# Rocket equation exhaust velocity question...

by randomking333
Tags: rocket equation
 P: 2 The rocket equation given by Delta V = Ve ln (Mo/M) says that Ve is the exhaust velocity of the rocket. The orbital velocity for Low Earth Orbit is about 7.8 Km/s. How then, is this velocity attained by a rocket which is fuelled only by a bipropellant rocket engine producing an exhaust velocity of 4.4 Km/s. Using conservation of momentum, the heavier mass of the rocket wont reach the velocity of the exhaust? Pls explain to me how this works.. THanks in advance
 Emeritus Sci Advisor PF Gold P: 2,352 In The rocket equation, Mo would be the Mass of ship+fuel and M would be the mass of the ship alone. Given that, we can solve for the fuel mass by: $$e^{\frac{\Delta V}{V_e}}= \frac{M_s+M_f}{M_s}$$ $$e^{\frac{\Delta V}{V_e}}= \frac{M_f}{M_s}+1$$ $$M_s e^{\frac{\Delta V}{V_e}}-1= M_f$$ With the given final velocity and escape velocity, this gives a fuel mass of 4.87 times that of the mass of the ship alone. So the mass of the fuel is far greater than the mass of the rocket, and there is no problem with conservation of momentum.
 Sci Advisor P: 2,470 Mass of the stage of the rocket that reaches orbital velocity is far, far smaller than total mass of the propellant. If you were simply applying a conservation of momentum, you'd end up with final speeds well above orbital. The reason you get only 7.8km/s is that propellant ends up traveling at speeds varying from -4.4km/s to +3.4km/s. The rocket formula you quoted is a result of adding up momenta of all these contributions in correct proportions.
P: 2

## Rocket equation exhaust velocity question...

 Quote by Janus In The rocket equation, Mo would be the Mass of ship+fuel and M would be the mass of the ship alone. Given that, we can solve for the fuel mass by: $$e^{\frac{\Delta V}{V_e}}= \frac{M_s+M_f}{M_s}$$ $$e^{\frac{\Delta V}{V_e}}= \frac{M_f}{M_s}+1$$ $$M_s e^{\frac{\Delta V}{V_e}}-1= M_f$$ With the given final velocity and escape velocity, this gives a fuel mass of 4.87 times that of the mass of the ship alone. So the mass of the fuel is far greater than the mass of the rocket, and there is no problem with conservation of momentum.

Yes.. but for any point in time taken.. say a span of 1 second early in the launch...

mv=mv

so the pass of the rocket at that time is much greater.. since the exhaust velocity is of the fuel being burned per unit time?.. so that mass of fuel is a lot less? .. or are we meant to consider even the fuel that is still in the chamber and not burnt (and hence not at that exhaust velocity)

Emeritus
PF Gold
P: 2,352
 Quote by randomking333 Yes.. but for any point in time taken.. say a span of 1 second early in the launch... mv=mv so the pass of the rocket at that time is much greater.. since the exhaust velocity is of the fuel being burned per unit time?.. so that mass of fuel is a lot less? .. or are we meant to consider even the fuel that is still in the chamber and not burnt (and hence not at that exhaust velocity) Thanks in advance
Let's take the first second of launch. First you have to overcome the force of gravity which is the equivalent of 9.8m/s². Which means that it is the same as accelerating to 9.8 m/s in one second. Using the rocket equation we find that it would take 2.2e-3 times the remaining mass of the ship to do this.( In this case, Mo is equal to the mass of the rocket plus the un-used fuel on board)

IOW, with an exhaust velocity of 4.4 km/s, If you burn about 2 one-thousandths of the ships total mass in the first second, you will overcome gravity. Anything over that will accelerate the ship upward.

In terms of momentum, if MV is the final momentum of the ship and mv is the momentum of the exhausted fuel, then at a v of 4400 m/s, at V of 9.8m/s:

9.8M= 4400m

m/M = 9.8/4400 = 0.0022

Which again gives an answer of about 2/1000 of the ships total mass needed to be burned in one second to overcome gravity.

If you carry out the above solutions to a few more decimal places you will find that the rocket equation gives a slightly larger value. This is because it factors in the fact that during the first part of the second, the rocket has to still lift the fuel it will burn later in the sec. That is the point of using the rocket equation, it takes into account the mass of the remaining unused fuel.
 HW Helper P: 6,924 Although the terminal exhaust velocity relative to the rocket is fixed, that exhaust accelerates both the rocket and the remaining on board fuel, so that as the rockets and on board fuel increases speed, so does the exhaust velocity (with repect to the initial velocity of the rocket).

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