Question

A certain rocket of initial mass mo carries 60% of its initial mass as fuel. What is its final speed if it accelerates from rest in free space, as it burns all of its fuel? Express your answer in multiples of the fuel velocity out of the rocket, a. b) Supos this rocketsul in Inth brms mass of 03 m f fel. then jettisons the first stage fuel tank with a mass of 0.1mo. Then, it burns the remaining 0.3mo of fuel. Find the final speed, assuming ver does not change.

Answer 1

Answer:

A) $\frac{\Delta v}{ve} = 0.91$

B) $\frac{\Delta v}{ve} = 1.05$

Explanation:

Tsiolkovsky rocket equation states that:

$\Delta v = ve * ln(\frac{m0}{mf} )$

Where

Δv: change of speed of the rocket after burning the fuel

ve: exhaust velocity

m0: initial mass of the rocket

mf: mass of the rorcket after completing the burn (if it burned all its fuel this is the dry mass)

Since the rocket starts from rest (v = 0) its final speed will be the Δv

To express this speed in multiples of the fuel velocity out of the rocket, we simply:

$\frac{\Delta v}{ve} = ln(\frac{m0}{mf} )$

In this case we know that 60% of the mass of the rocket is fuel, so:

$mf = m0 * (1 - 0.6) = 0.4*m0$

So:

$\frac{\Delta v}{ve} = ln(\frac{m0}{0.4*m0} ) = ln(\frac{1}{0.4} ) = 0.91$

If the rocket jettisons a first stage with 0.3*m0 of fuel, and 0.1*m0 of fuel tank, it will be doing two burns.

The first burn will go from mass m0 to m1

m1 = m0 - 0.3 * m0 = 0.7 * m0

The speed gain from this first burn would be:

$\frac{\Delta v1}{ve} = ln(\frac{m0}{m1} )$

Then it will jettison the tank, since jettisoning is done with small explosives the gain of speed can be ignored for being very small. However it would have lost some mass.

m2 = m1 - 0.1 * m0 = 0.7 * m0 - 0.1 * m0 = 0.6 * m0

Next it will activate the second stage, burning fuel and going from m2 to m3

m3 = m2 - 0.3 * m0 = 0.6 * m0 - 0.3 * m0 = 0.3 * m0

The speed gain of this second burn would be:

$\frac{\Delta v2}{ve} = ln(\frac{m2}{m3} )$

And both added together:

$\frac{\Delta v}{ve} = ln(\frac{m0}{m1} )+ ln(\frac{m2}{m3} )$

$\frac{\Delta v}{ve} = ln(\frac{m0}{0.7 * m0} )+ ln(\frac{0.6 * m0}{0.3 * m0} ) = 0.356 + 0.693 = 1.05$