So, who wants to win 1 million Euro?

[ka9q]

Space travel is similar to a voyage at sea (my speciality) and you need fuel to get from A to B.

Here we see your big mistake (other than thinking you know what you're doing when you don't).

Space travel is not like a voyage at sea, so your intuition is utterly out of place.

Just for starters:

1. Water exerts substantial drag on ships moving through it. This drag must be continuously overcome with propulsion or the ship will stop relative to the water.

Spacecraft operate in vacuum, without drag to overcome. They will keep moving at a constant velocity until acted on by an external force, so continuous thrust is not required. That's Newton's First Law. This makes its "velocity" totally dependent on your choice of reference frame. Choose any inertial reference frame, use it consistently and you'll get the exact same answers about propellant requirements, etc. If you don't, you're doing something wrong.

2. Ships create thrust by grabbing the water and pushing on it. In space, there's nothing to grab. If a spacecraft wants to change its velocity through means other than gravity, it must carry along something on which to push. That "something" is called "rocket propellant".

3. Most ships operate at constant or near-constant altitude where changes in potential energy are negligible compared to drag losses. Potential energy changes in spacecraft are substantial and very significant.

4. The fuel in a ship is typically a tiny fraction of its gross weight so changes in weight due to fuel consumption can usually be ignored. This is most decidedly untrue in space flight. Fuel (propellant) accounts for nearly all of any rocket's launch mass. Because this mass is ejected overboard to produce thrust, the continuous decrease in vehicle mass during thrusting must be properly accounted for in the calculations.

The key here is the Tsiolkovsky rocket equation to which both Jay and I have alluded. It is perhaps the most important equation in space flight after Newton's F=ma (from which it is derived). Since you obviously don't know it, you have absolutely no business making any kind of pronouncements on the topic of space flight.

[raven]

It's a little evil of me, but its nice to have someone to debate, for lack of a kinder word, with again here. Or, as I said on the other thread, a Fray, Rufferto, a Fray!

[ka9q]

Sailing in the interface air/water is evidently also in three dimensions and the forces applied to the sea going ship are much more complicated than those of a space ship.

This is actually the first halfway correct thing you've said -- the forces on a ship (or aircraft) are much more complicated than those on a spacecraft because the former involve fluid flows that can be very complex to model.

So your total ignorance of the physics of space flight is even less excusable.

[ka9q]

Evidently you do not need much energy to change the orientation of the moving space ship as you just rotate it around itself keeping an eye of the gyro.

Wow, this is actually correct. I'll give you that.

The problem is to change direction and velocity, particularly to change velocity from, e.g. 2400 to 1500 m/s at arrival the Moon. According my calculations you need >46 000 kg of fuel to do it.

And your calculations are dead wrong. The actual figures are as follows for Apollo 11 LOI #1 (first lunar orbit insertion burn):

Mass of CSM/LM at ignition: 96,061.6 lbm
Mass of CSM/LM at shutdown: 72,037.6 lbm
Propellant used: 96,061.6 - 72,037.6 = 24,024 lbm = 10,897.1 kg
Velocity at ignition: 8250 ft/s = 2514.6 m/s
Velocity at shutdown: 5479 ft/s = 1670 m/s
Velocity change = abs(8250 - 5479) =  2771 ft/s = 844.6 m/s

Now consider the Tsiolkovsky rocket equation:

delta-V = V_e * ln(mass_at_ignition/mass_at_shutdown)

We want to know if these numbers are reasonable for the rocket engine in use, so let's solve for V_e, the effective exhaust velocity of the rocket engine:

V_e = delta-V / ln(mass_at_ignition/mass_at_shutdown)
= 844.6 m/s / ln(1.33349)
= 2934.7 m/s

This corresponds to an I_sp of 2934.7 / 9.80665 = 299 seconds. This is just under the nominal I_sp for a large hypergolic rocket engine burning these propellants. (I expected a very small discrepancy because the altitude of the CSM/LM was not precisely constant during the burn.)

Note that the kinetic energy (in any coordinate frame) of the spacecraft doesn't even enter into it. Only the change in velocity matters, and it'll be the same in any inertial reference frame you choose. The kinetic energy won't be, and that alone should tell you that you've made a mistake by thinking it's important.

[ka9q]

It is very easy; Pls, try to get the basics right.

Please take your own advice.

Applying a force 1 N to a mass of 1 kg will accelerate that mass at 1 m/s² ... and no energy is required for that acceleration.

But you need energy to produce the force.

Exactly wrong!

You can produce a force forever with no energy at all when that force does not act through a distance. A magnet or a rubber band can apply a force forever to a stationary object without any energy at all. But there is also no acceleration.

But a 1N force applied to an unrestrained 1 kg object will accelerate it, so this will require a power that will depend on time.

Note, however, that forces never occur unopposed; that's Newton's third law of motion. Forces are always between objects. Applying a 1N force to an object in a certain direction absolutely requires the application of a 1N force on something else in exactly the opposite direction. For example, a car can only generate 1N of forward force on itself by applying 1N of rearward force on the earth; that's why you need good tires. Because the car will then begin to accelerate, the power required to maintain that 1N will increase with velocity. From a standing start, the power starts at 0. When the car reaches 1 m/s, it will take 1 watt to generate that 1N force. When it reaches 1,000 m/s, it will take 1,000 watts to maintain that same 1N of force, and so on without limit (ignoring drag, which would change the net forces on the car).

A rocket is no exception to the rule that forces never occur unopposed; indeed, rockets are one of the best examples of Newton's Third Law in action. But they push on their own exhaust, so they must lose mass to do so. Because it is not pushing on any external object (such as the earth), a rocket's velocity with respect to any outside object is completely irrelevant -- and so is its kinetic energy in that object's reference frame. That's why the kinetic energy of Apollo, in the moon's reference frame as it entered lunar orbit, is wholly irrelevant to the amount of propellant required. And it's why your claims are so utterly wrong.

[ka9q]

FYI, ships normally use a rudder to move starboard and port. Actually forces acting on the rudder move the ship sideways or transversly.

Have you found the rudders on the Apollo spacecraft yet? How about the propellers?

[gwiz]

To reduce the speed of a mass of 43 000 kg from 2 400 to 1 500 m/s you need 75.47 GJ brake energy! If 1 kg rocket fuel produce 1.63 MJ energy it seems you need 46 300 kg fuel for this maneouvre. You should wonder, where it was carried.

Afraid I'm rather late to this thread, but I am an aerospace engineer and I thought I should just chip in to say that several people have already pointed out exactly where you go wrong with this.  Basically, the change in kinetic energy is calculated from the change in velocity, not as you think the difference between the energies before and after the change.

This is because the initial and final velocities are completely arbitrary numbers which change with the frame you choose to measure them in. 

If you are this ignorant of the basics, then you are not a competent engineer.

[ChrLz]

Just curious.. Anders, do you acknowledge ANY of the (many and substantial) errors so far pointed out in your 'understanding'?

I thought this wasn't a troll, but that stuff about the LM-CM maneuver, and Walter...  Nah, sorry - I'm just sitting back with popcorn and watching the train wreck, now.  Not going to waste my time responding to the ever changing Gallop..

BTW, Anders, you have refused to prove the existence of the $1m, so I think we can take that as a lie.  How surprising..

[Heiwa]

(I mistyped force when I meant acceleration. I fixed my that post and I striked out my errors and fixed them in this post.)

Yes, that's why I explained to you what force is, so you could correct your post. A force causes change in velocity of the mass it is applied to and is required, when a spaceship like Apollo 11 arrives to the Moon and wants to orbit the Moon at reduced speed compared to arrival speed.

The only way for Apollo 11 to produce that braking force is to fire is SM rocket engine at the right time and in the right direction (i.e. opposite the one travelling, etc). The amount of force and its time of application must also be correct or you will not achieve your objective, i.e. miss it.
 
To produce this force by firing your rocket engine requires fuel. Say the rocket engine is a P-22KS rocket NASA engine with 97 400 N thrust that is fired for 357.5 seconds (or something close to it) to reduce the speed to 1 500 m/s from 2 400 m/s at 2.52 m/s² deceleration.

During the 357.5 seconds braking the space ship travelled about 774 000 meter with a brake force 97400 N provided by the P-22KS rocket engine at full blast.

To reduce the speed of a mass of 43 000 kg from 2 400 to 1 500 m/s you need 75.47 GJ brake energy! I assume we all agree to this - see discussion above.

If 1 kg rocket fuel produce 1.63 MJ energy as NASA suggests, it seems you need 46 300 kg fuel for this maneover.

My question is therefore - where did NASA store 46 300 kg fuel in the Apollo 11 SM?

[Heiwa]

Just curious.. Anders, do you acknowledge ANY of the (many and substantial) errors so far pointed out in your 'understanding'?

I thought this wasn't a troll, but that stuff about the LM-CM maneuver, and Walter...  Nah, sorry - I'm just sitting back with popcorn and watching the train wreck, now.  Not going to waste my time responding to the ever changing Gallop..

BTW, Anders, you have refused to prove the existence of the $1m, so I think we can take that as a lie.  How surprising..

I read most posts that are on subject at this forum, so do not worry. What errors are you talking about?

Re the money, it is in the bank evidently, so you do not have to worry about it. It is also OT.

[frenat]

At departure Earth the Command Module and the Service Module are together loaded on top of the Saturn rocket with the Lunar Module stored below the Service Module, actually below the rocket engine outlet of the Service Module.

After lift off and one orbit Earth the space ship is sent off towards the Moon and one way or another the Lunar Module is shifted to the top of the Command Module, so that later, in Moon orbit, two persons can enter it via the hatches. Can anybody explain how the transfer of the Lunar Module from below the Service Module to the top of the Command Module was done?

Getting trounced because you're using the wrong equations (because you haven't bothered to understand them) so you try to change the subject?

[frenat]

Of course I am real! Like my €1 000 000:- at my bank.

Monopoly money doesn't count.

Nobody here believes you have the money.  You have shown no proof of it.  Nobody here believes you to be anything more than a troll at this point because you have clearly shown you have no interest in learning and far more in handwaving.

[frenat]

I am also curious to know how Walter managed to shift the Lunar Module in space from one end to the other of the CSM! Do you know, how Walter did it?

[Heiwa]

... the initial and final velocities are completely arbitrary numbers which change with the frame you choose to measure them in. 

So you suggest that Apollo 11 orbited at an arbitrary number velocity around the Moon prior arrival at another arbitrary number velocity?

Sorry, I do not follow. OK, NASA in one document says Moon orbit speed for Apollo 11 is 3000 m/s and in another 1500 m/s and the latter seems correct (so I use it) and also the arrival speed 2400 m/s seems correct as NASA says Apollo 11 had to slow down to orbit. You know 2400>1500.

If you find errors in my presentation, please say what data, number, etc, is wrong and I will correct it and the associated calculations and conclusions. I doubt very much the final conclusion is wrong, i.e. Apollo 11 lacked stored energy/rocket fuel aboard to brake into and out of Moon orbit, but if you disagree, show it with numbers and not arguments of no value.

[frenat]

The problem is to change direction and velocity, particularly to change velocity from, e.g. 2400 to 1500 m/s at arrival the Moon. According my calculations you need >46 000 kg of fuel to do it.

Your calculations have been shown to be wrong repeatedly.  You have ignored it.  That alone shows you are not really interested in learning.