While calculating the Point Ahead Angle we work with Tangential Velocity vector component and to calculate the Doppler shift i.e changes in frequencies, we work with Radial Velocity Vector component.

Figure 1

Correct me if I am wrong, we calculate radial and tangential velocity like the picture above of it (Figure 1). But is this correct way to find out the radial velocity of both satellite while calculating the Doppler shift in changes in frequencies, because we also need to find out the relative radial velocities of both satellite, or the picture below (Figure 2) this is the correct one? where radial velocity component is just opposite of the line of sight of the both satellite.

Figure 2

Because, I looked at the picture below of it (Figure 3) where radial component Vp is as with Line-of-sight of both satellite, then I become confused how should I put my vector components?

Figure 3

Any sort of leads would be much helpful. Please let me know what I am missing here. Thanks in advance.


3 Answers 3


I'm not an expert on the subject, but here's an analysis based on basic physics.

Since you've used 2D diagrams where it looks like the two orbits are in the same plane, I'll stick with that as well, but remember that orbits are 3D and you'll need to calculate the radial and perpendicular velocities using the 3D velocity vectors of each satellite.

The vectors shown in your Figure 2 shows the correct way to calculate both the point-ahead angle $\theta_{PA}$ and the doppler shift.

I think you can call those two vectors $Vr$ for radial velocity, but where the radius is drawn from one satellite to the other, and $Vp$ for perpendicular velocity, which is the velocity perpendicular to the line connecting to the two satellites.

In this case, with the vectors drawn as shown, the Doppler shift will be related to

$$\frac{\Delta f}{f} \approx -\frac{Vr_1+Vr_2}{c}$$

and the look-ahead angle will be

$$\theta_{PA} \approx 2\frac{Vp_1+Vp_2}{c}$$

If you have proper orbital state vectors for the two spacecraft $\mathbf{r_1}, \mathbf{v_1}$ and $\mathbf{r_2}, \mathbf{v_2}$ then you can do the following:

caution: these state vectors can be from any inertial frame but I don't think this is appropriate for a rotating frame. Notice that the two velocities are subtracted; in the 2D drawings from the question, the arrows point opposite ways so the scalar velocities are added, but that's an artifact of working with the images having arrows pointing in the directions that they are.

$$\mathbf{\hat{r}} = \frac{\mathbf{r_2} - \mathbf{r_1}}{|\mathbf{r_2} - \mathbf{r_1}|}$$

$$\frac{\Delta f}{f} \approx -\frac{(\mathbf{v_2} - \mathbf{v_1}) \cdot \mathbf{\hat{r}}}{c}$$

$$\theta_{PA} \approx 2\frac{|(\mathbf{v_2} - \mathbf{v_1}) \times \mathbf{\hat{r}}|}{c}$$

Figure 2 modified

Here's a random image from the internet, taken from Space Laser Communications Systems, Technologies, and Applications:

enter image description here

  • 1
    $\begingroup$ Thanks for your answers. Yes, We are on the same path, I also have that paper (the link you have uploaded). Well, now the problem is I have the Radial and Tangential (what you are telling the perpendicular one) component of velocity vector what I have shown you on my uploaded picture Figure 1 (Thanks for putting those numbers). Now how to transform it according to your figure. Do I need to rotate with same angle as between, Earth-Satellite1-Satellite 2. Yes I am also aware of the 3D orbital components and I have coordinates for Velocity vectors in 3D and in that scenario, I dint show 3D $\endgroup$
    – JOY
    Commented Sep 26, 2019 at 0:07
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    $\begingroup$ @Jyoti no, that's just basic vector math. If you have some vector $\mathbf{v}$, the scalar magnitude of the component parallel to some vector $\mathbf{r}$ is $\mathbf{v} \cdot \mathbf{\hat{r}}$, and the scalar magnitude of the component perpendicular is $|\mathbf{v} \times \mathbf{\hat{r}}|$. (fyi this was meant as a comment, not an answer post; I used it to compose/render the MathJax but apparently posted it instead of copy/pasting here. $\endgroup$
    – uhoh
    Commented Sep 26, 2019 at 15:08
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    $\begingroup$ @Jyoti I've added a little bit more. I had it wrong, there needs to be a minus sign between the two velocities. The plus sign was left over from the 2D case (where the signs of the velocities were defined by opposite-pointing arrows). This is right now, and with the minus sign the Earth is subtracted from the problem. They could be Earth-centered state vectors or Heliocentric or any other reference point and it will still subtract. It just can't be a rotating frame of reference. $\endgroup$
    – uhoh
    Commented Sep 26, 2019 at 15:19
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    $\begingroup$ Thanks, I actually did with v2 - v1. Thank you so much for your help. Coming soon, with some new questions regarding orbit and all. Hahaha. :P $\endgroup$
    – JOY
    Commented Sep 27, 2019 at 15:52
  • 1
    $\begingroup$ I just did here. Thank you. $\endgroup$
    – JOY
    Commented Oct 2, 2019 at 8:32

If $\vec{R}_1$ and $\vec{R}_2$ are the positions of the two satellites in any consistent coordinate system, including an Earth-centered one, then $\vec{R}_1 - \vec{R}_2$ is what you’re looking for: the vector distance between them. You don’t need to worry about the motion of one vs the other, just how that vector lengthens/contracts and rotates.

For same-plane near-circular orbits it really doesn’t change length very much at all. Points on a circular orbit just follow each other at fixed distances.

  • $\begingroup$ @bobjacobson problem is not having the same plane and not a circular orbit. $\endgroup$
    – JOY
    Commented Sep 27, 2019 at 15:53
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    $\begingroup$ R1-R2 is still the value that matters, no matter what the case. Working with the individual positions just adds confusion and doubles the calculation. $\endgroup$ Commented Sep 27, 2019 at 16:05
  • $\begingroup$ @BobJackson Thanks man. But how do you suggest me to calculate Point Ahead Angle and Doppler Effect, for one satellite in LEO and another satellite in GEO case? Top of it, it's already answered and but do you suggest me the same way or you have any different approach to this problem? Thanks. $\endgroup$
    – JOY
    Commented Oct 1, 2019 at 8:47

You said,

While calculating the Point Ahead Angle we work with Tangential Velocity vector component and to calculate the Doppler shift i.e changes in frequencies, we work with Radial Velocity Vector component. $\newcommand{\d}{\partial} \newcommand{\r}{\vec{\mathbf{r}}} \newcommand{\R}{\vec{\mathbf{R}}} \newcommand{\v}{\vec{\mathbf{v}}} \newcommand{\V}{\vec{\mathbf{V}}}$

but that's not correct, or at least not when combined with the pictures you supplied.

The Doppler shift is given by the rate of change in the distance between the two satellites, which is the same as the difference in their total 3D velocities when projected onto the instantaneous line of sight. The "radial" and "tangential" velocities that matter have nothing to do with the local frame in use at each satellite, but rather the components parallel to or perpendicular to the position difference.

Let $\r_1$ be the position of the transmitting satellite, and let $\r_2$ be the position of the receiving satellite. Define the relative position $\R=\r_2-\r_1$ . Then, since taking derivatives commutes with subtraction, the relative velocity $\V$ equals both $\d\R/\d t$ and $\v_2-\v_1 = \d\r_2/\d t -\d\r_1/\d t$ .

The instantaneous range, $R$, is the square root of $\R\cdot\R$. Its time derivative, the "range rate", is $$ \frac{\d R}{\d t} = \frac{\d}{\d t}\sqrt{\R\cdot\R} = \frac{1}{2 R} \frac{\d}{\d t}(\R\cdot\R)=\frac{\V\cdot\R+\R\cdot\V}{2R} = \frac{\R\cdot\V}{R}$$ It has to be that way, because all we've done is restate the Doppler definition: $\R/R$ is the unit vector in the direction of the displacement, and dotting that with $\V$ is what projection means.

To discuss look-ahead, we have to become more precise in how we work with time. In particular, rather than use the instantaneous position difference as a function of just one time, $\R(t)=\r_2(t)-\r_1(t)$ , we need instead to define a more generic position difference as a function of two different times, $\R(t_1,t_2)=\r_2(t_2)-\r_1(t_1)$ . This gives us a formula to express the difference between where the transmitter used to be at $t_1$, and where the receiver is going to be at $t_2$. Then the formula $$\frac{\R(t_1,t_1)\cdot\R(t_1,t_2)}{R(t_1,t_1) R(t_1,t_2)}$$ gives us the cosine of the "point ahead angle". To get this exactly right, you need to wiggle things a bit, to maximize the self-consistency of your solution. That is, if you start with $R(t_1,t_1)$, then you could use $t_2 \approx t_1 + R(t_1,t_1)/c$ ($c$ is the speed of the transmission, which I assume is light) to compute a better answer for $R(t_1,t_2)$, but that changes the estimated $t_2$ at which you need to interpolate your ephemerides, which changes your best estimate of the time, which changes the position difference, etc. You run this procedure around several times, and hopefully iterate to convergence on a $t_2$ estimated to the same order of accuracy as your ephemerides.


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