2
$\begingroup$

While reading the NASA overview of Apollo 11, says:

Apollo 11 launched from Cape Kennedy on July 16, 1969, carrying Commander Neil Armstrong, Command Module Pilot Michael Collins and Lunar Module Pilot Edwin "Buzz" Aldrin into an initial Earth-orbit of 114 by 116 miles.

What does "114 by 116 miles" mean? Is this listing the semi-major and semi-minor axes? What are the two numbers supposed to be referring to?

$\endgroup$
3
  • 3
    $\begingroup$ Can't be the axes. Earth is somewhat bigger than that. Maybe the altitude of the spacecraft at perigee and at apogee? $\endgroup$ Jan 15 at 18:55
  • 2
    $\begingroup$ This question might be better suited to Space Exploration — there's some actual rocket scientists who hang out over there. $\endgroup$ Jan 15 at 18:56
  • $\begingroup$ I think this might be a dupe, but it's not something that's easy to search. $\endgroup$ Jan 16 at 0:30
3
$\begingroup$

The "X by Y" orbit convention is perigee and apogee altitude above some surface reference level, not the semi-major and semi-minor axes.

"114 by 116 miles" means the low point of the orbit is 114 miles above the surface, and the high point is 116 miles above the surface. You can add the appropriate Earth radius figure -- about 6378 km for equatorial radius -- to get the approximate perigee/apogee radius figures.

$\endgroup$
8
  • 1
    $\begingroup$ I'm pretty sure those numbers are referenced to Earth's nominal equatorial radius of 6378 km, not "sea level". See How is the altitude of a satellite defined, given that the Earth is not spherical? $\endgroup$
    – uhoh
    Jan 15 at 23:28
  • $\begingroup$ Fischer ellipsoid? $\endgroup$ Jan 15 at 23:35
  • 1
    $\begingroup$ Your answer there doesn't seem to be definitive; you say "roughly 6378 kilometers, or some reference radius". I'll envaguen my answer and remove the reference to sea level because I entirely don't care. $\endgroup$ Jan 16 at 0:00
  • $\begingroup$ @RussellBorogove if you can find an example of "AxB orbit" that has decimal values rather than rounded to the nearest integer kilometer or mile, then that should be revisited. However as far as I have seen, when orbits are described with this shorthand it's always rounded to the nearest integer, thus quibbling between 6378.137 km and 6378 km seems an exercise in quibbling. It doesn't matter if you care or not. This is Stack Exchange, we write, correct, and comment on answer posts for the benefit of future readers, not for ourselves. Our posts are not our personal blogs. $\endgroup$
    – uhoh
    Jan 16 at 1:03
  • 1
    $\begingroup$ @OrganicMarble no. They take periapsis and apoapsis and subtract 6378 km which is the radius of the smallest sphere that encloses Earth. If you made the reference surface anything more complicated, then you'd have to know several more orbital elements so that you could determine if the apses occurred near the equator or the poles making otherwise identically-dimensioned orbits appear to be different by 20 km. (KISS principle) $\endgroup$
    – uhoh
    Jan 16 at 1:10
-5
$\begingroup$

The following is a quote from the wikipedia article about the Apollo 11 mission

Apollo 11 entered a near-circular Earth orbit at an altitude of 100.4 nautical miles (185.9 km) by 98.9 nautical miles (183.2 km), twelve minutes into its flight.

It appears that orbit description in the form of 'X by Y' refers to near-circular orbit.

I gather from the wikipedia article that the first burn of the third stage completed insertion in a very low Earth orbit. I assume that if for any reason the mission would have to be aborted at that point in time the orbit would give the astronauts time to prepare for re-entry.

I assume that after checks were completed the go ahead was given for the second burn of the third stage, to bring the spacecraft on a trajectory to the Moon.

Once the spacecraft is on its trajectory to the Moon there is no turning back. The only way back to Earth is by going around the Moon. The trajectory to the Moon was calculated in such a way that the gravity of the Moon would swing the spacecraft around the Moon and back on course to Earth, with no need for a main engine burn, only course corrections by firing smaller thrusters.

$\endgroup$
1
  • 1
    $\begingroup$ This doesn't answer the question at all. I am asking what the numbers X and Y actually are. $\endgroup$
    – XYZT
    Jan 15 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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