Going through Kepler's Orbit Derivation.. having a problem

I imagine this was a simple task for many of you during undergrad studies, but I am a lowly engineer attempting to derive Kepler's Law of Orbits. I've tried to do it as cleanly as possible. I've omitted a lot of the derivations for the stuff involving spherical coordinates, angular momentum not changing with respect to time, dot product/cross product identities, etc.

Note that I'm using little m as reduced mass. In my head, I am just thinking of them as the sun (M) and earth (m). Given that the earth has little bearing on the orbit of the sun.

I'm using r hat and theta hat for the respective directions in polar coordinates.

The one problem I am having with the derivation is regarding the constant of integration that is yielded when taking the anti-derivative of v cross L with respect to time. It seems rather important that it should equal the eccentricity times cosine of the angle. But I have no idea where this reasoning comes in to play. If anyone has any advice that would be greatly appreciated.

I've based the derivation on this HyperPhysics page: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/keplerd.html

They have this sentence on the HyperPhysics page: "where we can argue that the vector constant of integration D must be in the plane of the orbit since the other two quantities lie in that plane." I have no idea how they got a Dcos(theta) out of this, since the overall direction of the integrated quantity is in the r-hat direction.

$$\vec{L}\cdot\vec{L}=L^2$$

$$(m\vec{r} \times \vec{v})\cdot \vec{L}=L^2$$

$$\vec{r}\cdot(\vec{v} \times \vec{L}) = L^2 /m$$

Find v cross L

$$\frac{d}{dt}(\vec{v} \times \vec{L})=\vec{a} \times \vec{L}$$

Since L is constant with respect to time

$$\vec{a} \times \vec{L} = m(\vec{a} \times \vec{r} \times \vec{v}) = m((\vec{a} \cdot \vec{v})\vec{r} - (\vec{a} \cdot \vec{r})\vec{v})$$ $$\vec{a} = a\hat{r}, \vec{r}=r\hat{r}, \vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$$ $$\vec{a} \times \vec{L} = -mar^2\dot{\theta}\hat{\theta}$$ Where, $$\vec{a} = \frac{-GM}{r^2}\hat{r}$$ $$\vec{a} \times \vec{L} = +GMm\dot{\theta}\hat{\theta}$$ $$\frac{d}{dt}(\vec{v} \times \vec{L}) = GMm\frac{d}{dt}\hat{r}$$ Where, $$\frac{d}{dt}\hat{r}=\dot{\theta}\hat{\theta}$$ $$d(\vec{v} \times \vec{L}) = GMmd\hat{r}$$

$$\int{d(\vec{v} \times \vec{L})} = \int{GMmd\hat{r}}$$

$$\vec{v} \times \vec{L} = GMm(\int{d1})\hat{r}$$

$$\vec{v} \times \vec{L} = GMm(1+B)\hat{r}$$

$$\vec{r} \cdot (\vec{v} \times \vec{L}) = GMmr(1+B) = L^2/m$$

$$r = \frac{L^2 / m^2}{GM(1+B)}$$

Let $$B = \epsilon cos\theta$$ (This is the part I don't understand)

Noting that $$r(\theta) = a\frac{(1-\epsilon^2)}{1+\epsilon cos\theta}$$

When you integrated $$\frac{d}{dt}(\vec{v} \times \vec{L}) = GMm\frac{d}{dt}\hat{r}$$ it appears that you assumed the constant vector, $$\vec{B}$$, was in the same direction as the unit vector, $$\hat{r}$$. However, $$\hat{r}$$ is not constant. With a constant vector of integration the result is $$(\vec{v} \times \vec{L}) = GMm\hat{r}+\vec{B}$$ Now taking the dot product with $$\vec{r}$$ gives $$\vec{r}\cdot(\vec{v} \times \vec{L}) = G M m r +\vec{r}\cdot\vec{B}$$ and $$\vec{r}\cdot\vec{B}=r B\, cos(\theta)$$, with $$\theta$$ the angle between $$\vec{r}$$ and $$\vec{B}$$. The rest should follow directly by identifying the eccentricity as $$e=\frac{B}{GM}$$.

• Mathematically this seems like the correct answer. I missed this entirely, and you are right that the constant of integration should actually be a general vector, and no reason for it to be in the r hat direction particularly. When dotting with r I get how it yields rBcos(theta). But why does this "theta" correspond to the "theta" that the smaller mass, m, makes with respect to one of the focii? I'm still struggling with the physical intuition behind this concept. It's almost as if the B vector (integration constant) makes the same angle with the origin as r. Dec 21 '20 at 19:35
• Actually, thinking about this again, you have stated it clearly that $\theta$ is the angle between $\vec{B}$ and $\vec{r}$. It seems an odd coincidence to me that it's the same angle that $\vec{r}$ makes with the x-axis in a stationary x-y plane (is such a plane completely arbitrary?) Dec 21 '20 at 20:01
• Well you already know that the vector. $\vec{B}$ must be in the plane of the orbit. Its direction provides a reference point on the orbit. When $\theta=0$, you can see from the equation of the ellipse that r is minimum, so the reference direction is toward periapsis. However, $\theta$ does not vary uniformly in proportion to time as the object moves around in its orbit. To determine position requires an initial condition for the object, and then Kepler's equation is solved for the time dependence of $\theta$, also called the eccentric anomaly. Dec 21 '20 at 20:16
• Thanks for going through this. So $\vec{B}$ is in the plane, I understand this more now, and the HyperPhysics quote I put above also makes sense to me now. However, I'm still not seeing a strong line of reasoning as to why the angle $\theta$ that $\vec{B}$ makes with respect to $\hat{r}$ is necessarily the same as the angle that $\hat{r}$ makes with the x-axis. Does this mean that the $\vec{B}$ must lie along the x-axis? Or that it is rotated by $\theta$ in the CCW direction away from $\hat{r}$? Sorry if I am sounding like a broken record here. Just trying to understand this Dec 21 '20 at 20:32
• Since this is a vector derivation, the angle $\theta$ is not a coordinate. It is simply the angle between two vectors, as defined by the dot product. As such, it is just a parameter of the orbit obtained from a constant of integration. Of course, once you have an equation for the ellipse, the last expression in your question, then you are free to assign coordinates for the vectors, rotate the ellipse, etc. Dec 21 '20 at 22:35

I find the presentation on that hyperphysics page rather odd.

It is stated:
"Kepler's Laws depend upon the principle of conservation of angular momentum"

In my opinion the following should be stated:
Kepler's second law, the law of areas, is equivalent to conservation of angular momentum.

The shape of the Kepler orbit is a logical consequence of the fact that gravity is an inverse square force. A derivation of the shape of the orbit will in one form or another make use of conservation of angular momentum.

Other than that, the form of that derivation is more abstract than necessary.

Take for example the wikipedia derivation
That derivation is explictly in terms of polar coordinates.

On that hyperphysics page the quantities $$\vec{r}$$ and $$\vec{v}$$ are expressed as vectors in the plane, without choosing whether to use (for actual calculation) polar coordinates or cartesian coordinates. The form of the derivation is abstracted away from choice of type of coordinate system. But it serves no purpose; further along in the derivation the notation does shift to explicit polar coordinates.

I recommend doing without that unnecessary level of abstraction, and to follow a derivation that is all the way in polar coordinates or all the way in cartesian coordinates.

• Thanks for the insight. I'll try to go through the Wikipedia derivation next. The logic does seem to jump around all over the place on the HyperPhysics page. Dec 21 '20 at 19:38