Many different sources (e.g. here, here, here, and here) say that Florida is the most common rocket launch site in the United States because it's the most southeastern part of the U.S. that is conveniently accessible (ruling out Puerto Rico), which allows rockets to be launched eastward over water and gain the maximum boost in kinetic energy from the Earth's rotation.

On paper, this makes perfect sense: if we let $M$ and $R$ be the mass and radius of the earth, then the necessary energy per unit mass required to reach an orbit of radius $r = x R$ is

\begin{align*} \frac{\Delta E}{m} &= \frac{E_f - E_i}{m} = \frac{E_f - (\mathrm{KE}_i + \mathrm{PE}_i)}{m} \\ &= -\frac{GM}{2 r} - \frac{1}{2} v_i^2 + \frac{GM}{R} \\ &= \frac{GM}{R} \left( 1 - \frac{1}{2x} \right) - \frac{1}{2} \left( \frac{2 \pi R \sin \theta}{T} \right)^2, \end{align*} where $T$ equals one day, the Earth's rotational period, and $\theta$ is the angle of the launch latitude as measured from one of the Poles. So indeed the required energy is lower the closer you launch to the Equator. But if you actually plug in numbers, you get that $$\frac{\Delta E}{m} = 6.3 \times 10^7 \text{ J/kg} \times \left( 1 - \frac{1}{2x} \right) - 1.1 \times 10^5 \text{ J/kg} \times \sin^2 \theta.$$

Cape Canaveral has a latitude that is 1.075 radians (about 57 degrees) from the North Pole. Getting to, say, the International Space Station, which orbits at altitude $x = 1.06$, from there requires an energy per unit mass of $3.320 \times 10^7$ J/kg. Getting there from, say, Virginia, whose latitude is 0.918 radians from the North Pole, requires an energy per unit mass of $3.321 \times 10^7$ J/kg - a $0.03\%$ increase. Going into higher orbits further decreases the relative energy boost from starting closer to the equator.

This energy boost seems pretty much completely negligible to me. (If anything, I suspect that the lower surface gravity near the equator due to the Earth's equatorial bulge might actually dominate the effect of the energy boost from the initial kinetic energy, although I haven't done the calculation.) It seems to me that the infinitesimal improvement in fuel requirements would be completely dominated by the fact that many parts of the East Coast of the U.S. are

  1. much more centrally located then southern Florida, and therefore more logistically accessible (at lower costs),
  2. closer to NASA Headquarters in Washington, DC,
  3. no more densely populated, and
  4. most importantly by far - not plagued with pretty much the worst possible weather for space launches.

With all due respect to the state of Florida, it actually seems to me like pretty much the worst possible place on the East Coast of the U.S. to launch rockets (other than the middle of a city). Are my calculations for the energy boost correct, and if so, then do these tiny gains really justify the big inconvenience of having to be in Florida?

  • 3
    $\begingroup$ Would space.stackexchange.com be a better home for this question? $\endgroup$ Aug 27, 2020 at 14:50
  • 2
    $\begingroup$ Remember that fuel requirements are exponential with respect to required delta-v: you need to carry more fuel to get more delta-v, but you just made yourself heavier so you need more fuel to offset the extra weight, but adding that extra fuel makes you even heavier, etc. So even a small savings in delta-v can lead to significantly lower fuel requirements. $\endgroup$ Aug 27, 2020 at 15:04
  • $\begingroup$ @probably_someone Well, $e^{0.0003}$ is still ... $1.0003$. (I know that energy isn't the same thing as $\Delta v$, but still.) $\endgroup$
    – tparker
    Aug 27, 2020 at 15:35
  • 1
    $\begingroup$ There is also the problem of finding 144,000 acres. Have you been there? The place is huge. $\endgroup$
    – JEB
    Aug 27, 2020 at 17:22
  • 2
    $\begingroup$ @tparker - to be fair, the densities were different back in the 1950’s. $\endgroup$
    – Jon Custer
    Aug 27, 2020 at 18:33

3 Answers 3


As you point out, the orbit of the ISS is not equatorial.

One aspect of spaceflight is that there is a very high propellant cost to changing the plane of orbit. Necessity to change orbital plane goes at the expense of payload.

As the Earth rotates the launch site of a rocket travels around the Earth's axis. Hence the launch site is in the ISS orbital plane about twice a day.

For a spacecraft to rendez-vous with the ISS the following procedure is used: The launch is timed to the second, such that as the second stage inserts the payload into orbit that orbit is in the same plane as the ISS orbital plane.

When inserted in orbit the spacecraft is not close to the ISS. The spacecraft may trail the ISS at very large distance. To make it to the ISS the spacecraft is inserted into orbit at a slightly lower altitude than the ISS. As a consequence of gravity being an inverse square force law: a smaller orbital altitude corresponds to a faster angular velocity.

Orbiting at lower altitude the spacecraft approaches the ISS, and when the spacecraft is slightly ahead thrusters are fired to raise the orbit of the spacecraft. That gives the rendez-vous. (It is not uncommon for a spacecraft that has been inserted in orbit to coast for more than a day to reach the ISS.)

Achieving the rendez-vous in this way avoids the cost of having to change orbital plane.

General rule: insertion in a specific non-equatorial orbit can be accomodated by precise timing of the launch.

There is of course a class of orbits for which the cost of having to change orbital plane is unavoidable: equatorial orbits.

It is specifically for equatorial orbits that it is a disadvantage to have the launch site not on the Equator. While many satellites go in non-equatorial orbit, some of the launches will be for satellites that will be inserted into equatorial orbit (I assume mostly satellites in geostationary orbit). For that class of satellites: the closer the launch site is to the equator, the better.

Overall: my understanding is that the idea of getting-kinetic-energy-boost-from-Earth-rotation is not relevant.

My understanding is that the decisive factor is propellant cost of having to change orbital plane.

[Later edit]
If Earth would only be slightly more massive (hence stronger gravity) then getting payload to orbit would effectively be impossible. With the current rocket technology: payload to orbit is about 2 or 3 percent of the take-off mass. That is, getting any payload to orbit is just barely possible.

The time from lift-off to insertion in low Earth orbit is in the order of minutes. That is such a small fraction of a day that the amount of Earth rotation during those minutes is not significant. Prior to launch the mass of the rocket is circling the Earth's axis. Upon insertion into orbit the spacecraft is orbiting the Earth's center of mass. That shift is crucial. If the launch site is at 30 degrees latitude then the most energy efficient orbit to insert into is an orbit with an inclination of that 30 degrees. Any other orbital inclination requires additional propellant.

A general law of acceleration is that efficiency is optimal if all the acceleration is in alignment with the existing velocity. Conversely, in order to change direction of velocity accelerating force is applied at some angle to the existing velocity. You can decompose that acceleration vector in a component aligned with the existing velocity and a component at right angles to the existing velocity. The component at right angles requires additional propellent.

  • 3
    $\begingroup$ The plane-change limitation is perhaps the biggest benefit of a low-latitude launch site. You can only insert into an orbit at your launch site inclination or higher, but never lower! Getting to lower inclinations takes an expensive plane-change burn. The ISS is in the orbit it is in so it can be serviced by the US and Russia without requiring plane-changes to lower inclinations when launched from Russia. $\endgroup$
    – tpg2114
    Aug 28, 2020 at 21:54
  • 1
    $\begingroup$ This answer is way too long for something quite obvious. -1 $\endgroup$ Aug 30, 2020 at 11:38
  • 1
    $\begingroup$ Do you have quantitative evidence? I mean, the math involved? This answer, although extremely long, doesn't show there is a significant difference between the two launch sites. Besides that, Cape Canaveral was already there wasn't even a thought yet about an ISS. Well, maybe a thought, yes, but the actual s**t wasn't present then. $\endgroup$ Sep 3, 2020 at 11:25
  • $\begingroup$ @descheleschilder For orbital mechanics our daily life intuition is not suitable. The subtleties of orbital mechanics are often underestimated, or even not suspected at all. This lengthened my answer; I felt I could not rely on the reader being familiar with orbital mechanics. Obviously, at the time of first development of a national launch site the ISS was not part of the thought process. Obviously the considerations were general: what launch site will be suitable to acommodate both launches to equatorial orbit (geostationary) and to non-equatorial orbit (presumably: spy satellites). $\endgroup$
    – Cleonis
    Sep 3, 2020 at 21:15

The main point is not the energy, but the speed. (In essence, that's @probably_someone's comment). It's best explained here, of course.

As your computation shows, the energy of the ISS is predominantly kinetic energy, about $3\times10^7\text{ J/kg}$, compared to about $4\times10^6\text{ J/kg}$ potential energy.

The required mass (of rocket plus fuel) is exponentially related to the desired change in speed, so that even a 35% increase in initial speed (from latitude of 48° to 24°) makes a significant difference (and of course, starting towards the east).

Note that, of course, various practical and political considerations are important, but it is no coincidence that not only Cape Canaveral is close to the equator, but also Baikonur and Kourou. In partiucular, Kourou is much less convenient to reach from Europe than destinations actually in Europe. Nevertheless, both the European Space Agency and the Soviet space program chose sites that were close to the equator, within the accessible range.

  • $\begingroup$ If I understand correctly, you're saying that if spacecraft were launched from a mass driver, then the kinetic energy boost from the Earth's rotation would indeed be completely negligible, but since they need to carry along their own fuel, kinetic energy is in practice more costly to achieve than a simple energetics analysis would suggest? $\endgroup$
    – tparker
    Aug 28, 2020 at 20:00
  • $\begingroup$ Baikonur is located at 45 degrees latitude (according to the wikipedia article you linked to.) The ISS orbit has a 51.64 degrees inclination. tparker provided the link to a Scientific American publication on the question: why launch from Florida? According to Smithsonian Air and and Space Museum curator Roger Launius: if the ISS orbit inclinatin would be closer to equatorial then it would be harder for missions from Baikonur to reach the ISS (hence less payload per mission). $\endgroup$
    – Cleonis
    Aug 28, 2020 at 21:17

I'm taking your equations and calculations for granted. I don't have any doubt (seen in the light of your reputation) that they are correct, because of the final result (made in an ideal situation) has about the same value with only a minor modification. The answers above are somewhat opinion-based.
The best place would be somewhere close to the equator, of course. The speed given to the rocket is optimal there. Of course, it depends where you want to go but you can take the launch time into consideration to get around this problem. Or launch the rocket non-vertically.

The situation is not ideal though which is not because of physical conditions.
So, what are the conditions for making places outside Florida?

I don't have evidence for this but I think it has to do with decisions made in the past. Political (near to Cuba, after the crisis in 1961, the site was built in 1963, although missiles with an atomic warhead could be launched from many locations, the one in Florida is practically in Cuba's backyard) as well as economic decisions (based on lobbying; which State doesn't want to have such a site within its boundaries?), which can't be turned back anymore once the launching site has been built.

That's why I doubt physics is involved in answering this question. A huge variety of launching sites could be chosen with even better characteristics for the desired result.

  • $\begingroup$ Why the downvote? $\endgroup$ Aug 30, 2020 at 11:41
  • $\begingroup$ I didn't downvote, but probably because there's not really any physics in this answer, just some unevidenced speculation. $\endgroup$
    – Nat
    Sep 11, 2020 at 10:42
  • $\begingroup$ @Nat Unevidenced? Lots of evidence, even an abundancy. Don't forget the cold war was at its height back then. $\endgroup$ Sep 21, 2020 at 10:54
  • $\begingroup$ It's pretty speculative. The speculation's a tad far-fetched, too, as it's unclear what military advantage could've even potentially been had. I'm not quite clear on what your specific speculation might be, but I get the vague impression it's about ICBM launchers at Cape Canaveral? What'd be the point of that against Cuba? $\endgroup$
    – Nat
    Sep 21, 2020 at 11:38
  • $\begingroup$ @Nat The Cuba crisis brought the world to the edge of a nuclear war with the USSR. In 1961. Three years later rhe rocket launch site was built in Cape Canaveral. A coincidence? I doubt it. Florida is close to Cuba. Maybe it deterred the Russians to do the same thing. $\endgroup$ Sep 21, 2020 at 14:15

Your Answer

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

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