Point Ahead Angle and Doppler Shift Frequency Changes Calculation with the tangential/radial velocity component for Inter-Satellite Link

Question

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.

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.

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?

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

Answer 1

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.

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

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

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

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

Answer 2

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.