If the Earth were spherical, I would agree with this. But the Earth is not spherical. And at the equator, the radius of the Earth is 21 km greater than at the poles. By convention, the altitude of the spacecraft is equal to the distance to the center of the Earth minus approximately 6378.137 kilometers. If the altitude of your target orbit is 200 km, then from the pole you need to rise to 221 km (+21 km relative to the equator).

Even taking into account the fact that you have to counteract the velocity of earth’s spin if you launch from the equator, I think that due to the differences in heights, launching from the equator will still be more effective. Сorrect me if I am wrong!

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    $\begingroup$ compare the energy difference in height against the energy difference in velocity for your example orbits $\endgroup$
    – Erin Anne
    Commented Jan 30 at 11:12
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    $\begingroup$ The whole argument is nonsense; it's always better to launch a satellite from wherever your established launch site facility is located, regardless of trajectory. $\endgroup$ Commented Jan 30 at 16:09
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    $\begingroup$ @RussellBorogove - for example SpaceX does not launch from the near the equator (other than Falcon 1 at 9° latitude) even though it would be "better" for many types of launches. However it seems that the intent of the question is about energy advantages, not overall operational logistics. A better wording would be something like, "is there an advantage to launching a satellite into polar orbit from the pole than from the equator? I think the OP could make this type of edit without affecting the existing answer. $\endgroup$ Commented Jan 30 at 17:51
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    $\begingroup$ @RussellBorogove And that's why no space agency ever established a second launch site? $\endgroup$ Commented Jan 31 at 4:47
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    $\begingroup$ the earth is not spherical because it deforms until the entire surface is equipotential when considering the sum of gravitational potential energy and rotational kinetic energy. The energy cost to go from polar ground height to equatorial ground height is equal to the energy cost to remove equatorial rotation speed in the absence of air resistance, mountains, etc. $\endgroup$ Commented Feb 1 at 20:18

3 Answers 3


When looking only at the required $\Delta v$, and modeling Earth as a uniform oblate spheroid, launching into a 90 degree polar orbit from the equator is indeed more expensive than doing so from one of the poles, despite the altitude advantage, but only marginally.

First, let's look at the cost of canceling the Earth rotation velocity when launching from the equator. At the equator, Earth's surface is moving 465 m/s to the east relative to its center. Meanwhile, for a target orbit altitude of 200 km, the required orbital velocity is about 7800 m/s, and we want to be going north or south to hit the 90 degree orbit inclination. To cancel the eastward motion, basic trigonometry shows we need to launch our rocket toward an azimuth about 3.4 degrees (atan 465/7800) west of north or south, with an ascent profile such that its final velocity relative to the launch site is 7814 m/s (by the Pythagorean theorem), as shown in this diagram:

Hypotenuse to cancel rotation

Thus, it costs about an extra 14 m/s of $\Delta v$ to cancel the rotation when launching to polar LEO from the equator.

Meanwhile, if we launch from one of the poles, we don't need to cancel any rotation, but we're about 20 km further below our target altitude. (Note: For the purposes of this crude calculation, the atmosphere is essentially the same at both locations. The troposphere is thinner at the poles, but that is not relevant, as its edge is determined by the temperature gradient. At all surface locations (assuming calm weather!), the atmosphere is in hydrostatic equilibrium, leading to a density profile that is practically the same for our purposes, the relevant portion of which is entirely inside the troposphere at both locations.)

So, let's assume we use the same ascent profile as before, but initially targeting an altitude 20 km lower, thus expending the same $\Delta v$, minus the 14 m/s. Then, let's use a Hohmann Transfer to climb the extra 20 km of altitude. (We can use this calculator, among others.) It costs a combined 12 m/s of $\Delta v$ to do the two burns of that transfer.

However, that's not the most efficient mission plan, since it's better to launch directly to the target altitude, and then circularize a single time once there, than to make a stop in a parking orbit partway up. So 12 m/s is an upper bound on the cost of the extra 20 km of altitude.

Since the equatorial launch required 14 m/s to cancel the Earth rotation, and the polar launch required at most 12 m/s to gain the extra 20 km of altitude, we can conclude that the polar launch is more efficient, when only considering $\Delta v$.

Of course, even if the difference was the full 14 m/s, that would not come close to compensating for the enormous logistical costs of launching from either pole. Practically speaking, if you want to get into a polar orbit, you launch from wherever is logistically convenient and has a clear range to either the north or south.

  • $\begingroup$ How would the choice of polar versus equatorial launch point affect the required thrust/mass ratio at take-off and the size of engines required to achieve it? $\endgroup$
    – supercat
    Commented Jan 30 at 22:01
  • $\begingroup$ Interestingly the OP did not specify a timeframe, so we don't know if they are asking about using current technology which is very labor intensive, or technology say 100 years in the future. Obviously the physics won't change in 100 years, but the technology certainly will, we just don't know how. It's easy to imagine space launches becoming much more heavily automated in the future, in which case the economics might tilt in favor of polar launches far up into the polar circles, even if not necessarily directly on the poles. And maybe more eastward launches from closer to their ideal latitude. $\endgroup$ Commented Jan 30 at 23:41
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    $\begingroup$ @supercat I'm not aware of anything that would significantly affect required TWR or engine size, assuming sea-level launches. (The ice surface right at the south pole is not at sea level, but I omitted consideration of that under the "uniform oblate spheroid" assumption.) $\endgroup$ Commented Jan 31 at 1:11
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    $\begingroup$ @PM2Ring You're right, that was clumsy phrasing. I wasn't meaning to say gravity was stronger at the pole, just suggest a frame of reference that isn't tied to a surface location. I've changed it to say "20 km further below the target altitude". $\endgroup$ Commented Jan 31 at 23:47
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    $\begingroup$ @uhoh It's not clear how to do a meaningful comparison since SSO can't be reached from a pole (without a plane change maneuver). Maybe you have in mind to compare with launching westward from (say) 83 degrees N/S? $\endgroup$ Commented Jan 31 at 23:57

When a rocket launches from the equator, it also has the velocity from Earth's rotation. Normally this would be a good thing because then a rocket does not need as much delta V. Since Earth is rotating at 1 670 km/h at the equator, that is a free 463,8 m/s of delta V.

enter image description here

However, a rocket launching into a polar orbit would have to use more fuel since it would have to counteract the movement of Earth's rotation. Earth rotates the slowest at the poles and the fastest at the equator. This means that the best place to launch into a polar orbit is the poles because there the Earth is rotating the slowest and the rocket requires less fuel to counteract the rotation of Earth.

The height difference plays a really small role. Rockets normally have to reach about 8km/s to reach orbit, maybe also another 2km/s to get above the Karman line. Starting 20km higher does not really change that much in terms of delta V. This video by the Everyday Astronaut explains the situation really well. The delta V saved by not having to counteract the rotation definitely is much more than the delta V saved by starting 20 km higher.

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    $\begingroup$ Mandatory xkcd. "getting to space is easy. The problem is staying there". $\endgroup$
    – jcaron
    Commented Jan 30 at 18:33
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    $\begingroup$ Other than the first few seconds of liftoff, there aren't gravitational "losses" to speak of. It's incorrect to think of an orbit as [Energy to gain altitude] + [Energy to gain horizontal velocity]. The horizontal velocity itself is what determines the altitude of an orbit, and the only reason we spend fuel going up first is so that the initial minutes of building that horizontal speed pass through less atmosphere (and landscape). $\endgroup$
    – Jay McEh
    Commented Jan 30 at 19:30
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    $\begingroup$ @JayMcEh: Only a tiny fraction of a mission's time will be spent during "the first few seconds of liftoff" , but the rate at which fuel is burned during that time will be so much greater than the rate of fuel burn later in the mission that the fraction of overall mission fuel will be significant. $\endgroup$
    – supercat
    Commented Jan 30 at 21:31
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    $\begingroup$ @supercat can you give an example of a booster where the "rate at which fuel is burned" (i.e., kg/s) is significantly greater at liftoff than a few seconds (or even a minute) later? $\endgroup$ Commented Jan 31 at 1:55
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    $\begingroup$ @JayMcEh The term "gravity losses" (some use "gravity drag" instead) reflects the fact that launch is not impulsive. A finite burn performed in a non-constant gravity field inherently requires more energy than does an impulsive burn that achieves the same end. $\endgroup$ Commented Jan 31 at 16:12

Not sure how much of a factor it is in total launch budget, but atmosphere thickness probably has an effect on launches between the two locations.

Atmospheric thickness varies, it is thickest at the equator and thinnest at the poles, meaning launching from the equator would encounter more drag during it's longer path through the thicker atmosphere.

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    $\begingroup$ What really matters is density. The polar air is more dense than tropical due to temperature differences. The thickness, while significant (6km vs 20km) is overshadowed by the rotational speed. $\endgroup$ Commented Jan 30 at 21:19
  • $\begingroup$ Launching from either pole has environmental and logistical issues - Cannot pollute in Antarctica, and the Arctic lacks stable land (its all ice) $\endgroup$
    – Criggie
    Commented Feb 1 at 4:23

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