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If a spaceplane like the Rockwell X-30 existed and was flying to a location in geostationary orbit, would it be substantially more efficient to take off from a runaway at the equator, or would it not make that much of a difference?

This question is not about craft designed and powered solely by rockets for an airless environment, it is about spaceplanes and functioning as an aircraft up to the Karman Line.

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  • $\begingroup$ You might want to ask this question on space. There a plenty of KSP fans over there. $\endgroup$
    – Aron
    Jun 18, 2019 at 3:03

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It's best to be right at the Equator for a launch to Geostationary Orbit. You also launch east to get some of earth's rotational speed.

When launching, the orbital plane's inclination will be the degree of latitude or more. With a 0° latitude, you can pick your inclination - 0° in this case. This saves very costly changes to the inclination (among the most rocket fuel consuming maneuvers. For reference, since Russia is really high in the north, they need a whole additional rocket stage just to change the inclination).

Since you want to launch east, you'll need some empty space in that direction, since people get mad if you launch over their territory and your launch fails.

Best places to launch are places like:

  • East coast of Brazil, near Macapá

  • Southern Somalia, around Jamaame

Both are at the equator and have a lot of empty ocean east of them. For political reasons I'd prefer Brazil.

Probably the best existing space port would be Kourou in French Guiana, the ESA's "best" spaceport.

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In addition to what Bilbo Baggins said, another major reason to use equatorial (or at least as close to equatorial as you can get politically) launches is that geostationary orbit is only possible at 0 inclination, i.e. directly above the equator. If you launch from any other latitude, you have to make a second burn of the engines once you're in orbit to correct your inclination, and the further from the equator your launch site is, the bigger this burn has to be. This is because your inclination can't be lower than your latitude. For instance, rockets launched from Cape Canaveral, Florida can't enter an orbit lower than about 28 degrees inclination in a single burn, because Cape Canaveral is about 28 degrees north of the equator. These burns can be really expensive as far as your fuel budget goes, and the burn generally has to be done in orbit, which drastically increases how much weight you have to lift into orbit in the first place. Look up "the tyranny of the rocket equation" for more about this.

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    $\begingroup$ I understand the issue of the fuel budget for rockets and inclination, but I was wondering about a craft that lifted off from a runway, and flew to the Karman line as an aircraft. I assumed the fuel budget for an aircraft to get from anywhere above or below 0° inclination to 0° inclination would not be as significant as that of a rocket. $\endgroup$
    – Bob516
    Jun 17, 2019 at 16:59
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    $\begingroup$ @Bob516 it depends on just how much better your space plane is than rockets. If you gain more delta-v from the space plane than you lose from the sub-optimal launch site, it will be worth it, in terms of delta-v budget anyways. Financially is best left to another question. $\endgroup$
    – Ryan_L
    Jun 17, 2019 at 17:08
  • $\begingroup$ I had assumed it would take an aircraft less fuel to change its inclination compared to a rocket because an aircraft is using lift from the airfoil to make the change compared to a rocket which can only change inclination through the expenditure of fuel. Am I missing something? $\endgroup$
    – Bob516
    Jun 19, 2019 at 4:21
  • $\begingroup$ Lift ultimately results in an expenditure of fuel though, because any d/v the turn takes has to come from somewhere. Drag cannot be avoided if you rely on lift, and any velocity lost to that drag will have to be made up for by running the engines longer. You might be better off flying south, low and slow to the equator, using your most fuel efficient engine settings, and then you turn east and head for orbit. Turns at low speed cause less drag than turns at high speed. This is definitely true if you don't need it to be an SSTO, and can use a carrier plane like Virgin Galactic. $\endgroup$
    – Ryan_L
    Jun 19, 2019 at 5:13
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The centrifugal force would not really make an effect. According to Wikipedia, objects weigh "about 0.3% less at the equator than at the poles."

However, the eastward rotation of over 1000 mph would help some. In fact, it is one of the reasons why NASA launched rockets from Cape Canaveral.

Also, launching from the equator will prevent having to angle back into the orbital plane of the earth to get into a stationary orbit. For better information about geostationary orbit costs, go here.

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    $\begingroup$ It's even more important for geostationary orbit. If you launch from anywhere other than the equator, it requires a second burn of the engines once you're in orbit to correct your inclination; geostationary orbit is only possible with 0 inclination. $\endgroup$
    – Ryan_L
    Jun 17, 2019 at 3:37
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    $\begingroup$ @Ryan_L, you nailed the main reason for launching from at or near the equator. That 2nd burn to turn an inclined orbit into an equatorial orbit, can be very "expensive" in terms of fuel usage (and hence overall weight). I'd suggest submitting it as an answer. $\endgroup$
    – JDM-GBG
    Jun 17, 2019 at 4:12
  • $\begingroup$ @BilboBaggins Welcome to the Worldbuilding Stack Exchange. You are correct that launching from the equator imparts the greatest initial speed from the rotation of the Earth. My question is focusing on reaching a location in geostationary orbit, not the most effective way to make use of the the rotation of the Earth. $\endgroup$
    – Bob516
    Jun 17, 2019 at 4:17
  • $\begingroup$ Would the spaceplane changing to a 0° inclination from one that is 10° off, while still in the atmosphere and operating as an aircraft, require that much extra fuel that it would not make sense to take off from that location? $\endgroup$
    – Bob516
    Jun 17, 2019 at 4:18
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    $\begingroup$ @Bob516 that depends on how much better than conventional rockets the space plane is. Take the figure you get from Bilbo's calculations and compare to how much better the space plane is. If the Space Plane gets more than 100% more d/v than the rocket, the space plane is better even with the poor launch site. If it's less than 100% more d/v, the space plane is worse, from a mass-to-GEO standpoint; it may still be better from a cost standpoint. I don't think that's a question any of us can answer. $\endgroup$
    – Ryan_L
    Jun 17, 2019 at 5:16
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Altitude provides a better return on launch efficiency because altitude intrinsically reduces distance of travel and air resistance.

We don’t build our launch sites on mountain tops because we do so few launches that infrastructure costs would be comparatively too high. But, if you are describing a planet or country with a thriving space based economy then the cost of the infrastructure to support getting passengers and cargo too and from Pikes Peak in a CO or Mt Denali in Alaska or Everest in Nepal (?) could be amortized across 10K or more launches, greatly reducing the operating costs of space flights

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    $\begingroup$ Altitude helps, but only slightly. The main thing you need to achieve orbit or escape Earth's gravity well is speed. That's why rockets begin turning on their side pretty quickly into a launch, and it's one reason we don't care to use balloons to lift rockets before firing them. I'm not convinced you can just dismiss an equatorial launch vs a high altitude one in this case. $\endgroup$
    – Harabeck
    Jun 17, 2019 at 14:01
  • $\begingroup$ @Harabeck, for every mile of altitude over sea level a rocket is launched from the work required to reach orbit is reduced by ~1.6%. And that doesn't account for reduced drag due to lower air resistance. Another answer sites a 0.3% boost for equatorial launch. $\endgroup$
    – EDL
    Jun 17, 2019 at 15:31
  • $\begingroup$ Bilbo's answer cites 0.3% as the reduction of weight of object at the equator vs the poles. That is not the total "boost" for an equatorial launch, which also includes ~1000mph of speed the rocket does not have to generate itself. That speed boost is reduced as you move away from the equator. And again, the speed is by far the most dominating factor. $\endgroup$
    – Harabeck
    Jun 17, 2019 at 15:47
  • $\begingroup$ The tangential velocity coming from the rotation of the earth is ~ 1Kmph*cos(latitude). Launching from Mt Everest at 25 degrees north reduces the tangent component by < 1% but reduces the Work 8%. A 100 PMH difference in initial tangental boost does not dominate altitude reductions due to Work and drag. The best launch place is a high altitude place near the equator and the worst is a sea-level point on the north pole. I'd say south pole, but it's mountainous $\endgroup$
    – EDL
    Jun 18, 2019 at 19:28

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