# Rocket Motion - calculating distance

I'm trying to solve a rocket problem and have found no clear answer on how to do this. If I have a rocket in space (no significant gravitational forces, no drag, etc.) with a known constant thrust (T), mass flow rate (mdot), exhaust velocity (Ve), total initial mass including propellant (Mi), and propellant mass (Mp), I know that I can easily find delta-V with the rocket equation and amount of burn time is simple enough, but what I can't seem to figure out is how to calculate how far it travels.

My initial naive attempt at this was to derive an equation for instantaneous acceleration as a function of time: a(t) = T / (Mi - mdot * t), and then integrate to get an equation for velocity, and then again for position. I'm guessing the fact that it's not a closed system is why this is wrong?

In any case, is there a reasonably straightforward solution to this problem? I know that for computing orbital trajectories and accounting for all the different forces during launch, like gravity, drag, etc., the real rocket scientists use numerical integration to solve the problem. But I'm hoping that for a such a simplified system, that won't be necessary.

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No one seems particularly interested in answering this. Perhaps if I restate the problem more formally?

Consider a rocket in space with the following characteristics:

total mass (including propellant): M0 = 30000 kg
propellant mass: Mp = 24000 kg
exhaust velocity: ve = 5000 m/s
thrust: T = 1.5 MN

Assume that the rocket is far enough away from any massive bodies such that gravitational effects are negligible.

a) How long does the rocket take to burn all of its fuel?
b) What is the rocket's final velocity?
c) How far has the rocket traveled when its fuel is expended?

Solution attempt:

a) If the thrust and exhaust velocity are constant, then the mass flow rate must be constant, so

mdot = T / ve = 1.5 MN / 5000 m/s = 300 kg/s

Since the rocket has 24000 kg of propellant, it takes 80 s to burn all of its fuel.

b) Using the rocket equation,

delta-V = ve ln(M0 / (M0 - Mp))
delta-V = 8047.2 m/s

c) I have no idea how to calculate this. As I mentioned above, I initially thought I could derive an equation for instantaneous acceleration and go from there, but I'm pretty sure that's wrong. Even though I know how long it takes to burn the fuel, and it's final speed, I can't see any simple answer to this.

folks probably think you're asking for homework help and this isn't the place for it.

folks probably think you're asking for homework help and this isn't the place for it.
This is the homework forum. I originally posted this in the general physics forum, but it was edited and moved by one of the moderators.

Any chance you could help with the answer?

Filip Larsen
Gold Member
c) I have no idea how to calculate this. As I mentioned above, I initially thought I could derive an equation for instantaneous acceleration and go from there, but I'm pretty sure that's wrong.

It should be possible to integrate the distance traveled more or less directly from the rocket equation.

It should be possible to integrate the distance traveled more or less directly from the rocket equation.

Right, of course. I thought about that originally, but didn't see a way to integrate wrt time. I forgot that the mass ratio is a function of time (duh).

It's a nasty looking integral, but I think I've got it. Thanks!

I figure I might as well complete this and show the solution if anyone else is interested.

First, get the rocket equation as a function of time, and then integrate that.

V = delta-V
Ve = exhaust velocity
mdot = mass flow rate
M0 = initial total mass including propellant
M = mass without propellant
M0 / M = mass ratio
D = distance travelled

(1) M = M0 - mdot * t

Substituting for M in the rocket equation

V = Ve * log(M0 / M)

(2) V = Ve * log(M0 / (M0 - mdot * t))

Integrating (2) over time:

D = Ve * (t - ((M0 - mdot * t) * log(M0 / (M0 - mdot * t))) / mdot)

Substituting M from (1)

(3) D = Ve * (t - (M * log(M0 / M) / mdot))

Rearranging (1),

t = (M0 - M) / mdot

and substituting for t in (3)

D = Ve * ((M0 - M) / mdot - (M * log(M0 / M) / mdot)

D = (Ve / mdot) * (M0 - M - M * log(M0 / M))

(4) D = (Ve / mdot) * (M0 - M * (log(M0 / M) + 1))

Using (4), you can calculate the distance a rocket will travel in "empty" space - away from gravitational sources - from the basic parameters of the rocket, namely, exhaust velocity, mass, mass ratio, and mass flow rate. It's straightforward to do some other substitutions and rearranging to come up with other forms. Two in particular may be of interest.

Letting Mp = initial mass of propellant

Mp = M0 - M

D = (Ve / mdot) * (Mp - M * log(M0 / M))

I found this other form of the this equation in a book after having derived the above equations. :grumpy:

D = Ve * (M0 / mdot) * (1 - (M / M0) * (log(M0 / M) + 1))

This flip side of this is that the instantaneous acceleration can be calculated by taking the derivative of (3) wrt to time. Perhaps more importantly, you can determine the maximum acceleration (and ensure that the astronauts are not splashed on the floor of their compartment).

A(t) = Ve * mdot / (M0 - mdot * t)

Amax = Ve * mdot / M

and, of course, thrust = T = Ve * mdot, unsurprisingly yielding

Amax = T / M

ileacus,

Multiple thanks. I'd been trying to find an answer to the same question for a couple of weeks. I am a complete dunderhead at math so I was unable to correct the unbalanced parentheses in your solution. Would you be so kind? Thank you.

Me again, almanddundersmithhead. ¿What formula would give D for a rocket decelerating to zero from v?