# Past a certain speed $v_\text{rocket} > v_\text{exhaust}$, why don't rockets slow down with more prograde thrust?

Let's say that you're a stationary observer in a zero-G environment. From your perspective, you measure the speed of a rocket at different points in time. In the first frame, $$t = 0$$, the rocket is stationary and, consequently, $$v = 0$$. Let's also say that in time $$dt$$ the rocket ejects a mass of fuel $$dm$$ from its fuel tanks at a speed of $$v_\text{exhaust}$$. In order to find the velocity of the fuel ejected from the rocket from your point of reference, and assuming that the rocket is traveling only in the +x direction, you would use the equation $$v_\text{fuel} = v_\text{rocket} - v_\text{exhaust}$$.

To calculate the velocity of the rocket from time t to time dt, you would use conservation of momentum as follows:

$$P_1 = P_2$$

$$P_1 = (M_\text{rocket} + M_\text{fuel}) \cdot v_\text{rocket}$$

$$P_2 = ((M_\text{rocket} + M_\text{fuel}) - dm) \cdot (v_\text{rocket} + dv) - dm \cdot v_\text{fuel}$$

Here's where my question comes in: Once the rocket gets to a speed $$v_\text{rocket}$$ that is greater than the speed $$v_\text{exhaust}$$, by the aforementioned equation for $$v_\text{fuel} = v_\text{rocket} - v_\text{exhaust}, v_\text{fuel}$$ would be a negative number. This means that

$$P_1 = (M_\text{rocket} + M_\text{fuel}) \cdot v_\text{rocket}$$

$$P_2 = ((M_\text{rocket} + M_\text{fuel}) - dm) \cdot (v_\text{rocket} + dv) + dm \cdot v_\text{fuel}$$

with a special emphasis on the plus sign before the last term. If we assigned an arbitrary value to $$P_1$$ and $$P_\text{fuel} = dm \cdot v_\text{fuel}$$, say $$500\text{ kg m/s}$$ and $$20\text{ kg m/s}$$ respectively, then

$$P_1 = 500\text{ kg m/s}$$

$$P_2 = ((M_\text{rocket} + M_\text{fuel}) - dm) \cdot (v_\text{rocket} + dv) + 20\text{ kg m/s}$$

$$P_1 - 20\text{ kg m/s} = 480\text{ kg m/s}$$

meaning that the momentum of the rocket from $$P_1$$ to $$P_2$$ went down in value. This leads me to believe that after a certain speed, prograde thrust from a rocket actually slows down the vehicle, which from intuition doesn't seem right at all.

Did I make a mistake somewhere in my math or is there actually something here?

• The fuel and the exhaust were the same thing initially. How are you making a distinction between $V_{\text{fuel}}$ and $V_{\text{exhaust}}$? And how would that mean "v-fuel = v-rocket - v-exhaust"? And how does the momentum change when fuel is converted to exhaust (i.e., expelled from rocket)? Commented Feb 14, 2022 at 1:14
• @josephh I think that $v_{\mathrm{fuel}$ and $v_{\mathrm{exhaust}}$ are supposed to be the velocity of the exhaust with respect to the observer and rocket, respectively, although there does appear to be a sign error in the equation relating them. The question seems to be asking about the significance of the fact that, eventually, the velocity of the exhaust will be in the same direction as that of the rocket. Commented Feb 17, 2022 at 4:42
• @Sandejo Ah, I see. Thanks. Commented Feb 17, 2022 at 8:33

Rocket's exchaust is 3 km/s, rocket's speed is 8 km/s when in orbit, so at least from this your idea isnt holding

I assume error is in ignoring mass change when assuming that momentum is all there is to care about

Before fuel is expelled, it initial momentum was higher than after it turned to exchaust. This different in momentum is added to the rocket. But rocket's mass is now less, so less momentum is needed to keep it going fast.

Rocket speed 7 km/s

Exchaust speed 3 km/s relative to a rocket

Exchaust speed 4 km/s in stationary frame of reference

Exchaust momentum is 4 km/s kg in stationary frame of reference - here I think you assume that positive momentum is bad.

Exchaust momentum is 3 km/s kg in rockets frame of reference - this is how much the rocket is accelerated. So that rocket's speed increases.

Consider the case with 0 exchaust speed and significant forward speed, probably that would help to see how it works. Speed is not changing even when expelled material has a significant forward momentum. Because mass of the rocket is reduced.

TLDR: even if rocket's momentum decreases, it moves faster and thats what we care about usually

• Oh yeah, I forgot that the mass of the rocket going down means that the velocity could go up. Commented Feb 14, 2022 at 1:10

You state that: $$v_{fuel} = v_{rocket} - v_{exhaust}$$ That means you take $$v_{rocket}$$ to be positive in one direction, and $$v_{fuel}$$ positive in the same direction (and $$v_{exhaust}$$ positive in the opposite direction). In the beginning, at low $$v_{rocket}$$ values, $$v_{fuel}$$ will have a negative value. Therefore due to conservation of momentum you will get: $$P_1 = P_2 = m_{rocket}v_{rocket} + m_{exhaust}v_{fuel}$$ and you should always use the your equation with the plus sign $$𝑃_2 =((M_{rocket}+M_{fuel})−dm)⋅(v_{rocket}+dv) + dm⋅v_{fuel}$$ just noting that the $$v_{fuel}$$ has a negative value when $$v_{rocket}$$ is small, and the value of $$v_{fuel}$$ is positive when $$v_{rocket} > v_{exhaust}$$

In your calculation, it is unclear how you get to the result of 480kg m/s. Because you don't specify how large $$dm$$ is, you are probably ignoring that in the calculation.