3
$\begingroup$

I want to understand what the revisit rate is. If the payload is required to make the revisit rate of 5 hours. The orbit is LEO (altitude less than 2000 kms). What does this mean?

Revisit time - the time gap between the satellite to revisit the same area again. What is the revisit rate?

$\endgroup$
4
  • 3
    $\begingroup$ Usually rate = 1 / period. If it revisits every 12 days, then the revisit rate is 1/12 = 0.08333 day$^{-1}$. $\endgroup$
    – uhoh
    Sep 13 at 1:56
  • 3
    $\begingroup$ What is the orbit that you are considering ? SSPO ? Or other ? Please specify that also. Please provide context to the question also. Where did the number 5 hours a day come from ? $\endgroup$
    – AJN
    Sep 13 at 12:15
  • 2
    $\begingroup$ Revisit can be expressed as @uhoh did for sure. In addition, it highly depends on your communication cone. In our case, revisit time, with respect to our ground station, is between 4 and 5. As AJN mentioned, your question needs to be improved with your requirements. $\endgroup$ Sep 15 at 6:34
2
$\begingroup$
  • This terminology needs clarification, otherwise there will be many ways to compute it.

Loosely speaking, it is an indicator of how often a satellite revisits a given site on Earth. While it is self-evident that it depends on the orbit of the satellite, less obvious is that it depends on the latitude of the observing site.

Furthermore, it is non-ambiguous only when you pin down what “often” and “visit” mean. For example, you may wish to clarify that “how often” means “how many times in a day” (note that "in 48h", "in a week", are perfectly plausible options too). Likewise a visit means that the satellite appears above the horizon, at least above 10° elevation (or 50° above).

Still, another less obvious aspect is the implicit assumption that the events are more or less periodic. In other words, there would be a quasi-constant “period” between visits.

  • Let’s examine an example. Say we want to know how often, PER DAY, we have the ISS (ZARYA) appearing above New-York (~ 40.74°N,~ 73.95°W) horizon, at least above 10° elevation.

Referring to heavens-above: 17Sep= 7 visits; 18Sep= 5; 19Sep= 6; 20Sep= 6; 21Sep= 7; 22Sep= 6; 23Sep= 6; 24Sep= 6;

You may hastily make the “short-cut” by generalizing that “revisit rate” is 6 per day, on average, equivalently, one visit every 4 hours (about). But this is not correct, the different visits are irregularly spaced (look at the rise times data in details), although the number of visits per day is fairly constant.

Let’s change our definition of “visit” by requiring that the satellite is at least 50° above the horizon. The number of visits per day then becomes (look at the “Highest points” data): 17Sep= 1; 18Sep=2; 19Sep=1; 20Sep=1; 21Sep=1; 22Sep=2; 23Sep=2; 24 Sep=0.

Now, change the observation site’s latitude to 51°N. The result for 50° above horizon: 17Sep= 3; 18Sep=3; 19Sep=3; 20Sep=2; 21Sep=3; 22Sep=3; 23Sep=2; 24 Sep=2.

This shows the strong dependency to the latitude of the observing site.

  • Concluding remarks:

From these examples, for a single satellite, there is in general insufficient “periodicity” to warrant the use of “revisit rate” and/or “revisit period”.

A particular case for a regular visit pattern is when the orbit is equatorial (or near-equatorial) AND the observing site is near the equator. Another particular case leading to regular visits is when the orbit is polar (90° inclination), or near-polar, and the observing point is near-polar.

Nevertheless, with a constellation of many satellites, it is possible to make the visiting pattern (visits from several satellites) almost regular for almost any points on Earth, while minimizing the number of satellites. This is however a complex optimization problem, requiring sophisticated genetic algorithms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.