# How do we describe the radial velocity in elliptical orbits?

When I look at the velocities of elliptical orbiting satellites the radial velocity (k in the figures) increases from zero magnitude at periapsis, to a maximum at the latus rectum, then back down to zero at the apoapsis. This describes a rate of increase opposite the direction of gravity that changes over time.

From periapsis to the latus rectum, the satellite to planet distance increases at an increasing rate. Then from the point of the latus rectum to the furthest point, apoapsis, this distance increases at a decreasing rate.

The change in distance describes a net acceleration away from the planet until the satellite reaches the latus rectum where the net acceleration reverses direction - in the same direction as gravity. This occurs despite the fact that gravity is in one direction and is decreasing in magnitude throughout the periapsis to apoapsis satellite journey.

What is the math that we use to calculate the radial velocity?

When the satellite crosses the latus rectum, the radial velocity k decreases. Fig 4 reveals that the rate of change in distance from the planet decreases at that point which agrees with the change of radial velocity k. There is a point of inflection at the latus rectum which means there would be a mathematical change of sign. Is there math out there that describes this change of sign?

What is the physics we use to describe why the radial velocity does what it does? I’m not looking for a geometrical answer here. We should be able to describe this physically like we do with any other thing in motion experiencing a force and acceleration. I just can’t find any mention of it in my searches.

• Have you checked out Wikipedia? Jul 15, 2020 at 14:24
• Nothing about the radial component of the velocity there. Jul 15, 2020 at 15:47
• $\dot r=\text dr/\text dt$ is the radial component of the velocity. Jul 15, 2020 at 15:49
• You can understand the behavior of $\dot r$ using the concept of the effective potential. The final two equations in this article are the important ones. Jul 15, 2020 at 16:33
• I have nothing to add to BioPhysicist’s answer. You seem to have a misconception of radial acceleration in two and three dimensions, where it is not simply the rate of change of the radial velocity. Jul 15, 2020 at 18:52

You are making a common mistake in assuming that the radial acceleration is the time rate of change of the radial velocity. In Cartesian coordinates we can apply this reasoning because the unit vectors are constant. However, in polar coordinates the unit vectors depend on the spatial coordinate.

For planar motion using polar coordinates, the radial acceleration actually has two terms: $$a_r=\ddot r-r\dot\theta^2$$

where $$\ddot r$$ is the time rate of change of $$\dot r$$, which is the time rate of change of the radial coordinate $$r$$, and $$\dot\theta$$ is the time rate of change of the polar coordinate $$\theta$$.

As you can see, $$\ddot r$$ is not the whole picture. Even if $$\ddot r>0$$ this does not mean that $$r\dot\theta^2<\ddot r$$. In fact, if $$\ddot r$$ is positive then $$r\dot\theta$$ has to be larger than $$\ddot r$$, because the net acceleration has to always point in the negative $$\hat r$$ direction if the only force acting on the satellite is gravity which points inwards towards the planet.

To help you further, since we know that $$a_r=-GM/r^2$$, we can express the rate of change of $$\dot r$$ as $$\ddot r=r\dot\theta^2-\frac{GM}{r^2}$$ This should show you how the radial velocity can oscillate, as these two terms will change in magnitude throughout the orbit, thus changing the sign of $$\ddot r$$.

When examining the gravitational acceleration along a line from planet to satellite we see an increase in distance that describes an acceleration away from the planet - proven by the radial velocity change - until the satellite reaches the latus rectum. If you had a rocket, with its thrust, accelerating directly away from the planet until it reaches a certain distance where it reverses the thrust direction it could mimic exactly what we see along that planet to satellite line described here.

These aren't the same scenario. Just because we can choose to focus just on $$r$$ does not mean it is the only relevant coordinate. The satellite is still orbiting around the planet; the satellite is not moving along a 1D path described by $$r(t)$$ like you are proposing with your rocket. We are dealing with polar coordinates and vectors, and care needs to be taken before you simplify the analysis by considering scalar, Cartesian values instead.

• Thanks @BioPhysicist I appreciate your points about how I only focus on r, but doesn't gravity only accelerate along r at any instant? I wouldn’t think gravity does anything different whether the satellite is stationary or is in orthogonal motion relative to r. There must be some physical threshold at the latus rectum when r changes the direction in its rate of change. The radial velocity k goes from increasing in magnitude to decreasing. There must be a simple physical explanation. Gravity only decreases gradually through the window of time observed in Figure 2 and 3. Jul 16, 2020 at 20:23
• @Nectac Yes, gravity acting on the satellite always points towards the planet, so $a_r=\ddot r-r\dot\theta^2<0$ always. Yes, $\dot r$ does change sign during elliptical orbits. When the satellite is getting closer to the planet $\dot r<0$, and when the satellite is moving farther away from the planet $\dot r>0$. This also means that yes, there are points where $\dot r=0$. Jul 17, 2020 at 12:28
• I added a figure that graphs the distance of the satellite from the planet through time from periapsis to apoapsis. Also included is the graph of gravity (dashed) through the same time period. The distance graph shows the angle of the slope increasing until it reaches the latus rectum where it begins decreasing down to zero at apoapsis. Gravitational acceleration graph shows nothing that produces this effect we see at the latus rectum. There must be a physical explanation for this change. Jul 18, 2020 at 19:52
• @Nectac I'm still not sure about what you are confused about here. "Gravitational acceleration graph shows nothing that produces this effect" what do you mean by this? What graph are you referring to. What about my answer still doesn't make sense to you? Why do you keep ignoring the information in my answer? Jul 18, 2020 at 20:42
• I can find a geometrical and mathematical description of the change in radial velocity k. That’s how I provided the values for k. My question asked for the physical explanation. When something that has a velocity that is changing we should always be able to explain why, physically. In this example, gravity is always directly opposite the radial velocity. Why does the velocity increase then decrease? BTW I added a figure yesterday that is a graph. It shows where the change occurs at the latus rectum while gravity is constant in its decline. Jul 19, 2020 at 20:51

What is the math that we use to calculate the radial velocity?

With the usual nomenclature:

$$G$$ = universal gravitational constant

$$M$$ = mass of primary

$$m$$ = mass of secondary

$$a$$ = semi-major axis of the elliptical orbit

$$e$$ = eccentricity of the elliptical orbit

$$\theta$$ = true anomaly

$$r$$ = radius vector (position vector from de focus)

The mathematical expression to calculate the radial velocity is:

$$v_r=\sqrt{\frac{G (M+m)}{a(1-e^2)}} \ \ e \sin \theta$$

And the expression for the velocity component perpendicular to the radius vector is:

$$v_{\theta}=\sqrt{\frac{G (M+m)}{a(1-e^2)}} \ \ (1+e \cos \theta)$$

Naturally, it follows that:

$$v=\sqrt{v_r^2+v_{\theta}^2}$$

That operating can be converted into the familiar expression:

$$v=\sqrt{2 G (M+m)\left ( \frac 1 r - \frac 1{2a} \right )}$$

$$v=\sqrt{2 \mu \left ( \frac 1 r - \frac 1{2a} \right )}$$

Where $$\mu=G(M+m)$$ is the gravitational parameter.

The mathematics required to detect the maxima and minima of a function consists of calculating the places where the derivative of the function is zero.

The derivative of the radial velocity with respect to the true anomaly is:

$$\frac{dv_r}{d\theta}=\sqrt{\frac{G (M+m)}{a(1-e^2)}} \ \ e \cos \theta=0 \rightarrow \cos \theta =0$$

$$\displaystyle \theta=\frac{\pi}2 \qquad or \qquad \theta=\frac{3\pi}2$$

Which corresponds to the two "semilatus rectus".

$$\theta=0 \longrightarrow v_r=0$$

$$0<\theta <\dfrac{\pi}2 \longrightarrow v_r>0$$ increasing positive value

$$\theta=\dfrac{\pi}2$$ maximum positive value of $$v_r$$

$$\dfrac{\pi}2 <\theta < \pi \longrightarrow v_r>0$$ decreasing positive value

$$\theta=\pi \longrightarrow v_r=0$$

$$\pi<\theta <\dfrac{3\pi}2 \longrightarrow v_r<0$$ increasing module of negative value

$$\theta=\dfrac{3\pi}2$$ maximum module of the negative value of $$v_r$$

$$\dfrac{3\pi}2<\theta < 2\pi \longrightarrow v_r<0$$ decreasing module of negative value

$$\theta=2\pi \longrightarrow v_r=0$$ And we have made a complete turn

I attach, as an example, a table with the radial velocity, the perpendicular velocity and the total velocity (metres per second) of the Earth in its elliptical orbit around the Sun, as a function of the true anomaly (radians)

Best regards

• Thanks @Albert. When the satellite crosses the latus rectum, the radial velocity k decreases. Fig 4 reveals that the rate of change in distance from the planet decreases at that point which agrees with the change of radial velocity k. There is a point of inflection at the latus rectum which means there would be a mathematical change of sign. Is there math out there that describes this change of sign? Sep 7, 2022 at 11:49
• Hi @Nectat , I have added new information to the post to try to answer your new questions. Sep 7, 2022 at 13:54
• Thanks @Albert. As we can see, there is a max positive and negative radial velocity of 497.7822 at both semi latus rectums. Before then after, the radial velocities are increasing then decreasing. At the furthest (apoapsis) and closest (periapsis) the radial velocities change signs. This change in sign is easily understood by gravitational acceleration over time. Though, at the latus rectum, there clearly appears to be a change in net acceleration. A change in direction. Do you know of any acknowledgement of this change in net acceleration? Sep 8, 2022 at 13:04