Shifting of pass times due to orbital maneuvers

19 Apr 2024 in Astrodynamics

If you execute orbital maneuvers and have scheduled satellite passes in advance, you will experience passes shift. Here’s why.

1. Introduction
2. A satellite pass over a ground station
3. Case #1: Going to a higher orbit (increasing the semi-major axis)
4. Case #2: Going to a lower orbit (decreasing the semi-major axis)
5. Calculating the actual drift
   1. For Case #1 (increasing the semi-major axis)
   2. For Case #2 (increasing the semi-major axis)

Introduction

In practical applications, we like to schedule satellite passes for a given ground station in advance. Usually, pass times are computed for a given timeframe in the future (usually a few days up to maybe a few weeks) and this process is repeated once the timeframe has passed. We are interested in computing the acquisition of signal (AOS) of the pass, the loss of signal (LOS) of the pass and the elevation of the pass. This information is used to prepare the ground station in combination with a TLE so that tracking and communications with the satellite can take place.

However, this process can be interrupted by orbital maneuvers. By changing the geometrical properties of the orbit, the properties of the pass and most importantly the AOS and LOS of the pass can be severely affected to the point where the satellite cannot be properly tracked and passes are shifted. Let’s explore why this happens and how we can quickly adjust the pass times.

A satellite pass over a ground station

Let’s assume a satellite in orbit around the earth. To simplify things a bit, let’s also assume a circular orbit around the planet at an altitude of $h=400\text{ km}$.

All passes scheduled in the future have been considered using this initial orbit.

Case #1: Going to a higher orbit (increasing the semi-major axis)

Let’s now assume that we want to perform a station-keeping maneuver and raise the orbit of the satellite by 10 km to a new circular orbit with an altitude of $h=410\text{ km}$. Before calculating the actual shifting of the passes, let’s quickly examine what is happening. From orbital mechanics, the speed of the satellite for a circular orbit will be: $$v_{circ} = \sqrt{\frac{GM_{Earth}}{r}}$$

From this equation, we can immediately determine that the satellite will be moving slower compared to the original orbit since it’s at a higher altitude.

Raising the orbit of the satellite. The satellite is slower at the new orbit.

If $t_{0}$ refers to the AOS time of the pass planned for the initial orbit and in the new orbit the satellite is moving slower, it is evident that the satellite will require some more time to reach the point over the horizons which is considered the AOS of the pass, and thus the new AOS will now occur at a time $t_{new} = t_{0} + c$, meaning that the pass will occur later than predicted (pass shifting forward in time).

Case #2: Going to a lower orbit (decreasing the semi-major axis)

On the opposite, in case the orbit of the satellite needs to be lowered by 10 km to a new circular orbit with an altitude of $h=390\text{ km}$, the same approach can be used.

Raising the orbit of the satellite. The satellite is faster at the new orbit.

In this example, the satellite will be moving faster compared to the original orbit since it’s at a lower altitude. If $t_{0}$ refers to the AOS time of the pass planned for the initial orbit and in the new orbit the satellite is moving faster, it is evident that the satellite would have already crossed the point over the horizon which is considered the AOS of the pass, and thus the new AOS will now occur at a time $t_{new} = t_{0} - c$, meaning that the pass will occur earlier than predicted (pass shifting backward in time).

Calculating the actual drift

In this simplified example, we are dealing with circular orbits. We can thus use the definition of the mean anomaly and the mean motion of the satellite to do our computations. Let’s express the mean motion in degrees per day computed from: $$n=\frac{360^{\circ}}{T}$$

Where $T$ is the period of the satellite. The difference of the mean motion between the two orbits can be computed with: $$\Delta n = n_{initial} - n_{final}$$

Based on the definition we used, we can use these values to compute the difference in mean anomaly for a single day. We can then use the radius of the final orbit to compute the corresponding arc on the final orbit using the simple formula $s=r\Delta Μ$, where $\Delta M$ is the difference in the mean motion. This arc corresponds to far ahead or behind the satellite is moving (and corresponds to the along-track error) in $\left[ km/day\right]$. We can then divide that by the speed of the satellite to see how much the passes will drift in $\left[ sec/day\right]$.

For Case #1 (increasing the semi-major axis)

At the initial altitude $h=400 \text{ km}$, $T=5553.456\text{ sec}$ or $T=0.064276110\text{ days}$

At the final altitude $h=410 \text{ km}$, $T=5565.751 \text{ sec}$ or $T=0.064418409 \text{ days}$

Computing $\Delta n$ for this case we obtain $\Delta n = 12.3721 \text{ deg/day}$, which means the difference in Mean Anomaly would be $12.3721\text{ deg}$ after a day. Computing the arc length (pay attention with degrees and radians here), we can see that the difference in along track rate would be $1465.76\text{ km/day}$. Dividing that with the speed of the satellite $v=7.662984862\text{ km/sec}$, we see that the passes would drift by $191.28\text{ sec/day}$ or $3.19\text{ min/day}$.

For Case #2 (increasing the semi-major axis)

At the initial altitude $h=400 \text{ km}$, $T=5553.456\text{ sec}$ or $T=0.064276110\text{ days}$

At the final altitude $h=390 \text{ km}$, $T=5541.170\text{ sec}$ or $T=0.064133917\text{ days}$

Computing $\Delta n$ for this case we obtain $\Delta n = -12.4178 \text{ deg/day}$, which means the difference in Mean Anomaly would be $-12.4178\text{ deg}$ after a day. Computing the arc length (pay attention with degrees and radians here), we can see that the difference in along track rate would be $-1466.84\text{ km/day}$. Dividing that with the speed of the satellite $v=7.674298887\text{ km/sec}$, we see that the passes would drift by $-191.14\text{ sec/day}$ or $-3.19\text{ min/day}$.