kwan3217 - thank you very much! very useful info!:thumbup:
ok i'll try not mixin units anymore
so one question more:
in the Mars travel we've got along with trans-mars injection (TMI) and mars parking orbit insertion and trans earth injection and earth orbits insertion and mid-course corrections as well.
so how calculate how much fuel we need for all entire trip (all deltaVs are known)?
it confused me little bit cause in rocket equation we got just mass ratio:
deltav=Isp*g_0*ln(m_0/m_1) (1a)
m_0 - is spacecraft dry mass+cargo + all propellent
m_1 - is spacecraft dry mass+cargo + propellent left for next maneuvers
so i need to find "propellent left for next maneuvers" - how can i calculate this?
For concreteness, we'll use some numbers. 20 metric ton (20000kg) spacecraft (dry+cargo). Say that TransX says you need 4km/s to depart Earth parking orbit (7.5km/s to 11.5km/s), 3km/s to enter Mars orbit, and 3km/s to depart Mars, and 4km/s to return to Earth parking orbit. You then decide to budget 100m/s for trajectory correction. This all presumes no aerobraking, which you could use to almost eliminate the fuel you need to enter orbit at each end.
To get the total DeltaV needed, you just add up the DeltaV of each maneuver, so 14.1km/s (14100 m/s).
Now, you know the DeltaV, effective velocity, and final mass of the spacecraft with no fuel at the end of the mission (20 tons, as mentioned above). How much fuel do you need?
First, solve the rocket equation for the missing value, m_0
deltav=v_e*ln(m_0/m_1) (1)
deltav/v_e=ln(m_0/m_1)
exp(deltav/v_e)=m_0/m_1
m_1*exp(deltav/v_e)=m_0 (2)
Let's say you are using a currently achievable v_e of 4500m/s (Space shuttle main engines are this efficient). So, plug the numbers:
m_0=20000kg*exp(14100m/s / 4500m/s)
m_0=459007kg (459 tons!)
If the starting mass is 459 tons and the final mass is 20 tons, then this mission requires 439 tons of fuel. This is roughly equivalent to a fully fueled Shuttle external tank. However, a Shuttle external tank weighs over 30 tons by itself.
We begin to see how difficult a trip to Mars is... What if we use the DeltaGlider engine efficiency, 44000m/s?
m_0=20000kg*exp(14100m/s / 44000m/s)
m_0=27555kg
This time it only takes 7.5 tons of fuel.
You can do it the other way also. If you have a 20-ton delta glider with 10 tons of fuel, how much DeltaV are you carrying? Just run equation 1 directly:
m_0=30000kg
m_1=20000kg
v_e=44000m/s
deltav=v_e*ln(m_0/m_1)=44000*ln(30000/20000)=17840m/s
Now to answer the question you started with. This time we solve (1) for m_1
m_0=30000kg (20ton vehicle+cargo and 10ton fuel)
v_e=44000m/s
deltav=14100m/s (Earth->Mars->Earth mission)
deltav=v_e*ln(m_0/m_1) (1)
deltav/v_e=ln(m_0/m_1)
exp(deltav/v_e)=m_0/m_1
exp(deltav/v_e)/m_0=1/m_1
m_0/exp(deltav/v_e)=m_1 (3)
30000/exp(14100/44000)=m_1
m_1=21775kg
so you have m_fuel=m_1-m_dry=21775kg-20000kg=1775kg of fuel left after the mission. You can break the mission down by each maneuver, using the deltav of that maneuver only as the required deltav, and the m_1 of the previous maneuver as the m_0 of the next.
Note that as long as you only use fuel, never re-load fuel, and never drop stages, you can consider your tanks to be full of DeltaV rather than fuel. We calculated above that a delta glider with 10 tons of fuel has 17.84km/s of deltav. After the Mars mission uses up 14.1km/s, you have 17.84-14.1=3.74km/s of deltav left. You used 83% of your fuel mass on the mars mission, but only 79% of your available DeltaV. The last bit of fuel is more efficient at producing DeltaV since it doesn't have to push around the first bit of fuel, while the first bit of fuel does have to push around the last. The lower the fuel v_e, the bigger this effect gets.
