Did the moon lander have enough fuel to get off the moon?

[dr_g]

Yeah, "Dead Man" area (a term I am told is not preferred by the "professionals"' so, sorry about that!) was the altitude where it was too late for a stage separation. Idea then would be to loose any forward (or lateral) movement ad drop as quickly as possible for a soft(ish) landing. Then a stage sep could occur safely and they could ascend if necessary.

Now we know why Charlie Duke, Apollo 11 landing CapCom, sounded like his pants were a bit too tight when he said "30 seconds!" into the mike.

As I recall, the popular version was that the descent stage was down to about 17 seconds of hover/landing time, but later the estimate was rounded up to more like 30-40 secs, as the anti-slosh baffles had compromised the low-fuel reading. Heck, my Chevy does the same thing, tho the results of power loss are bit less dramatic.
As dramatic was the ice fuel plug in a descent stage fuel line that started building up pressure in the fuel system after touchdown. I don't think they were clear on whether or not it would result in a burst helium disk or a real explosion, but the latent heat in the descent engine melted it out and all was well. But there were some heart-stopping seconds on a couple of mission control consoles.

Re the original poster, I too welcome newbie questions, but a bit of research beforehand never hurts. I suspect that LM graphic was off of Wikipedia. and much of what is clarified here would be clearly seen there after a careful reading.

Sorry to diverge from the thread topic and follow this zepher but I heard the landing recounted first hand at dinner one night with one of the crew. He said they were steady at 50 feet altitude tracking down range trying to find a suitable landing site , as the original one was strewn with large boulders. I was told there were two seconds of fuel remaining but did not think to qualify if that was absolute or the time ( in fuel cap) before go - no go for landing. I was left with the impression that it was 2 seconds left until abort. However your 17 seconds minus 2 sounds about right. 15 seconds with 50 feet of altitude was probably the cut off point.
Nice add about the iced fuel line!

 

[MasterMiend]

Another factor not mentioned is that much of the weight of the original launch vehicle was to support the massive engines and fuel tanks and so on under earth gravity.

On the moon, much less structural strength is required and a simple 1-stage launch vehicle is possible.

Also thinking back on the propellant, both the fuel and oxidizer are propellant, since they both react and their mass produces thrust. In that case, a large portion of the mass shown in the diagram is propellant.

Engineers had considered using those propellants on rockets in the past, but minor leaks created enormous explosions. Many of those videos of exploding rockets are the failed attempts. Hydrazine fuel was abandoned except for attitude control (the reason astronauts quickly vacate a landed shuttle and the area is locked down). Satelites also use hydrazine fuel (or hypergolics as they are also known), although small, safer ion thrusters are replacing them.

About the abort on descent, I think that it would be very difficult to release half of your vehicle while descending. If it did not detach simultaneously all around, the mass tugging unevenly on one side could wreck your vehicle as it spins over.

How do you control a lower stage that is ignited while you separate? Will it continue firing and ram back into you when the weight of the upper stage is lifted? What is your ascent vector?

It would be safer, if given sufficient reserve fuel, to burn everything in the lower stage to increase altitude. When the fuel burns out, separate and then ignite the ascent stage when clean separation is confirmed.

Also consider, the moon landing was a powered descent, keeping the engines burning so they did not approach the ground too fast. In the final approach is when most of the fuel would be burned to bring the vehicle to a stop. So before that you are going pretty fast. That last 300 feet is pretty terminal. All the fuel is needed to come to a stop. Someone mentioned momentum before. That is part of it. You are going to impact. It is too late to change events once you are committed. The upper stage has enough fuel to go from stationary on the ground to orbit and not much fuel for anything else. So to de-orbit, come to a stop and then to ascend again would be out of the question. Those vehicles were designed with very little margin for error.

On the ground, your position is fixed. You can time your ascent to coincide with your mothership's orbit so you can meet up. On the other hand, if you are in the middle of a descent and you abort, how will you get back to the mothership to dock with the fuel you have? That may also be a factor.

 

[Mighty]

I'm pretty sure the low fuel limit was the go/no-go abort limit. They would power up to a safe altitude to separate from the descent module.

But, I've read that Neal and Buzz had decided together that they were probably going to push past that limit if they thought they saw a good landing spot.

 

[wtrix]

This is a simple physics question. The LEM must obtain a minimum kinetic energy of (.5)(mass of LEM)(escape velocity of moon ^ 2) to escape the moon and get back to the command module. this is just an estimation, since the LEM actually becomes lighter as it burns fuel, so the total KE needed to achieve orbit would be lower than the figures stated below if the LEM's engines were firing. (It would require calculus to get a more exact answer). These are a "worst case scenario" figures based on a LEM of unchanging mass...

Changes in mass produce a linear increase in fuel demand. Changes in velocity require an exponential increase in fuel demand. The basic formula is KE = 1/2 M V^2 .

For example, the 4670KG loaded LEM would require the following energy:

KE= (1/2) * 4670KG * (2380 m/s)^2 = 13226374000 Joules = 1.322 * 10^10 J

To launch the same mass from the surface of the Earth to orbit:

KE= (1/2) * 4670KG * (11,200 m/s)^2 = 2.92 * 10^11 J

In other words, it takes about 5% as much fuel to achieve lunar orbit than Earth orbit for the same given mass.

And that's not all because the surface gravity on Moon is 1.622 m/s² vs. surface gravity on Earth which is 9.78 m/s². In rocket dynamics You first have to achieve zero weight (means that the forces of the rocket push and the gravity pull are equal) and then You can start to accelerate the objects mass (4670kg, which in acceleration terms is unrelated to gravitational pull - http://en.wikipedia.org/wiki/Mass_versus_weight) to the speed necessary. Thus to get the lunar module to afloat is 6 times easier on Moon compared to doing this on Earth as of instead of 4670kg you are lifting only 778kg i.e. the rocket has to lift 778kg, but it must accelerate the full 4670kg multiplied by acceleration.

A mistake is also made on above cause the Ascent Module does not have to be accelerated to lunar Escape velocity (2380 m/s), but only to orbital velocity, which is 1022 m/s. Escape velocity is achieved by Control and service Module (when the lunar ascent module is discarded). Plus You have to carry less fuel (cause You have to lift less weight and make the mass move slower. And the mass is lower because you carry less fuel), don't have to fight aerodynamic forces (which at the speed of sound and above are considerable) and You don't have to get as high above the surface before you enter the orbit. And last, but not least - all rocket engines are roughly 25% more efficient in vacuum than in atmosphere end even more so if they are designed only for operating in vacuum.

Put that all together and You have multiple magnitudes less of fuel required.

 

[Anonymous_John]

At least the discussion of launch aborts is interesting.
Nobody has mentioned the faulty switch that could have resulted in an abort of the Apollo 14 LM. I'm curious about what would have happened then. Reaching orbit is one thing, docking with the CSM quite another.
The Shuttle only has a short launch window to dock with the ISS. I can't quite see why launching from the Moon and docking in lunar orbit should be any different.

 

[marktaff]

A mistake is also made on above cause the Ascent Module does not have to be accelerated to lunar Escape velocity (2380 m/s), but only to orbital velocity, which is 1022 m/s. Escape velocity is achieved by Control and service Module (when the lunar ascent module is discarded). Plus You have to carry less fuel (cause You have to lift less weight and make the mass move slower. And the mass is lower because you carry less fuel), don't have to fight aerodynamic forces (which at the speed of sound and above are considerable) and You don't have to get as high above the surface before you enter the orbit. And last, but not least - all rocket engines are roughly 25% more efficient in vacuum than in atmosphere end even more so if they are designed only for operating in vacuum.

I'm so glad someone brought up the issue of orbital velocity versus escape velocity. Saves me the trouble.

 

[drwayne]

Very tangetially related point. The Apollo spacecraft did not reach Earth escape velocity either, it was
in an orbit that took it out to the moon, and for many missions, mostly let the moon do the rest.

Wayne

 

[marktaff]

At least the discussion of launch aborts is interesting.
Nobody has mentioned the faulty switch that could have resulted in an abort of the Apollo 14 LM. I'm curious about what would have happened then. Reaching orbit is one thing, docking with the CSM quite another.
The Shuttle only has a short launch window to dock with the ISS. I can't quite see why launching from the Moon and docking in lunar orbit should be any different.

They are two very different things. To begin with, the space station is essentially motionless with regards to the shuttle docking--that is, the station will not adjust its orbit to try and get to the shuttle. Basically, the shuttle slows down until its orbit intersects the station's orbit when their separation is zero meters.

In the case of an emergency abort to orbit during an Apollo lunar landing, the command module would be able to adjust its orbit to help the LM dock with it, if needed.

Secondly, when the shuttle launches, it has an "orbit" defined by a point on the surface of the Earth (the launch pad) as the Earth revolves. That orbit is *far* different than the station's orbit. For any shuttle to station mission, the shuttle needs to be able to spend x days closing with the station, y days on station, and z days returning home, plus some safety margin. Given the energy requirements to change the shuttle's orbit, plus the limitations of the shuttle (fuel, waste, electricity, air), there are limited windows when all of the parameters are met.

In the case of the Apollo emergency abort to lunar orbit, the LM begins in the *same* orbit as the CM. It then fires its rockets to change its orbit to one that intersects with the lunar surface. If it aborts, it only need to use enough energy to undo its initial deorbit burn, to account for any translation, and to adjust its orbit so that it *just misses* the lunar surface. This new orbit can then be circularized into an orbit that eventually dock with the CM.

Because of the moon's low gravity, and especially the lack of an atmosphere, this is fairly easily done. This would be a far more energy-intensive option if we were talking about Mars. I suspect an abort to orbit during a Mars landing wouldn't even be feasible, at least not with chemical rockets, or without a rocket fuel gas station in Mars orbit.

 

[Mighty]

One big difference in Moon vs Earth orbital discussions is the difference in the rate of rotation of the two bodies.

The reason the shuttle has only a five minute window to get to the station is that changing orbital planes is *very* expensive in fuel. Kennedy Space Center is at about 28 degrees above the equator. Therefore, the "natural" orbital plane from there is also 28 degrees. The ISS is at 51 degrees, because that's what the Russians needed to launch from Baikonour. Getting from 28 degrees to 51 degrees is right at the edge of what the shuttle is capable of. And then only when starting from the ground, where it's speed is the rotational speed of the Earth. Somewhere under 1/4 mile/second at Florida. Once in orbit there's no way to change the orbital plane even a couple of degrees.

The Moon, OTOH, rotates only once every 30 days. The orbital plane at the start of descent is chosen to make it efficient to land at the intended spot. IOW, it's pretty much a straight shot down. In the case of an abort, getting back up and into the orbital plane of the CSM doesn't take much extra fuel. Slowing down to the rotation rate of the moon and then getting back out to orbit simply doesn't put it very far out of plane.

Take for example Apollo 14. They actually stayed in orbit for one extra orbit and still had the fuel necessary to get over to where the original landing site had rotated to.

Once you're in-plane, then catching up or slowing down to match the target requires nothing but time. Raise the orbit a little to slow down and let the other guy catch up. Or lower the orbit a little to speed up. Depending on how quickly you need to close the distance, it can take only ounces of fuel.

On Earth, there are two times each day, 12 hours apart, where the orbit of the station passes over KSC. It takes longer than 12 hours to reset for a launch attempt, so the shuttle ends up with just one launch attempt per day. The reason the shuttle can have multiple landing attempts is that it uses the atmosphere to provide a lot of change in direction necessary. On launch, all that change has to come from the fuel.

 

[odiedb]

I thought the Lunar Lander used Dimethyl Hydrazine and Nitrogen Tetroxide as fuels... These combust when allowed to come in contact with each other... The Command Module utilized the same... These were called the old faithful fuels...The question asked is misleading in that the whole of the lander could not get back off the Moon, only the upper stage. Remember on the first landing they only had what 10 seconds of fuel in the descent stage left when touch down came.... Each section carried just enough to do what it was supposed to and not much more......................

 

[odiedb]

Oh yes, and on Apollo 10, which was my favorite, even though a landing never happened, the LEM went out of control 10 miles above the surface of the Moon! The emergency manuver of separation between the upper stage and the lower landing stage took place. I remember the Astronauts voices well over the airways: To quote: "Son of a *****!" They at the time did not know what had happened. Fear was in their voices as they tried to bring the space craft under control!

 

[mortisthewise]

This is a simple physics question. The LEM must obtain a minimum kinetic energy of (.5)(mass of LEM)(escape velocity of moon ^ 2) to escape the moon and get back to the command module. this is just an estimation, since the LEM actually becomes lighter as it burns fuel, so the total KE needed to achieve orbit would be lower than the figures stated below if the LEM's engines were firing. (It would require calculus to get a more exact answer). These are a "worst case scenario" figures based on a LEM of unchanging mass...

Changes in mass produce a linear increase in fuel demand. Changes in velocity require an exponential increase in fuel demand. The basic formula is KE = 1/2 M V^2 .

For example, the 4670KG loaded LEM would require the following energy:

KE= (1/2) * 4670KG * (2380 m/s)^2 = 13226374000 Joules = 1.322 * 10^10 J

To launch the same mass from the surface of the Earth to orbit:

KE= (1/2) * 4670KG * (11,200 m/s)^2 = 2.92 * 10^11 J

In other words, it takes about 5% as much fuel to achieve lunar orbit than Earth orbit for the same given mass.

And that's not all because the surface gravity on Moon is 1.622 m/s² vs. surface gravity on Earth which is 9.78 m/s². In rocket dynamics You first have to achieve zero weight (means that the forces of the rocket push and the gravity pull are equal) and then You can start to accelerate the objects mass (4670kg, which in acceleration terms is unrelated to gravitational pull - http://en.wikipedia.org/wiki/Mass_versus_weight) to the speed necessary. Thus to get the lunar module to afloat is 6 times easier on Moon compared to doing this on Earth as of instead of 4670kg you are lifting only 778kg i.e. the rocket has to lift 778kg, but it must accelerate the full 4670kg multiplied by acceleration.

A mistake is also made on above cause the Ascent Module does not have to be accelerated to lunar Escape velocity (2380 m/s), but only to orbital velocity, which is 1022 m/s. Escape velocity is achieved by Control and service Module (when the lunar ascent module is discarded). Plus You have to carry less fuel (cause You have to lift less weight and make the mass move slower. And the mass is lower because you carry less fuel), don't have to fight aerodynamic forces (which at the speed of sound and above are considerable) and You don't have to get as high above the surface before you enter the orbit. And last, but not least - all rocket engines are roughly 25% more efficient in vacuum than in atmosphere end even more so if they are designed only for operating in vacuum.

Put that all together and You have multiple magnitudes less of fuel required.

You are mostly correct, but keep in mind that mass never changes, even when you are outside a gravity well . The formula for kinetic energy depends entirely on mass and velocity. (That is to say, kinetic energy is a property of an object's inertia, its "resistance to acceleration", and not upon gravity at all). Possibly you are thinking of Gravitational Potential Energy, which must also be overcome as part of the "work done" on the way to orbit. (Moving to a higher orbital position requires an increase in GPE, also paid for by the rocket fuel.)

Nevertheless, I agree with you that it requires much, much less fuel to leave the moon, far less than my spitball estimate, but I was just looking to show that a minute amount of fuel can do a lot more work on the moon than on Earth. The exact orbital velocity required depends on the altitude of the command module, so it should be something like 1022 m/s < velocity < 2380 m/s .

 

[Duewester]

I'm no mathmetician but apparently they did have enough fuel to get off the moon. A full escape was accomplished 6 times and a half hearted (Appolo 10) escape was conducted once.
Say good night gracie.

 

[Zubenelgenubi]

Seriously? This canard has been floating around for years, and responded to many times before.

Important points:

The Saturn V launcher has to get into Earth orbit not just the LEM upper stage, but the lower stage, the CSM and all of the fuel required to escape both from Earth orbit on the way there and from Lunar orbit on the way back. The mass involved is very different in each case.

The ratio of initial mass (ie with fuel) to the final mass (ie after the fuel has been burned) is an exponential function of the required delta-v, assuming no air resistance. Hence, the fuel required to get a given mass to Earth orbit is much, much larger than to get the same mass to Lunar orbit. Multiply that by the ratio of final mass after you've taken that into account.

Then add in the fuel burned dealing with air resistance in Earth's atmosphere. That's kind of a big deal when you need to get to 17,000 miles per hour...

Bottom line: yes, obviously the LEM ascent stage had enough fuel to get to rendezvous orbit. It did it several times.

 

[wtrix]

And that's not all because the surface gravity on Moon is 1.622 m/s² vs. surface gravity on Earth which is 9.78 m/s². In rocket dynamics You first have to achieve zero weight (means that the forces of the rocket push and the gravity pull are equal) and then You can start to accelerate the objects mass (4670kg, which in acceleration terms is unrelated to gravitational pull - http://en.wikipedia.org/wiki/Mass_versus_weight) to the speed necessary. Thus to get the lunar module to afloat is 6 times easier on Moon compared to doing this on Earth as of instead of 4670kg you are lifting only 778kg i.e. the rocket has to lift 778kg, but it must accelerate the full 4670kg multiplied by acceleration.

A mistake is also made on above cause the Ascent Module does not have to be accelerated to lunar Escape velocity (2380 m/s), but only to orbital velocity, which is 1022 m/s. Escape velocity is achieved by Control and service Module (when the lunar ascent module is discarded). Plus You have to carry less fuel (cause You have to lift less weight and make the mass move slower. And the mass is lower because you carry less fuel), don't have to fight aerodynamic forces (which at the speed of sound and above are considerable) and You don't have to get as high above the surface before you enter the orbit. And last, but not least - all rocket engines are roughly 25% more efficient in vacuum than in atmosphere end even more so if they are designed only for operating in vacuum.

Put that all together and You have multiple magnitudes less of fuel required.

You are mostly correct, but keep in mind that mass never changes, even when you are outside a gravity well . The formula for kinetic energy depends entirely on mass and velocity. (That is to say, kinetic energy is a property of an object's inertia, its "resistance to acceleration", and not upon gravity at all). Possibly you are thinking of Gravitational Potential Energy, which must also be overcome as part of the "work done" on the way to orbit. (Moving to a higher orbital position requires an increase in GPE, also paid for by the rocket fuel.)

Nevertheless, I agree with you that it requires much, much less fuel to leave the moon, far less than my spitball estimate, but I was just looking to show that a minute amount of fuel can do a lot more work on the moon than on Earth. The exact orbital velocity required depends on the altitude of the command module, so it should be something like 1022 m/s < velocity < 2380 m/s .

I even provided the link about the differences between mass and weight. To be said once more. You have to lift the weight and accelerate the mass. Regarding the mass you're correct, but first you have to counteract the gravity.

 

[Zubenelgenubi]

I even provided the link about the differences between mass and weight. To be said once more. You have to lift the weight and accelerate the mass. Regarding the mass you're correct, but first you have to counteract the gravity.

I think mortisthewise was objecting to this:

...instead of 4670kg you are lifting only 778kg ...

Which isn't correct, even though the point you were making is understood: you still have to lift 4670kg, it's just that 1kg doesn't weigh as much on the Moon as it does on Earth. If the example was in pounds, then the division would work, but kilograms are a unit of mass, not weight...

 

[wtrix]

I even provided the link about the differences between mass and weight. To be said once more. You have to lift the weight and accelerate the mass. Regarding the mass you're correct, but first you have to counteract the gravity.

I think mortisthewise was objecting to this:

...instead of 4670kg you are lifting only 778kg ...

Which isn't correct, even though the point you were making is understood: you still have to lift 4670kg, it's just that 1kg doesn't weigh as much on the Moon as it does on Earth. If the example was in pounds, then the division would work, but kilograms are a unit of mass, not weight...

Mybad. Point taken. The right unit instead of pounds however (SI units in this context) would have been kgf or kilopond as stated sometimes. The correct saying, thus, is: ...instead of 4670kgf you are lifting only 778kgf, but you're still accelerating 4670kg.

 

[MeteorWayne]

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Meteor Wayne