# Velocity of rocket

1. Nov 28, 2014

### Samuelriesterer

I am stuck on how my teacher got to the last equation in #4.

Problem Statement:

A 2,000 kg rocket carrying 3000 kg of fuel is drifting in space when it fires its engine. While operating, the rocket engine burns 20 kg of fuel per second and the exhaust gasses leave the rocket at an exhaust velocity of Ve =300 m/s relative to the rocket.

Relative equations:

p = mv

Work done:

1) At some moment while the engine is firing, the rocket has mass M and is moving at velocity v, as is all the fuel it contains. After a short time Δt, the rocket has burned an amount of fuel Δm which leaves the rocket at velocity ve relative to the rocket. Write expressions for the total momentum before Δm is burned and after Δm is exhausted.

pi = MVi
pf = (M - ∆M)(Vi + ∆V) + ∆M(Vi –Ve)

2) Set the two expressions in (1) equal and expand any products of sums and differences.

(M - ∆M)(Vi + ∆V) + ∆M(Vi –Ve) = MVi →
M Vi - Vi∆M + M∆V +∆M∆V + ∆M Vi - ∆MVe = M Vi →
- Vi∆M + M∆V +∆M∆V + ∆M Vi - ∆MVe = 0 →
M∆V +∆M∆V - ∆MVe = 0

3) Delete any terms which are a product of two factors of Δm, Δv in any combination (they will be much smaller than other terms as Δt shrinks to zero).

M∆V +∆M∆V - ∆MVe = 0 →
M∆V - ∆MVe = 0 →
M∆V = ∆MVe

4) Now take a limit as Δt shrinks to zero and integrate from t = 0 to t to find an expression for the velocity of the rocket as a function of time.

M∆V = ∆MVe →

(M∆V)/(M∆t) = (∆MVe)/(M∆t) →

∆V/∆Vt = (∆MVe)/(M∆t) →

∆V/∆Vt = (Ve/M) (∆M/∆t) - Not entirely sure how he came to this or to the next equation:

V = Kt + Vi

5) Integrate from Δm = 0 to 3000 kg (t= 0 to [3000 kg/20kg/s]). to find the change in the velocity of the rocket.

2. Nov 28, 2014

### haruspex

It's a simple rearrangement of terms:
$\frac{\Delta M V_e}{M \Delta t} = \frac{V_e \Delta M }{M \Delta t}= \frac{V_e}{M} \frac{\Delta M }{ \Delta t}$
But don't you mean
$\frac{\Delta V}{ \Delta t} = \frac{V_e}{M} \frac{\Delta M }{ \Delta t}$?
Are you sure you've quoted that correctly? What is K here? That equation looks like uniform acceleration, which is not what will happen.

3. Nov 28, 2014

### Samuelriesterer

Sorry, the K I believe is a constant

4. Nov 28, 2014

### haruspex

Well, as I said, that'd be wrong.
You have $\frac{d V}{d t} = \frac{V_e}{M} \frac{d M }{d t}$
What do you get by integrating that?

5. Nov 28, 2014

### Samuelriesterer

$tV + C = (tVe)/(M) + C$ ?

6. Nov 29, 2014

### haruspex

No. You have
$\int \frac{d V}{d t}dt = \int \frac{V_e}{M} \frac{d M }{d t}dt$
Simplify both sides.

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