# What changes the velocity perpendicular to radius in an elliptical orbit?

I'm working currently on a problem that asks to justify that angular momentum and kinetic energy conserves for a planet in an elliptical orbit. Although I've been taught that angular momentum should conserve when there is no external torque present in the system, so the conservation seems to be logical, I still tried to do mathematical proofs, which makes me more confused. I tried to do like this:

$$L = r\times mv$$ where $$L$$, $$r$$, and $$v$$ are all vectors. For an elliptical orbit, we should be able to split the velocity vector into two components, $$v_r$$ (velocity parallel to radius) and $$v_p$$(velocity perpendicular to the radius), so $$L = r \times m(v_r + v_p)$$ which I split to become $$(r\times mv_r) + (r\times mv_p)$$, as $$v_r$$ is always parallel to radius the cross product would come out as zero, so it leaves $$L = r\times mv_p$$ So it seems to me that $$v_p$$ should keep constant as there is no force acting to change its magnitude or direction(the only force present in the system would be gravitational forces, which are forever perpendicular to $$v_p$$). This means $$L$$ would change as $$r$$ increases and decreases, making the angular momentum not conserved.

However, when I searched online basically every piece of information implies that this $$v_p$$ should change. They are using angular momentum conservation to justify $$v_p$$ should change, but I just don't see what drives the change in $$v_p$$, especially when my proofs tell me angular momentum, bizarrely, should not conserve.

A simple example to specifically point out your error here. In general $$v_p$$ can change even if there's no force at all!

So it seems to me that $$v_p$$ should keep constant as there is no force acting to change its magnitude or direction(the only force present in the system would be gravitational forces, which are forever perpendicular to $$v_p$$).

Imagine a planet at distance $$r_0$$ and velocity $$v_0$$ perpendicular to the radius. Imagine gravity is zero.

Initially $$r=r_0$$, $$v_p=v_0$$ and $$v_r=0$$. The planet keeps going on a straight line with constant speed, getting further and further away from the sun, thus the direction of its velocity will get closer and closer to the direction of its radius. Therefore $$v_p \rightarrow 0$$ and $$v_r \rightarrow v_0$$.

There's no force at all affecting the planet, yet still the $$v_p$$ and $$v_r$$ change. Therefore it's clear that it's wrong to say that "$$v_p$$ doesn't change because there is no force component parallel to $$v_p$$".

• Thank you for actually answering the question asked. Yes, the point is that cylindrical reference frame is not inertial. There are "fictitious" forces (centrifugal, Coriolis) induced by the choice of coordinates that make it so that components of velocity can change even though there is no "real" force acting in that direction. Commented Jun 11 at 12:40
• Thanks for the explanation. So the two velocity vectors simply seem to convert into one another as the radius's direction with respect to the motion changes over time, which is also the case for elliptical orbit. So actually the perpendicular component seems to change because in both straight line motion ignoring gravity and elliptical orbit, I'm constantly changing the definition of radius vector, or constantly shifting reference frames, as the respective position of radius to the line of motion changes constantly, which is not the case for circular motion. Correct me if I'm wrong, thank you. Commented Jun 12 at 0:43
• You are not constantly shifting reference frames. Rather, you have chosen a fixed reference frame $(r, p)$ that is not inertial. This means, as you say, that the velocity components can change without any real external force. The "forces" that cause this change are the centrifugal force and the Coriolis force. These forces arise because your basis vectors are rotating as you move, so your acceleration has a "real" contribution from your changing velocity and a "fictitious' contribution from your changing basis. Commented Jun 12 at 13:34
• Seems like mathematical artifice, but here's a classic problem that should convince you we really need to be careful when not working in inertial reference frames. Think about a bead on a stick. Assume there is no friction between the bead and the stick. Spin the stick around. The bead will slide off the end. But wait, there's no force along the stick. The stick only applies a force on the bead perpendicular to itself, so how does it move the bead along its length? Commented Jun 12 at 13:36

You can prove that $$\mathbf{L}$$ is constant performing time derivative and remembering that the force is central,

$$m \ddot{\mathbf{r}} = m \dot{\mathbf{v}}= - G\frac{M \, m}{r^3}\mathbf{r}$$

i.e. aligned with the position vector (taking the origin of the space coordinates in the mass producing the gravitational field).

Thus $$\frac{d}{dt}\mathbf{L} = \frac{d}{dt} \left( m\mathbf{r} \times \mathbf{v} \right) = m \underbrace{ \mathbf{v} \times \mathbf{v}}_{= \mathbf{0}} + m \mathbf{r} \times \dot{\mathbf{v}} = - \frac{G M m}{r^3} \, \underbrace{\mathbf{r} \times \mathbf{r}}_{\mathbf{0}} = \mathbf{0} \ .$$

• Yes, it should. Typo, corrected. Thanks Commented Jun 16 at 16:37

System: Planet $$B$$

$$\dots$$ kinetic energy conserves for a planet in an elliptical orbit is not a correct statement.

The force $$F$$ on body $$B$$ due to body $$A$$ is central and has a component in the direction of body B's velocity $$v$$ which will cause an acceleration in that direction.

The fact that the force $$F$$ is central, ie $$\hat F = -\hat r$$, immediately means that the torque about $$A$$ exerted by that force, $$\vec F\times \vec r$$, is zero.

For the body $$A$$ and body $$B$$ system in the positions shown in the diagram you can say the the speed of body $$B$$ is increasing hence its kinetic energy is increasing whilst the gravitational potential of the system is decreasing at the same rate but the total mechanical energy of the system is constant.

• Yes. The gravitational potential energy increases as the planet gets further away from the sun/star and, since total energy is conserved, at the same time the kinetic energy reduces. Commented Jun 11 at 17:39
• Sorry that was a typo. I meant total energy conserves, but thank you still for the explanation. That wasn't my main concern though, as I thought clearing my understanding for angular momentum conservation(which is clearly causing great confusion for me) would help me solve the conservation of total energy as well. Commented Jun 12 at 0:07
• @RChen You should mention that typo in your question. It's too late to fix it now, since the existing answers have (probably) been written with that error in mind (even if they don't explicitly mention it), and question edits shouldn't invalidate existing answers. Commented Jun 12 at 0:26

Angular momentum is a conserved vector, by your calculation, it is composed of two potential variable:

$$\vec{L} = \vec{r} \times m\vec{v}_p$$, $$m$$ is a constant.

In an elliptical orbit, $$\vec{r}$$ is constantly changing, which means $$\vec{v}_p$$ is also constantly changing. If $$\vec{v}_p$$ is a constant, it gives that $$\vec{L}$$ will not be a constant, which contradicts that it is supposed to be a constant.

• It is exactly what I'm confused about. I don't know what is driving the velocity component to change, and it seems to be changing without any force exerted, and I do not understand why angular momentum conservation overrules this fact when it is decisive in determining whether angular momentum should be conserved. Commented Jun 11 at 4:26
• @RChen, It could be about inertial force, but I have to do more research on this to tell you the answer. Commented Jun 11 at 4:38
• @Rchen, the origin of the velocity vector is changing, so the vector perpendicular to radius vector is also changing in direction Commented Jun 11 at 5:18

$$v$$, the velocity vector of the body, is not constant in magnitude because in a generic elliptic orbit it is not perpendicular to the radius. It is, however, in circular orbit which indeed has constant speed. In the formula for $$L$$ you wrote, $$r$$ is changing as well in general, so you cannot derive $$v_p$$ is constant. Moreover, as many correctly pointed out, $$v_p$$ generally varies regardless of the acceleration, because the axis on which you project $$v$$ to get $$v_p$$ is moving as the body moves, as it is orthogonal to r, so $$v_p$$ will change direction and magnitude even if $$v$$ is constant. Indeed in case of no forces we have $$v$$ constant and angular momentum constant as well. But the angular momentum is referred to a fixed point $$O$$ and so the radius $$r$$ is changing. What else is changing? The angle $$\theta$$ between the body direction and the radius, in a way such that the change of $$sin\theta$$ ( or equivalently the projection of $$v$$ on the axis orthogonal to r you prefer) is compensating for the change in $$r$$.

We can take advantage of the fact that the force is central, hence the acceleration will always be directed towards the center, to decompose the acceleration into the radial component and the component orthogonal to it (0). If we express the body position as $$(rcos\phi,r\sin\phi)$$ and differentiate twice with respect to time, we will get again the conservation of angular momentum:

Quoting from Analytical mechanics (Fasano, Marmi, 2006)

Introduce the radial unit vector $$e_r = (cos \phi, sin \phi)$$ and the orthogonal unit vector $$e_{\phi}= (− sin \phi, cos \phi)$$. The equation of motion $$m\ddot{\boldsymbol r} = f(r)\frac{\boldsymbol r}{ r }= −V'(r)\frac{\boldsymbol r}{r}$$ can then be written componentwise as follows: $$\frac{1}{r}\frac{d}{dt}(r^2\dot\phi) = 2\dot r \dot \phi+ r \ddot \phi= 0,\\ m\ddot r − mr\dot{\phi}^2= −\frac{dV}{dr}$$ and the first equation simply expresses the conservation of $$L_z$$.

$$r$$ and $$\phi$$ are polar coordinates, $$\boldsymbol r$$ is the position vector and $$r$$ its magnitude.

• Well, one thing is that I defined vp to be the perpendicular component with respect to the radius of the velocity vector, whatsoever it should always stay perpendicular to the radius, so is this way of splitting velocity vector literally incorrect? Another thing is I'm pretty bad at math and especially vector math, so could you kindly explain what orthogonal is and what the "phi" represents? Since I cannot comprehend the change between the first and second equations. Thank you. Commented Jun 11 at 7:44
• @RChen i think that in your formula r is changing as well, in general, and that explains why L stays constant, as the change in r is compensated by the change in vp. i will remove the subscript in my answer to make it clearer. orthogonal in this context means that the dot product is 0. will add more info to my answer later if you find it useful :) Commented Jun 11 at 7:54
• @RChen, it is very weird that you decompose the velocity vector like that, usually we decompose it so one vector is parallel to the major axis and one is perpendicular to the major axis of the orbit. Commented Jun 11 at 16:38
• @Polaris5744 It can be useful to decompose the vectors into parallel & perpendicular radial components. See gregegan.net/SCIENCE/LRL/LRL.html But of course one has to be careful when working with such a rotating coordinate system. Commented Jun 12 at 0:20

Why does $$v_p$$ change when there is apparently no force in that direction? Roughly, because polar coördinates are tricky.

It helps to have some concrete examples and the answer from JiK gives a good absolute simplest case: suppose at t = 0 we have r(0)=1, θ(0)=0, vr(0)=0, vp(0) = 1, and no force at any time, F(t) = 0. Since there is no force, the particle continues in uniform motion in a straight line r(θ) = sec θ, which in time looks like, $$r(t) =\sqrt{1 + t^2},\\ \theta(t) = \tan^{-1}(t),\\ v_p(t)=\frac1{\sqrt{1+t^2}}. \tag{1}$$ (Or, so I claim. The control questions below will prompt you to derive these.)

Why is this vp changing? Well, it's a funny story, when you take a velocity vector and extract out the components like that, those are components relative to basis vectors. But in polar coördinates, those basis vectors are a function of position, and position is changing! When you move around in space the definition of “the radial direction” and “the perpendicular direction” both change. In vector notation we would say, $$v_p(t) = \mathbf v(t)\cdot \hat\theta(r(t),\theta(t)).$$

In comments you asked to go light on the vector notation, so we'll leave this equation be: but let's still talk about components and such. If (x, y) are our usual Cartesian coördinates, the usual convention is for polar coördinates to relate to them as x = r cos θ, y = r sin θ. (Control question: derive the r(t) and θ(t) from (1), starting from this mapping with x = 1, y = t. Then use sin² + cos² = 1 to prove r(θ) = sec θ.)

There are a couple of ways to see how this affects the components of velocity. One is to just imagine that you rotate the coördinates by -θ so that the point (x, y) is now on the +x-axis, where vr coïncides with vx and vp coïncides with vy. And if you work out trigonometrically how to do this, it is \begin{align} v_r&=~v_x\cos\theta+v_y\sin\theta,\\ v_p&={-v_x}\sin\theta+v_y\cos\theta. \end{align}\tag{2}(Control question: in JiK’s example, sin and cos are y/r and x/r. Use (2) to derive the vp from (1), and vr with it. What do their squares sum to, and why?)

Let's now apply some calculus. First in the forwards direction: if r, θ are changing in time then by the product and chain rules,\begin{align} v_x &= \frac{\mathrm d\phantom{t}}{\mathrm dt}(r\cos\theta) \\ &= \dot r\cos\theta-r\dot\theta\sin\theta,\\ v_y &= \frac{\mathrm d\phantom{t}}{\mathrm dt}(r\sin\theta)\\ & = \dot r\sin\theta+r\dot\theta\cos\theta. \end{align} \tag{3} With some work, $$v_r=\dot r,~~ v_p=r~\dot\theta\tag{4}$$ (Control question: prove this. Then use these to derive the vr,p from the previous control question more directly.)

Now in the reverse direction, if we take derivatives of (2) with respect to time, we use the product rule and the chain rule, but each expression contains two sorts of terms: acceleration terms with $$\dot v=a$$, which have to do with the actual forces involved, and terms with $$\dot \theta$$, which have to do with the curvilinear coördinates changing out from underneath our feet. So we'll group those together. In the full expansion we have, \begin{align} \dot v_r&=a_x\cos\theta+a_y\sin\theta\\&~~-v_x\dot\theta\sin\theta+v_y\dot\theta\cos\theta\\ &=a_r + v_p~\dot \theta,\\ \dot v_p&=a_p - v_r~\dot \theta. \end{align}\tag{5}(The second derivation is like the first, so left as an exercise.) Putting together (4) and (5) gives $$\frac{\mathrm dv_p}{\mathrm dt} =a_p -\frac{v_r v_p}{r},\tag{6}$$ only the first term $$a_p$$ describes a “real” acceleration due to forces, the rest is due to coördinate shifting. (Control question: confirm that (1) works this way with ap = 0.)

So, polar coordinates are annoying and tricky because definition of the directions is changing while you are moving about the plane. You have to pay this vp vr/r cost as a result of these changing definitions, because the direction that θ refers to has not remained constant if vp ≠ 0.

(Extra practice for the liefhebbers: combining with (4), $$\dot v_p=r\ddot\theta+\dot r\dot\theta,$$ it looks like there's a sign error in (6)! Similarly $$\dot v_r=\ddot r$$ looks like it's flat out missing $$v_p^2/r$$. Use this to derive the centrifugal and Coriolis forces that would be invented by someone who is instantaneously corotating and doesn't realize it—so they think the particle is only moving in their x'-direction which is the r-direction, with no initial velocity in the y' direction which is the θ-direction.)

• I do get some grasp of the problem now, synthesizing all explanations received. However my vector and polar math is awful, and I do have some questions regarding the derivations. 1) My understanding about dot product is that it measures the projection, so does the general form you provided for vp mean the projected length of the velocity vector at the angle of unit vector theta? 2) May you explain the formula for unit vectors r and theta? I have no idea where they come from. 3) I understand the dr/dt and d(theta)/dt relationships by calculus, but how do they contribute to the whole derivation? Commented Jun 12 at 1:20
• 4) I am also a bit confused about the final dvp/dt equation. It seems to be a big jump for me, so may you do some explanation regarding the derivation or give some clues regarding which equation I should implement product rule on so that I can try to derive by myself? I would be very grateful if you would answer the above four questions. Commented Jun 12 at 1:21
• @RChen rewritten to make the derivations a bit more easy to follow and to thread some practice through so that you can confirm you understand each step. Hope that helps! Commented Jun 15 at 20:18

Suppose on Monday I decide that my favorite planet is Mars. Then my favorite planet is moving at 24 km/s. Suppose on Tuesday I change my mind and decide that my favorite planet is Mercury. Now my favorite planet is moving at 47 km/s. So in one day, the speed of my favorite planet went from 24 km/s to 37 km/s. How much force is needed to accomplish that change in speed?

The answer is: none. There was no force involved. The reason that the speed of my favorite planet changed is not because of a force, it's because what "my favorite planet" referred to changed. What "the speed of my favorite planet" means changed from Monday to Tuedsay.

Similarly, when you talk about "the component of velocity perpendicular to the radius", what that phrase means changes as the planet moves in its orbit. As the planet moves, the direction of the star changes, and so the direction that is "perpendicular" changes. We have $$|v_p|= |r \times v|/|r|$$. So $$|v_p|$$ can change from $$v_p$$ changing (from a force) or from the direction of $$r$$ changing. If the direction of $$r$$ changes, then what "perpendicular to $$r$$" means changes.