orbital mechanics question: time for Apollo transfer to the Earth's Moon

[munacademy]

Not sure if this forum is the right place to ask this question since it relates to the real world Earth/Moon and not to KSP bodies.

I've been calculating with some orbital mechanics. Per [1], the time required for a Hohmann transfer is:

t_H = pi * sqrt((r_1 + r_2)^3 / 8μ)

So if I want to transfer from a Low Earth Orbit of

r_1=191.2 km (this was the Apollo parking orbit from what I found on Wikipedia, anyway it doesn't really matter since it's negligible compared to r_2)

to a circular orbit equal to the Moon's semi-major axis:

r_2=384748 km (source: [2])

and considering the Earth's gravitational parameter of

μ = 398600.44 km^3 * s^(-2) (source: [3])

then the time for the transfer should be:

t_H = pi * sqrt((r_1 + r_2)^3 / 8μ) = 4.86 days

However, the *actual* time Apollo took to transfer to the moon was only about 3 days and 1 hour. Where does the discrepancy come from?

Wikipedia says that TLI only "approximates" a Hohmann transfer [4]. There are two ways which I can think of in which it only approximates the classical Hohmann transfers:

(a) The Moon's orbit around the Earth isn't circular but slightly elliptic (see [2])

( Apollo didn't transfer into an empty orbit, but into one and such a time that they intercepted the Moon. The Moon's gravitational pull would probably have accelerated the journey time somewhat.

The factor (a) is nowhere near large enough to shorten the journey from 4.86 days to 3.04 days even when the Moon is at perigee. Is the factor ( causing the discrepancy? Or is there some other consideration that I haven't considered, or maybe a calculation error or fundamental misunderstanding?

[1] http://en.wikipedia.org/wiki/Hohmann_transfer_orbit

[2] http://en.wikipedia.org/wiki/Orbit_of_the_Moon

[3] http://en.wikipedia.org/wiki/Standard_gravitational_parameter

[4] http://en.wikipedia.org/wiki/Trans-lunar_injection

[maltesh]

You're using orbital altitude instead of radial distance from the center of the earth for r_1. WHich makes the hohmann trip a little longer, actually.

That said, Apollo didn't do a true Hohmann to the Moon. They burned harder to get to the distance of the Moon faster, which both cut down the trip time and lowered life support requirements, and put them in the free-return trajectory for safety.

[LordOfTheOrbits]

You're using orbital altitude instead of radial distance from the center of the earth for r_1. WHich makes the hohmann trip a little longer, actually.

That said, Apollo didn't do a true Hohmann to the Moon. They burned harder to get to the distance of the Moon faster, which both cut down the trip time and lowered life support requirements, and put them in the free-return trajectory for safety.

Actually they didn't use a free-return trajectory, they just used a trajectory which could be changed to a free-return trajectory with minimum effort.

[munacademy]

Right, thanks for pointing that error out. When I add the Earth's radius of 6378km to r_2 (since TLI needed to happen when approximately on the other side of the Earth as the Moon) then I end up with a transfer time of 4.98 days.

Regarding "they burned harder to get the distance of the Moon faster", wouldn't such a burn have needed to occur mid-course? I can't remember reading about such a burn, but maybe I missed something.

[maltesh]

You can do the burn at from your parking orbit, you just need to time your burn differently than if you were doing a true Hohmann.

For instance, in Kerbal Space program, if you do a Hohmann from LKO to the distance of the Mun, it will take you about 7 1/2 hours to go from LKO to the 11,400 km altitude of the Mun's orbit.

If instead, you burn a little harder (about 80 m/s) to put your apoapse at the distance of Minmus, your spacecraft will cross the Mun's orbit about 3 1/2 hours after the LKO burn. The higher orbital energy of the larger orbit means that the time-of-flight from the common periapse to any altitude the two orbits share is shorter for the larger orbit.

[Kasuha]

Apollo used nearly escape velocity for the transfer:

This was followed by a Trans-Lunar Injection (TLI) burn of the S-IVB third stage for 318 seconds, accelerating the 63,531 lb (28,817 kg) spacecraft from an orbital velocity of 25,567 feet per second (7,793 m/s) to the injection velocity of 35,505 ft/s (10,822 m/s),[20] which set a record for the highest speed, relative to Earth, that humans had ever traveled.[23] This speed was slightly less than the Earth's escape velocity of 36,747 feet per second (11,200 m/s), but put Apollo 8 into an elongated elliptical Earth orbit, to a point where the Moon's gravity would capture it.[24]

The apogee of Apollo 11 translunar trajectory was at 566 thousand km, almost twice the distance to the Moon.

Edited March 23, 2014 by Kasuha

[munacademy]

Thanks folks for the responses so far! The 10.822 km/sec speed doesn't seem to be higher than what would be required for a Hohmann transfer though, unless I'm making a mistake in my calculations, so the explanation doesn't seem to add up:

At a 191km parking orbit (r_1=6569km above center of Earth), the orbital velocity would have been sqrt(μ/r_1) = 7.79km/sec.

For a Hohman transfer to an orbit equal to the Moon's semi-major axis, they would have needed to get up to a speed of

sqrt(μ/r_1) * sqrt((2*r_2) / (r_1 + r_2))

= 7.79km/sec * sqrt((2*384748) / (6569 + 384748))

= 10.923 km/sec

which is actually very slightly larger than the 10.822 km/sec quoted in the text as actually having been used.

If I assume that the Moon was at perigee (362600km rather than 384748km) which I haven't checked whether that was the case, then that slightly reduces the velocity need for Hohmann transfer to 10.9178km/sec for the TLI which is still higher than the 10.822 figure mentioned in the text.

So the 10.923km/sec doesn't seem to be consistent with the theory that they "burned harder" to get to the Moon in 3.04 days rather than 4.8 days or so? So maybe it was the Moon's gravitational attraction that made the trip this much shorter after all? Or am I missing something else?

[munacademy]

Thanks heaps for that link! That page looks very interesting. I'm heading to bed now, but will try to use that page to work through Apollo 11's trajectory more accurately tomorrow.

[munacademy]

PS. There must also be some error in my calculations. The page at

http://www.braeunig.us/apollo/apollo11-TLI.htm

specifically states that

"It should be noted that all the calculations in this web page treat the orbit as a simple two-body program, i.e. Earth and the spacecraft. Apollo 11 is assumed to coast freely in space under the gravitational influence of only Earth. The gravity of the Moon, and all other perturbing forces, are ignored."

however the table immediately following that statement has the spacecraft travelling at only 10.8343 km/sec initially (at altitude r_1=6712km, very similar to my numbers) yet still calculates the spacecraft as shooting past 395,369km in spite of not accounting for mid-course correction nor for the gravity of the moon.

So it seems to me that I started with the right velocity numbers, but my formula/calculation must have been wrong in some way such as to yield too low a target altitude.

PPS. I also wanted to mark this thread as "answered" to save you all the trouble of looking through this while I sleep or spend time to double-check my calculations, but couldn't figure out how.

Edited March 23, 2014 by munacademy

[Kasuha]

Thinking about it...

r_1=191.2 km

Didn't you forget the Earth diameter there? It may not matter for the transfer time (to apoapsis), but it may change the apoapsis position a lot. Especially when near escape speeds.

Edited March 23, 2014 by Kasuha

[Specialist290]

I've taken the liberty of moving this thread to the Science Labs, which is where threads for questions about real-world science and orbital mechanics should go. Even though it appears the question's already resolved, we don't want other people coming along later and getting the wrong idea about what goes where.

[munacademy]

Thanks folks for the responses so far! The 10.822 km/sec speed doesn't seem to be higher than what would be required for a Hohmann transfer though, unless I'm making a mistake in my calculations, so the explanation doesn't seem to add up:

At a 191km parking orbit (r_1=6569km above center of Earth), the orbital velocity would have been sqrt(μ/r_1) = 7.79km/sec.

For a Hohman transfer to an orbit equal to the Moon's semi-major axis, they would have needed to get up to a speed of

sqrt(μ/r_1) * sqrt((2*r_2) / (r_1 + r_2))

= 7.79km/sec * sqrt((2*384748) / (6569 + 384748))

= 10.923 km/sec

which is actually very slightly larger than the 10.822 km/sec quoted in the text as actually having been used.

If I assume that the Moon was at perigee (362600km rather than 384748km) which I haven't checked whether that was the case, then that slightly reduces the velocity need for Hohmann transfer to 10.9178km/sec for the TLI which is still higher than the 10.822 figure mentioned in the text.

So the 10.923km/sec doesn't seem to be consistent with the theory that they "burned harder" to get to the Moon in 3.04 days rather than 4.8 days or so? So maybe it was the Moon's gravitational attraction that made the trip this much shorter after all? Or am I missing something else?

I think I spotted my mistake. My r_1 was slightly too small, even after taking the Earth diameter into account. When calculating with r_1 = 6711.94 km (which takes into account the radius of the Earth under where the spacecraft was at TLI) rather than 6569km, I get a required velocity of 10.804km/sec for the Hohman transfer rather than 10.923km/sec. 10.804km/sec is indeed smaller than the 10.834km/sec actually used.

With r_1=6711.94km and v_1=10.834km/sec, we'd get to an apogee of ((v_1 * r_1)^2 / (2mu - v_1^2 * r_1)) = 566,380km, far greater than the semi-major axis of the Moon's orbit.

It's amazing that by just reducing r_1 a tiny bit to my incorrect original r_1=6569km, then even by increasing v_1 to my v_1=10.923, the apogee radius still is only 383,050km.

If I leave v_1 at 10.834km/sec (the correct figure) but keep r_1 at my incorrect r_1=6569km, I get an apogee of 193610km.

So just picking an r_1 that's 2.1% too small, I get an apogee that's 66% smaller than it should be. Wow!