Rocket equation

[Jairo]

I was trying to calculate the delta V of the ascent stage of lunar module by myself, to check for some people that it could actualy leave the Moon.

I know the full (4,691kg) and dry (2,333kg) mass, and I know the specific impulse of the engine (3,050Ns/kg). Using the rocket equation, this gives me a delta V of 2,120m/s, wich is indeed higher than the speed required for orbit (1,845m/s).

But I suppose this equation aplies only to absence of gravity. Is it wrong? How do I calculate that the right way?

Thanks.

[Bob B.]

Your delta-V number sounds about right, though I calculate 2,130 m/s and I've seen the actual number published as 2,220 m/s. However, your figure for orbital velocity sounds too high. Using Apollo 17 as an example, the LM was inserted into a 9.4 nm X 48.5 nm orbit. This makes the velocity at perilune equal to 1,688 m/s.

You are correct in saying the delta-V calculated by Tsiolkovsky's rocket equation does not take into account losses due to gravity (or drag). This is why your launch vehicle must produce more delta-V than your actual orbital velocity. For instance, orbital velocity in low Earth orbit is about 7.8 km/s, but your typical launch vehicle must produce about 9 km/s.

There is really no easy way, of which I'm familiar, to calculate the actual delta-V required. JayUtah and I had some discussions about this a couple years ago and neither one of us could come up with a really good way to demonstrate it. Jay ended performing some sample calculations for his Web page but it involved some short cuts that produce a ballpark estimate only. Perhaps he still has it and can provide a link. The only way I've ever been able to produce a calculation with reasonable accuracy was by writing a computer simulation. Unfortunately, this method is far more complex than I'd be able to explain here.

[Count Zero]

It certainly would be easier to calculate a lunar liftoff than a terran one - no atmospheric drag.

How 'bout working out the initial velocity required to lift the ascent stage straight up 9.4nm (i.e. a ballistic trajectory with no lateral component with an apex at 9.4 nm), then adding that to 1,688 m/s? Will that get you in the ballpark?

[Bob B.]

countzero said:

How 'bout working out the initial velocity required to lift the ascent stage straight up 9.4nm (i.e. a ballistic trajectory with no lateral component with an apex at 9.4 nm), then adding that to 1,688 m/s? Will that get you in the ballpark?

This is basically the method Jay used, though I don't know the link to his work. Here are my own calculations:

I'll use SI units, so 9.4 nm is about 17.4 km. The LM's APS has a thrust of 15.57 kN and an specific impulse of 311 s. Therefore the propellant mass flow rate is,

q = F/(Isp*g) = 15570/(311*9.807) = 5.105 kg/s

Traveling straight upward, let's burn the engine for 93.3 seconds. Our velocity at the end of the burn is,

V = Ve*LN(M/(M-qt)) - gt
V = 3050*LN(4691/(4691-5.105*93.3))-1.62*93.3
V = 175.4 m/s

(Note, g for the Moon is 1.62 m/s^2)

The distance traveled during the burn is given by,

D = Ve*[ t + t*LN(M/(M-qt)) + M*LN((M-qt)/M)/q ] - gt^2/2
D = 3050*[93.3+93.3*LN(4691/(4691-5.105*93.3))+4691*LN((4691-5.105*93.3)/4691)/5.105]-1.62*93.3^2/2
D = 7,910 m

The LM will now continue rising until reaching its apex, where its vertical velocity will be zero. The time to reach the apex is,

t = v/g = 175.4/1.62 = 108.3 s

The distance traveled from engine shutdown to the apex of the trajectory is,

D = gt^2/2 = 1.62*108.3^2/2 = 9,500 m

Thus the altitude of the apex above the surface is,

D_total = 7,910 + 9,500 = 17,410 m

(Note, the 93.3 s duration for the engine burn was solved by trial and error. I gave the answer at the start just to speed things along.)

We now must accelerate horizontally to 1,688 m/s. We have,

Mo = 4691-5.105*93.3 = 4,215 kg

V = Ve*LN(Mo/Mf)
Mf = Mo/EXP(V/Ve)
Mf = 4215/EXP(1688/3050)
Mf = 2,423 kg

The empty mass of the LM is 2,333 kg, so we have 90 kg of propellant left, or about 3.8% of the total load.

[Bob B.]

If anyone is interested, I've created an Excel spreadsheet simulating the launch of the Apollo 17 LM. The results worked out almost perfectly demonstrating quite nicely that the data published by NASA is exactly as needed to achieve orbit. Unfortunately it might not be too convincing to your typical HB since they probably won't understand the physics and math. Nonetheless, I'll make the file available to whoever wants it.

[Jairo]

I want it, please.

[Bob B.]

jairo said:

I want it, please.

If you send me your address in a personal message I'll email the file to you.