# Why did Apollo moon flights take 3 days?

Ok, riddle me this, Batman:

I was too young to witness the moon landings in 1969 - 197?, but as a latter-day afficionado, one thing has always puzzled me. We have always been told that it took three days for the Apollo spacecraft to reach the
moon. For example, Apollo 11 lifted off on July 16, 1969, and the “Eagle” landed more or less three days later on July 20th.

Now, the moon is supposed to be something like 230,000 miles from the earth. I’ve read that the modern Space Shuttle orbits the earth at something like Mach 25, which, even if figured at its lowest possible speed (with Mach 1 being 750 mph), would be 18,750 miles per hour. This fits in with accounts of orbits being approximately 90 minutes, given that the earth is approximately 24,000 miles in circumference at the equator.

Anyway, to leave orbit, we can surmise that the Apollo vehicles must have accelerated to a speed greater than 18,750 mph. 230,000 divided by 18,750 is twelve and a quarter hours. Why, then, did the trip take three
days?

This was YOUR generation – Please, give me the straight dope, Cecil.

Best regards,

p.s. I realize that the Apollo missions pointed ahead of the moon’s position, much in the same way that you “lead” with a shotgun while duck hunting, in the hopes of the spacecraft arriving in the same place as the
moon at the same time. But six times longer… ???

The Apollo flights essentially “coasted” to the Moon. Remember that it’s all “uphill” from Earth surface (the Apollo missions never went into LEO), so that the speed at the crossover point (where terrestrial and lunar gravities cancel out, the “highest” point on the geodesic between Earth’s and Moon’s gravity wells) was much less than escape speed.

“Kings die, and leave their crowns to their sons. Shmuel HaKatan took all the treasures in the world, and went away.”

My Guinness Book lists the highest speed for a manned spacecraft as 24,791 mph achieved by the command and service module of Apollo X at an altitude of 400,000 feet (while returning to earth) on May 27, 1969.

This graphic gives a good representation of the flight path of the Apollo missions; the spacecraft had to cover a lot more “ground” than just the distance between the moon and earth.

http://www.nasm.edu/APOLLO/FIGURES/Fig1a.jpg

Think of it this way. If you develop a super rocket and run full speed directly at the Moon, you have to have enough fuel left to slow down (there is no friction in space, so reversing your thrust is the only way to stop), or you’re going to zip right past the Moon, or smack into it. Then you have to have enough fuel to do it again on the return.

So, NASA made the logical choice. Escape Earth’s gravity on a trajectory that allows the Moon’s gravity to pull for half of the trip. The chosen trajectory has an added advantage, in that it wraps into a Lunar orbit, so no fuel is spent (aside from minor adjustments) decellerating. The Lunar module goes down and does its thing and then blasts back into a Lunar orbit. Since the return trip begins from the Moon’s orbit, it only takes a nudge to move back into (a rapidly decaying) Earth orbit.

## So, until we develop an antimatter drive, the three day trip is the only way to go.

Stephen
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Clearly both Earth and Moon exert gravitational pull throughout the trip, as did the Sun, etc.

What you meant, of course, was something like: “…They [NASA] planned a fuel-saving trajectory that made use of the Moon’s gravitational attraction to partially counter Earth’s, with the understanding that at some “crossover” point the Moon’s attraction would be larger than Earth’s. The optimum trajectory, then, was one in which an initial burn allowed the vehicle to reach this point with exactly the velocity (speed + trajectory) needed to achieve Lunar orbit with only gravitational influences continuing to act…”

I only point this out because there are others whose lack of physics knowledge would cause them to take your shorthand references literally.

And of course, while they did damn well, as far as I know the optimum trajectory/initial burn ‘goal’ was not quite achived, and minor course-corrective burns were also required.

All orbital transfers currently in use require two burns of the engine. The engine is fired once to move the spacecraft from its current orbit into a transfer orbit, and the engine is fired again to change from the transfer orbit to the new desired orbit. The most economical way to do this is to use a Hohmann transfer orbit. A Hohmann transfer is an elliptical orbit which intersects the two circular orbits you are trying to move from/to. It is frequently used for low-earth-orbit (LEO), i.e. space shuttle range, to high-earth-orbit (HEO) transfers of satellites. However, it is also the slowest transfer orbit. That usually doesn’t matter if you are dealing with unmanned spacecraft.

Other transfer orbits are possible but all require more fuel than the Hohmann transfer. By far the greatest expense in space flight is getting to LEO. And transfer fuel is just additional payload up to that point. So minimizing the transfer fuel is a very important requirement.

Consulting the chapter on lunar trajectories in Fundamentals of Astrodynamics, by Bate, Mueller and White, we discover that the maximum time-of-flight (i.e. a Hohmann transfer) for an earth to moon transfer is 120 hours: “If we try to take longer by going slower, we will never reach the Moon at all.”

The authors then provide a plot of lunar flight time vs. injection speed (injection speed is roughly equivalent to fuel used) and make the following observation:

## So the 72 hours was a trade between the cost of extra fuel to get their faster and the cost of additional life-support required for a longer trip.

“If ignorance were corn flakes, you’d be General Mills.”