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

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.

## 2 Answers

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.

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

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

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

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

• 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
– JOY
Sep 26 '19 at 0:07
• @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.
– uhoh
Sep 26 '19 at 15:08
• @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.
– uhoh
Sep 26 '19 at 15:19
• 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
– JOY
Sep 27 '19 at 15:52
• I just did here. Thank you.
– JOY
Oct 2 '19 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.

• @bobjacobson problem is not having the same plane and not a circular orbit.
– JOY
Sep 27 '19 at 15:53
• 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. Sep 27 '19 at 16:05
• @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.
– JOY
Oct 1 '19 at 8:47