# I Some confusion about Kepler’s formula

1. Jan 10, 2018

### JohnnyGui

Hello,

If I understand correctly, according to Kepler, a planet orbiting a star should cover a constant surface area of its ellipsoid or circular orbit per time unit, regardless of the distance of that planet to the star.

I have tried to prove this by using a scenario in which a planet is orbiting in a perfect circle around a star, so that the distance is the radius $r$. I would expect that changing the distance $r$ would still make the planet cover a constant area of the circle per time unit through changing its orbital velocity.

The time period $P$ in which this planet orbits a full circle around the star is defined by:
$$\sqrt{\frac{4\pi^2\cdot r^3}{G(M+m)}}=P$$
Dividing the circumference of the circle $2\pi r$ by that formula would give the orbital velocity per second. Divding that orbital velocity again by the circumference of the circle would give the fraction of the circumference that the planet covers per second. Multiplying that fraction by by the surface area of the circle $\pi r^2$ would give the surface area that the planet orbits per second.

Rewriting all this in the formula and simplifying would eventually give me:
$$\sqrt{\frac{r \cdot G(M+m)}{4}} = A$$
In which $A$ is the covered surface area per second. This formula still shows that $A$ is dependent on the distance $r$. From what I understand, this shouldn’t be the case since a planet revolving around a star should cover a constant area regardless of distance.

I have a feeling that this "constant area per unit time" rule only applies within the orbit of the planet itself, not when changing its distance. Within a circular orbit, $A$ would indeed be constant.
Furthermore, if the orbit of a planet $1$ is an ellips with a semi-major axis $a_1$, then the planet only sweeps equal areas per time unit within that particular ellips with that semi-major axis $a_1$. An equally massive planet $2$ orbiting in an ellips but has a different semi-major axis $a_2$ would not sweep the same area per time unit as planet $1$, but it will sweep equal areas within its own elliptical orbit with semi-major axis $a_2$.
Is this the cause of my misunderstanding?

2. Jan 10, 2018

### Orodruin

Staff Emeritus
It will cover a constant area, but this area does not need to be the same as that of the lower orbit. The area covered generally depends on the angular momentum and Kepler's third law ultimately boils down to conservation of angular momentum. If you have two orbits with different angular momentum, the area covered per unit time will be different.

For a circular orbit the third law is a trivial statement just saying that the orbital velocity remains constant.

This is a misunderstanding of Kepler's third law. It does not say that this area is independent of the orbit. It says that given a particular orbit, the area covered per time will be constant.

Essentially, yes.

3. Jan 10, 2018

### stefan r

picture from wikipedia:

4. Jan 11, 2018

### JohnnyGui

@Orodruin : Thanks for your explanation. I know now where my misunderstanding was. @stefan r : Thanks for the illustration.

One other question; I know that 2 massive objects, e.g. a star and a planet, orbit around a center of mass, the center of mass almost being in the center of the star if that star is way more massive than the planet. In case of an elliptical orbit, that center of mass is in one of the 2 foci of the ellips. So the true distance between the planet and the star is not actually the same as the distance between the planet and the focus of the ellips.
Does the planet sweep equal areas per unit time with respect to its distance to the star or with respect to its distance to the center of mass? I assume it's the latter?

5. Jan 11, 2018

### jbriggs444

Right. The two bodies orbit the combined center of mass in lock-step, each in its own elliptical path. The mutual attraction always has a line of action through the center of mass, so angular momentum of either about the center of mass is conserved and Kepler's law upheld.

6. Jan 12, 2018

### stefan r

7. Jan 13, 2018 at 1:36 PM

### JohnnyGui

Got it. Thanks for the confirmation.
Thanks again @stefan r for the helpful illustrations and info page!

How about the formula for orbital velocity $\frac{G(M+m)}{r}=v^2$ and gravitational force $\frac{G \cdot M \cdot m}{r^2} = F$.
Is it correct that the $r$ in both formulas represent the true distance between the 2 masses and not the distance to the center of mass?

8. Jan 15, 2018 at 12:18 AM

### stefan r

That statement reads ambiguous. You have, for example, the center of Pluto's mass, the center of Charon's mass, and the center of the Pluto-Charon system's mass.

$r$ is the distance between "2 point masses" acting on each other. The gravity of a sphere(or spherically symmetric solid) will act like the gravity of a point mass.

$r$ is not the distance between the closest point on the surface of 2 planets. $r$ is the distance between the center of mass of object #1 and the center of mass of object #2.

9. Jan 15, 2018 at 7:52 PM

### JohnnyGui

Apologies, with "true distance" I meant the distance between the centers of 2 masses. I'm leaving the barycenter out of this.

If $r$ is the distance between 2 point masses like you say, then in case of a circular orbit like the following in which the masses don't differ too much from each other...

...then the constant of Kepler's formula $\frac{p^2}{r^3}$ would be different from when $M$ is way more massive than $m$, correct?
This is because the distance $r$ between the 2 masses would have to be transformed into the distance between mass $m$ to the barycenter, $r_{bc(m)}$ in order to calculate its period $p_m$ correctly, by dividing the circumference of its circular orbit with radius $r_{bc(m)}$ by its orbital speed $v_m$.

So I'd reckon that in case of a circular orbit between mass $M$ and mass $m$ that don't differ a lot from each other (like in the above illustration) you'd have to calculate the distanc between mass $m$ to the barycenter $r_{bc(m)}$ first from $r$, in order to deduce Kepler's constant correctly from Newton's centrifugal force formula.

I deduced that $r_{bc(m)}$ can be calculated through $r$ by:
$$r_{bc(m)} = r \cdot \frac{M}{M+m}$$
This means that the period $p$ of mass $m$ should be its circular orbit's circumference with radius $r_{bc(m)}$ divided by its orbital speed calculated through Newton's centrifugal force formula:
$$p_m = \frac{2\pi \cdot r \cdot \frac{M}{M+m}}{\sqrt{\frac{G(M+m)}{r}}}$$
This eventually gives me:
$$\frac{p_m^2}{r^3} = \frac{4\pi^2 \cdot (\frac{M}{M+m})^2}{G(M+m)} = \frac{4\pi^2 \cdot M^2}{G(M+m)^3}$$
Here, $r$ is the distance between the centers of the 2 masses. Is this formula then correct?

Last edited: Jan 15, 2018 at 8:26 PM
10. Jan 16, 2018 at 3:52 PM

### JohnnyGui

Please scratch what I said above, I didn't realise that Newton's formula for orbital velocity was already corrected to get the distance to the barycenter So I kind of double corrected it.

However, I have another problem. I verified that $r \cdot \frac{M}{M+m} = r_{bc(m)}$.
The equation between gravitational force and centripetal force in case of $M$ being way larger than $m$ is:
$$\frac{GMm}{r^2} = \frac{mv^2_m}{r}$$
Here, the $r$ is the distance between $M$ and $m$. If these 2 masses don't differ a lot from each other, the $r$ in the right equation needs to be converted to the distance of $m$ to the barycenter; $r_{bc(m)}$. Substituing $r$ with $r \cdot \frac{M}{M+m}$ gives:
$$\frac{GMm}{r^2} = \frac{m \cdot v^2_m (M+m)}{r \cdot M}$$
Rewriting it to get $v_m$ gives me:
$$\sqrt{\frac{GM^2}{r(M+m)}} = v_m$$
This conclusion is correct according to this link . However, this link shows a different relationship for calculating $v_m$, saying that $\sqrt{\frac{G(M+m)}{r}} = v_m$.

Which one is correct?