Let me after about 785 posts remind you about topic, i.e. my Challenge about safety of space travel and associated fuel consumption. You have to demonstrate how to travel to the Moon and back to win the prize (€1M). It is not easy.
Say that your space ship has mass 32 676 kg excluding fuel and that you must slow down from 2 400 to 1 500 m/s velocity to insert into Moon orbit. Your space ship has a P-22 KS rocket engine with 97 400 N thrust (at full blast). How much fuel do you require to carry out the braking maneuver?
If you suggest, e.g. 10 898 kg, you must support your answer with proper calculations to win the prize (€1M). I have a feeling you need >80 000 kg.
OK, let's work through that. This will be simple math as - once again - I'm not an engineer and my math skills aren't that deep. Despite your comments, Tsiolkovsky gives us all we need.
BTW, without a time frame thrust is pretty much irrelevant - since we're talking about Apollo, I'm going to assume a
P22K AJ10-137 rocket engine with an exhaust velocity of 3079 m/s.
So. The Rocket Equation:
Delta-V = EV * ln(m
0/m
1)
Where m
0 = Total mass before the burn
m
1 = mass after the burn
ln = natural logarithm
EV = exhaust velocity
So, let's plug in the figures and solve for the difference between m
1 and m
0, which will be the fuel used to produce the desired change in velocity.
Delta-V = 2400 - 1500 = 900 m/s change in velocity.
900 m/s = 3079 m/s * ln(m
0/32676) ; divide both sides by 3079 (units cancel out)
0.2923 = ln(m
0/32676) ; take the inverse ln of each side
1.3395 = m
0/32676 ; multiply both sides by 32676
43769.5 = m
0 ; which gives us the mass before the burn...
43769.5 - 32676 = 11093.5 ; subtract the mass of the spacecraft and
voila! ; we get the amount of mass expended.
And there's your answer, arrived at using the figures you provided and some ninth grade math.
Some of you professional number-crunchers want to check my work, please?