# 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

### 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

### 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

### 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
10. Jan 16, 2018

### 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?

11. Jan 21, 2018

### JohnnyGui

UPDATE: Ok I learned that one formula represents the relative velocity between the 2 masses and the other one the velocity with respect to the center of mass. I was able to derive the one from the other since the relative velocity w.r.t. the masses is the sum of both velocities of $m$ and $M$ w.r.t. the center of mass.

I am now trying to find a way how one can derive the period of a mass in an elliptical orbit. I think I have found a way to deduce this.

I learned that one can calculate the velocity in an elliptical orbit, as a function of distance between the masses $r$ and the semi-major axis $a$ of the ellips, as follows:
$$v = \sqrt{GM(\frac{2}{r} - \frac{1}{a})}$$
I took a first step and thought of a very large $M$ w.r.t. $m$ so that $M$ is basically in one of the focus points of the ellips ($r$ is thus the distance from $m$ to the focus point). I think that it's possible to calculate the period of $m$ by calculating a small piece of the length of an ellips $dL$ and using the velocity formula to know how much time the mass $m$ would do over that piece of $dL$. Doing this for every piece $dL$ of the ellips and adding all those time intervals would give the total period for $m$.

Since the graph formula of half an ellips is $y=\sqrt{b^2(1-\frac{x^2}{a^2}}$, I figured that the distance from $m$ to the focus point $r$ is the hypotenuse of a triangle in the graph with its 2 short sides $y$ and $[ae - x]$ where $e$ is the eccentricity. In that case, $r$ for e.g..the domain $x ≥ 0$ is then calculated with Pythagoras theorem as:
$$r=\sqrt{y(x)^2 + [ae - x]^2}$$
$$x ≥ 0$$
I reckon that calculating two $r$'s (e.g. $r_1$ and $r_2$) merely differing from each other by $dr$, and then finding the x,y-coordinates of their intersections with the ellips, would get you a triangle with sides $dy$ and $dx$ and its hypotenuse $dL$ which is a small length piece of the ellips. By using the Pythagoras theorem this would mean that $\sqrt{dy^2 + dx^2} = dL$.
Then, using the velocity formula above for an ellips, one can calculate how long $m$ would to over that piece of $dL$ at a distance $r$ that is in between $r_1$ and $r_2$ (half of $d\theta$ subtended by $r_1$ and $r_2$ from the focus point). Doing this for each piece of $dL$ of half an ellips for the domain $-a ≤ x ≤ a$, adding their corresponding time intervals, and then multiplying that total time duration by 2 (to get the time duration for the other half of the ellips) would give the period of mass $m$.

This is probably a very inconvenient way to calculate the period of $m$. My question is though, would one theoretically deduce the correct period of mass $m$ with my procedure if $M$ is almost at an ellips focus point? (very large $M$)?

Last edited: Jan 21, 2018
12. Jan 23, 2018

### JohnnyGui

I have made an illustration to clarify my idea how one could calculate the circumference of an ellips.

The x and y coordinates of every distance $r$ and $(r - dr)$ from the focus to any point on the circumference of the ellips can be calculated using the formula for the ellips $y = \sqrt{b^2(1-\frac{x^2}{a^2})}$ and Pythagoras theorem. The difference in x- and y coordinates $dx$ and $dy$ can then be used to calculate a piece of the circumference $dL$ through Pythagoras theorem again. I am exaggerating the magnitude of $dL$ in the picture above.

Dividing each calculated $dL$ by the formula for the speed $v = \sqrt{GM(\frac{2}{r} - \frac{1}{a})}$ in which $r$ is the midpoint between $r$ and $r - dr$ would give the small time interval over $dL$. Doing this for each piece of the circumference of the ellips and adding all those time intervals together would give the time duration for mass $m$ to travel half the ellips as shown in the picture. So multiplying this by 2 would give the period of $m$. This is again if $M$ is so large that it is basically in the focus point of the ellips (green dot in the picture).

I was wondering if this is a correct technique to calculate the correct time period for a mass $m$. My technique sounds like a summation instead of an integration but I'm not sure regarding this.

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