equation for a period is (r=a) --> T^2=4pi^2/GM) x a^3.

2. Feb 21, 2016

### Staff: Mentor

m(v^2/r)=GMm/r^2 assumes that the radial acceleration is zero - the gravitational force is exactly the centripetal force necessary to keep the object in a circular orbit. That assumption is not true any more if the orbit is not circular.

3. Feb 21, 2016

### SteamKing

Staff Emeritus
The eccentricity of most planetary orbital paths in the solar system is so low that their orbits are almost circular. Mercury and Pluto do not have orbits which can be approximated so.

https://en.wikipedia.org/wiki/List_of_gravitationally_rounded_objects_of_the_Solar_System

4. Feb 23, 2016

### DanHelzer

So then why can r be replaced with a if the orbits are not circular? mv^2/r, as you stated, applies to a circular orbit. How then is it possible to simply replace "r" with "a" in the law of periods equation of Kepler's third law?

5. Feb 23, 2016

### DanHelzer

6. Feb 23, 2016

### SteamKing

Staff Emeritus
For most orbits, the eccentricity is negligible.

In the general case for an elliptical orbit, the velocity and radius of a planet from the central body, in this case the Sun, changes with time. At each instant during the orbit, the centripetal force on each planet is balanced exactly by the gravitational force exerted by the Sun on the planet. That's why the planet doesn't go flying off into the interstellar void or come crashing into the Sun.

https://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

7. Feb 24, 2016

### mpresic

The eccentricity that "most" orbits demonstrate is a red herring. I believe what DanHelzer is getting at is : Can we justify using T^2=4pi^2/GM) x a^3 for elliptical orbits (easily obtained in the special case of a circular orbit with the radius r identical to the semimajor axis a) for the general (elliptical) case by just substituting the semimajor axis a for the radius r? (I read this as his question addressed in his 1:02 post)

It is true that the equation for the period squared, T^2=4pi^2/GM) x a^3 is valid for any motion in an inverse square field, (whether the eccentricity is small (negligible) or not), so long as the eccentricity is less that 1 (i.e. elliptic orbit). That is, the equation above is true for all bound orbits in the inverse square field. I think it is OK to use the short-cut of using T^2=4pi^2/GM) x a^3, for the elliptic orbit case, if you are in a hurry.

In so far as using the centripetal force. This may be only my interpretation, but I think of centripetal force as "inward seeking" force which is most useful when the tangential acceleration is zero, and the radial acceleration tells the whole story. In an elliptic orbit, only two points meet this criterion. Namely these two points are apogee and perigee (for the Earth); aphelion and perihelion (for the Sun). However the analysis at these two points may be sufficient: It is evident that the distance from Sun to Planet at apogee added to the Sun to the Planet at perigee divided by two is the semi-major axis a, and considering the centripetal force (alone) may be an adequate justification for the replacement of r with a, above.

However, I think it is more proper to examine and if possible understand and reproduce the formal techniques introduced in the upper undergraduate and graduate level textbooks for a complete justification. This strongest justification involves the conservation of angular momentum, and energy to demonstrate the orbits in a inverse square field is an ellipse where the energy is a function of G, the product of the masses involved, and the semi-major axis, alone. This is done in Symon, Marion and Thornton, Goldstein. I find the formal justification from these textbooks as most convincing.

8. Feb 24, 2016

### Staff: Mentor

You can derive Kepler's laws in different ways, some of them work for elliptic orbits as well so the problem I mentioned does not apply there.
The fact that the period depends on the semi-major axis but not on the eccentricity is highly non-trivial.

9. Feb 24, 2016

### Janus

Staff Emeritus
You have to take into consideration a couple of things.
First, the total energy (kinetic energy+ gravitational potential energy) of any orbit is a constant and at any point of the orbit can be found with:

(1)$$E = \frac{mv^2}{2}-\frac{GMm}{r}$$

Where m is the mass of the orbiting body (and in this case, much smaller than M)
Thus if you take the above formula and use the perihelion distance (rP) and aphelion distance (rA) for r, you get two equations that equal each other, with this, and knowing that rP+rA= a(the semi-major axis), you can work out that the total energy of an elliptical orbit can also be found by
(2)$$E= -\frac{GMm}{a}$$
And by equating eq(1) and eq(2), we can solve for v:
(3)$$v = \sqrt{GM \left ( \frac{2}{r}- \frac{1}{a}\right )}$$

Another aspect of elliptical orbits is that they sweep out equal volumes in equal times(Kepler's 2nd law)
Now the "Areal" velocity(the "area" it sweeps out at any given instant) of the object at any point of its orbit is.
$$v_A = \frac{rv \cos \phi}{2}$$
$\phi$ is the angle between the object's velocity vector and the line perpendicular to the line joining the object and the focus of the orbit.
This, due to Kepler's 2nd law is also a constant throughout the orbit, Thus we can get the Areal velocity for all points of the orbit by finding it at any point of the orbit
Two particular points of interest are aphelion and perihelion, where $\phi=0$ and thus $\cos \phi = 1$, so it makes sense use one of these points as we continue.
For any orbit of eccentricity e, the perihelion distance is equal to
(4)$$r_P=a(1-e)$$
Thus at perhelion:
(5)$$v_A = \frac{a(1-e)v_P}{2}$$
Where vP is the velocity at perihelion
And since this is a constant and the orbit sweeps out the entire area of its orbit in one orbital period and area of an ellipse is:
(6) $$\pi a^2 \sqrt{1-e^2}$$
We can find the period of the orbit with by dividing the area of the ellipse by the Areal velocity.
(7) $$T=\frac{\pi a^2 \sqrt{1-e^2}}{\frac{a(1-e)v_P}{2}}$$
(8) $$T=\frac{2\pi a \sqrt{1-e^2} }{(1-e)v_P}$$
(9)$$T=\frac{2 \pi a \sqrt{\frac{1+e}{1-e}}}{v_P}$$

To find vP we go back to eq(3) and substitute a(1-e) for r and reduce:
(10) $$v_P = \sqrt{\frac{GM}{a} \frac{1+e}{1-e}}$$
(11) $$v_P = \sqrt{\frac{GM}{a}} \sqrt{ \frac{1+e}{1-e}}$$
Substitute for vP in eq(9)
(12)$$T=\frac{2 \pi a \sqrt{\frac{1+e}{1-e}}}{ \sqrt{\frac{GM}{a}} \sqrt{ \frac{1+e}{1-e}}}$$
(13)$$T=\frac{2 \pi a}{ \sqrt{\frac{GM}{a}}}$$
(14)$$T= 2 \pi \sqrt{\frac{a^3}{GM}}$$

As you will note, e drops out of the equation completely leaving only a, and the equation is the same as for a circular orbit with a for r. (technically, a circle is an ellipse with an eccentricity of 0, and in this case, the semi-major axis is equal to the radius)

10. Feb 24, 2016

### DanHelzer

Thank you all for the responses. I am in high school (AP physics) and its no wonder why I do not understand this if you're pointing me in the direction of graduate books.

11. Feb 25, 2016

### mpresic

As you can see DanHelzer, physics is a intriguing subject, when a first year college level physics question can generate so many involved answers. My hat's off to you for looking beyond the presented subject matter in AP physics which probably only considers circular orbits in detail