5
$\begingroup$

Rough estimates shows that having around one hundred satellites would be enough to observe a single spot on Earth continuously and via peer to peer communication deliver that information to an earth-based single station in real-time.

But is it possible to deliver those satellites to the right orbits in one shot?

$\endgroup$
3
  • 1
    $\begingroup$ For some applications, doing something like this using balloons in the stratosphere may be a good solution - Google X is investigating this for internet service under the name of Project Loon. Aside from the difficulties with this mentioned in answers below, also consider that multiplying the number of satellites needed for coverage like this would be an invitation to Kessler Syndrome if it was widely practised - debris buildup in orbit to levels so high it destroys everything. $\endgroup$
    – kim holder
    Commented Apr 17, 2016 at 21:04
  • 1
    $\begingroup$ @Dmitra Kyianytsia I suggest changing the title from GTO to GEO. In GTO one would still need a ring of satellites to ensure ground coverage of a single point. The OP appears to really refer to GEO, Geostationary Earth Orbit, which is synchronous and must be approximately circular. GTO by contrast means Geostationary Transfer Orbit. It is highly elliptical and is a staging post on the way to GEO. Whilst most GEO satellites are separated from the launcher in GTO and make their own way to GEO, the delta-V requirement for the last step is large, 1500 to 2000 m/s. $\endgroup$
    – Puffin
    Commented Apr 17, 2016 at 23:29
  • 4
    $\begingroup$ 100 sats to LEO is a poor choice where the Molniya orbit allows for view of a specific point of Earth (and well off equator!) with a constellation of only 3 satellites, all deployable in a single launch, and considerably less delta-V than a GEO deployment. $\endgroup$
    – SF.
    Commented Apr 18, 2016 at 1:43

2 Answers 2

2
$\begingroup$

In principle yes, some caveats.

  • Possibility 1: Let us imagine that the target orbit has zero inclination, in this case all the satellites are launched to that orbit and must spread themselves out, by being in slightly higher or lower orbits until they reach equidistant spacing.
  • Possibility 2: The target orbit has a non zero inclination, perhaps because the "single spot" on the Earth has a latitude sufficiently far from the equator to be reached from a zero degree orbit. Now the 100 satellites will need to be distributed over different planes, possibly 100 different planes. The inclination of the planes could be the same for all, but they would each have a different Right Ascension of the Ascending Node, RAAN. By launching in "one shot" we are now causing most of the satellites to have to perform a plane change. This isn't a show stopper in principle but it can be extremely expensive in delta-V and thus propellant terms. It may well put such a demand on the satellites mass in propellant capacity that they can no longer all be launched together.

In both of the foregoing examples I have assumed that the phasing along the orbit, or between planes, is carried out by the individual satellites after separation. This isn't at all necessary as the launch vehicle could perform this instead. The energy requirements would be about the same though.

As an aside the starting premise that one would need 100 satellites covers many assumptions itself - such as the beam width from ground terminals looking up, the beam width from the satellites looking down etc.

$\endgroup$
0
5
$\begingroup$

In one launch? That depends on the mass of the satellites. Most rockets can deliver (roughly) 2x as much payload to LEO as they can to GEO. So your LEO satellites would need to be 1/50 the weight of your original GEO satellite. So you'd go from 5 tons to 100 kg per satellite.
For Earth observation, being 100x closer (360 km vs. 36,000 km) also means your optics get a lot smaller (100x?). But other components of the satellite won't get smaller by that much.
Having 100 times more satellites increases redundancy, but it also increases the number of failures.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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