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.