Curiosity got me wondering. How much easier is it to get an object to orbit the earth when it is launched from the equator, and thus able to use the Earth’s rotation speed to assist on exit velocity.
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Well, there are at least two dopers in particular who’ll be along shortly to let you know in detail, but in the meantime, you can see for yourself by playing around with the Silverbird Launch Vehicle Performance Calculator.
I think it’s a wash … but others may correct me … sure, you’ll get a 1,000 mph head start launching from the equator … but the atmosphere is thicker so there would be more drag …
The location of the launch site may well have more to do with shipping costs … Surinam is generally ice free most of the year … Antarctica not so much …
If you wanted a polar orbit, then that 1,000 mph advantage is actually a drawback and needs to be corrected. But for most of the usual orbits, that 1,000 mph is a big advantage.
Is the air notably thinner over the poles?
(If you want really tiny effects, because of the equatorial bulge, equatorial launches begin with a free 20 km of altitude.)
Launching from one of the poles is going to bring its own set of problems with it. Ice buildup becomes an issue. The design of the rocket and the materials used in both the rocket and the launch equipment becomes an issue. One of the reasons the Challenger disaster happened was that the o-ring that failed didn’t perform its job as well at lower temperatures. Without the cold temperatures that day, they might have had only a partial o-ring burn through (as had happened on previous flights) instead of the complete failure that led to the disaster.
The atmosphere at the poles is thinner but it’s also colder and more dense. I don’t know what that means for rockets though.
As I understand it the drag from the atmosphere is much less important than the speed boost from the rotation at the equator; that’s why they don’t go to the effort of launching rockets from mountaintops or aircraft.
The What If section of xkcd had an interesting article on the matter that I read recently that explains things better than I can. But basically, the hard part isn’t getting into space: the hard part is getting up to enough speed to stay there.
What kind of orbit do you want?
The boost from the equatorial velocity is one thing–but changing your inclination is quite another. That is to say, what is the angle that the plane of your orbit makes with the plane of the equator?
From any given latitude, you can “easily” launch to inclinations greater than or equal to than the latitude–you just wait until the orbit intersects your launch site, and then launch in that direction. But it is very difficult to launch to inclinations less than that value–you have to head in the “wrong direction” first, and then change direction when you intersect the orbit. Changing direction in space is very hard.
So as others said, polar orbits would be easy (ignoring the practical considerations of launching in that environment), but any low orbit that’s somewhat equatorial would require a huge amount of extra delta-V. For example, if my napkin vector math is correct (probably suspect…), launching from a pole and hitting the ISS orbit (51.6 degrees) is roughly 5.3 km/s of extra delta V. In contrast, 3.8 km/s puts you on a transfer orbit to Mars.
It’s a bit easier if you launch to geostationary orbit. Way up there, orbital speeds are lower and it’s easier to change inclination. You can also make use of an effect where raising your apogee further allows reduced delta V. Again, if my mental math is correct, you can do this with 1.4 km/s, although you have to wait quite a long time.
In short, although the equatorial boost is beneficial, inclination is the far more significant factor if you want anything but a polar orbit. Near-equatorial launch sites are beneficial for both reasons. That said, the US does operate a launch facility in Alaska–used for polar and high-inclination orbits (even so, it has a latitude of only 58 degrees).
Speed for low earth orbit is about 17,400 MPH.
Rotational speed of the earth’s surface, at the equator, is about 1042 MPH.
Kinetic energy scales with the square of velocity, so by launching at the equator, you’ve already got (1042/17,400)^2) = 0.36 percent of the kinetic energy needed for low earth orbit.
0.36% doesn’t sound like much of a head start. At first blush, you might think that means you can only reduce your propellant load by 0.36%. But of course there’s a whole snowball effect going on: you reduce your propellant load by 0.36%, now you can reduce your tank volume by the same amount. Now your whole rocket weighs less, so you don’t need as much fuel to get it to orbit. So you can shrink the whole thing even more. Our resident rocket experts can probably give you a good estimate of what the final reduction in vehicle size/mass/cost might be by launching at the equator intstead of the pole, but my guess is that it would be a few percent - or alternatively, for a given rocket size, you can launch more payload. Just one percent of a million pounds would mean 10,000 pounds of extra payload.
Note that Cape Canaveral is at about 28 degrees north latitude, so the rotational velocity there is about 88% of what you’d have at the equator.
On the other hand, most satellites don’t particularly care about their inclination. The main reason that low-inclination orbits (i.e., equal to launch latitude, not greater) are favored is simply that they’re cheaper. If something were to happen that made high-inclination orbits cheaper, then a lot of satellites would just switch to higher-inclination orbits.
That’s probably not the clearest way of approaching the problem, though the math is correct. One way of explaining why that 0.36% has such an outsized effect is that it’s relative to the starting launch mass of the vehicle, which is enormous compared to the final delivered payload mass.
You can look at it at the other end; (17,400)^2/(17,400-1042)^2 = 113%. 13% more KE is nothing to sniff at; the reason there’s no conservation-of-energy problem is because that’s 13% of the final (tiny) payload vs. 0.36% of the (huge) whole vehicle mass.
But ultimately you just have to run the rocket equation if you want to know the effect on your delivered payload. A lot of this is dependent on the exact characteristics of your rocket.
Only true for the upper stage of your rocket. On a generic 2-stage rocket like the Falcon 9, savings on the first stage translate to about 20% on the payload. That’s part of why SpaceX can afford to put landing legs and such on their first stage, but not the second; extra first-stage mass has a much smaller effect on payload than second-stage mass.
I think that’s overstating the case somewhat–obviously, geostationary orbits require a specific inclination, and systems like GPS require specific relative inclinations between their different orbits (although the system as a whole could have any inclination). But you’re definitely right in that low inclinations would be designed around to the extent possible if they required an orbital inclination change.
Equatorial LEO in particular would be avoided since it would be the costliest, and because it’s not that interesting an orbit to start with due to the small ground track (imaging satellites, etc. want a reasonably high inclination so that they get visibility over a large part of the earth).
Isn’t that why theESA’s launch site is in Guiana?
That, and because Europe isn’t a great place for launch sites, which generally need lots of uninhabited area to the east (ocean, Siberia, etc.).
I did a quick search of low-inclination LEO satellites and out of 1,672 active satellites, only three had inclinations of <10 deg and apoapsis <2000 km: RAZAKSAT (launched from Omelek island, near the equator), HETE-2 (launched from the airplane-based Pegasus launch system), and NuSTAR (also launched from a Pegasus). So yeah, it’s a pretty boring orbit. That number goes up to 22 for inclinations <30 deg, and 84 for <50 deg. OTOH, there are 535 satellites currently in geostationary orbit.
Wait, only 84 satellites out of 1,672 with inclinations less than 50º? That can’t be right.
And with altitudes <2000 km. LEO (incidentally, almost half of these are Orbcomm birds). In contrast, there are 529 satellites with inclinations of 80 to 100 degrees in LEO. Or, if I relax the altitude restriction, I get 754 total with inclinations <50 deg.
The full list:
inclination: 0-9 deg
gso: 535
heo: 6
leo: 3
meo: 7
inclination: 10-19 deg
gso: 110
heo: 5
leo: 4
meo: 1
inclination: 20-29 deg
gso: 2
heo: 1
leo: 15
inclination: 30-39 deg
leo: 18
meo: 1
inclination: 40-49 deg
heo: 2
leo: 44
inclination: 50-59 deg
gso: 5
heo: 1
leo: 92
meo: 56
inclination: 60-69 deg
gso: 1
heo: 10
leo: 57
meo: 77
inclination: 70-79 deg
heo: 1
leo: 45
meo: 1
inclination: 80-89 deg
gso: 1
leo: 185
meo: 7
inclination: 90-99 deg
heo: 2
leo: 344
inclination: 100-109 deg
leo: 25
meo: 4
inclination: 130-139 deg
heo: 4
In terms of drag losses it is essentially a wash; the most significant drag losses occur in the supersonic regime where energy is lost in compression of the shockwave and heating of the air. And while local weather conditions can certainly have an impact on the reliability and performance of the vehicle, it isn’t really germane to what is essentially a question of orbital mechanics.
As Dr. Strangelove said, it really is largely a matter of what orbit you want to achieve. If you are looking for an orbit with low inclination (relative to the Earth’s equator) then it is definitely advantageous to launch from a lower latitude because changes in inclination are very expensive. Large telecommunications satellites launched into geostationary orbit, for instance, need to get in line with the equator. On the other hand, most satellites used for Earth surveillance, two way telecommunications, or observation missions tend to go with orbits that are somewhat inclined and often highly eccentric to increase their “dwell time” (the time they can be seen from a specified location or region the ground) for use. Satellites that need to return to the same position above a fixed location on a daily basis often occupy a heliosynchronous orbit.
One of the major drivers on what orbits you can put a satellite in, however, are governed largely by where your launch facilities are located. Each space-fairing nation only has a handful of locations developed for space launch and the range facilities to support them, and for types of launch vehicles these are the only locations from which they can launch. The permissible azimuth from these is not only governed by the latitude and vehicle capability for a given payload but also where there are lanes for permissible overflight. Flying out of Cape Canaveral Air Force Station (CCAFS), for instance, usually means flying in a somewhat southerly direction to avoid shipping lanes and not overfly Europe, but not so far south-pointing as to threaten the Bahamas. The International Space Station is at the 51.6 degree inclination because that is basically the lowest latitude that the Russian Soyuz crewed and resupply rockets can reach from Baikonur Cosmodrome, even though the now-retired US Space Transportation System (“Shuttle”) was near the limits of performance to achieve that inclination from CCAFS.
You have to remember, a lot of the satellites in LEO are now smallsat forms (<500 kg) with limited operating durations that are primarily doing some kind of Earth surveillance or thermospheric science missions, as well as ‘amateur’ satellites (e.g. CubeSats) launched primarily for the exercise of learning how to build and operate a satellite in an academic setting (although to be fair, the larger multiple-U CubeSat form has now been adopted by a number of companies seeking to do lower cost telecommunications or Earth observation). The smaller satellites often get to orbit as a rideshare with multiple payloads and so they often end up in some compromise of orbit with the co-payloads, and the 1U and 2U CubeSats are typically are just ridealongs to a primary payload and so get deposited somewhere along whatever trajectory the primary payload is taking (which was essentially the point of the CubeSat standard and the P-PODS dispenser, allowing them to hitch a ride on any space launcher with some excess capacity).
When we do near polar orbit or retrograde orbit launches in the United States we have to launch either from Vandenberg Air Force Base near Lompoc, CA, or the Kodiak Launch Complex on Kodiak Island, AK. It is also possible to launch small payloads from an air-launched rocket (currently only the Orbital ATK Pegasus, but if Stratolaunch Systems or Virgin Galactic’s LauncherOne gets up and running there will be other air-based orbital launch capability) or from an ocean platform (the defunct SeaLaunch) to a polar orbit.
Here is the Union of Concerned Scientists Satellite Database for those who are more than mildly interested.
Stranger
Yup. One of the common destinations is the ISS itself. My TLE parser gives this list of satellites with the same inclination as the ISS:
name=ISS (ZARYA), inc=51.6, apo=404.2 km
name=CHIBIS-M, inc=51.6, apo=155.3 km
name=RAIKO, inc=51.6, apo=140.9 km
name=FITSAT 1, inc=51.6, apo=169.9 km
name=TECH-ED-SAT, inc=51.6, apo=170.4 km
name=F-1, inc=51.5, apo=140.2 km
name=WE-WISH, inc=51.6, apo=162.3 km
I checked a few and they were all either deployed from the ISS or via a resupply craft.
As you know, cubesats have fairly rapidly decaying orbits, and so all of these are at a lower orbit than the ISS, and will fairly quickly decay completely. Our cubesat, SkyCube, was one of these. The complete list would be much larger, but this is just a snapshot of the ones that are currently up there.
BTW, that page says 1,419 satellites–the reason for the discrepancy between that and my 1,672 number is because I failed to remove duplicates. My list sometimes has multiple entries for the same satellite if the orbit changes rapidly–in particular, the ISS has many entries because it has so much atmospheric drag. So they project out a ways based on the drag model and have an entry for different time steps. Removing the dupes, I get 1,416 entries. Doesn’t really change the overall picture, though.
If my bracketed understanding/condensation is correct, my mental picture of this doesn’t compute: map the leading face visual cone (track field) on a sphere (Earth): is not the resultant ellipse area equivalent?
- off to check elementary math *