Question

A 1970 kg rocket has 4620 kg of fuel on board. The rocket is coasting through space at 95.7 m/s and needs to boost its speed to 319 m/s; it does this by firing its engines and ejecting fuel at a relative speed of 727 m/s until the desired speed is reached. How much fuel is left on board after this maneuver

Answer 1

To solve this problem we will apply the propulsion equations given by Tsiolkovsky. The Tsiolkovsky rocket equation, classical rocket equation, or ideal rocket equation is a mathematical equation that describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself using thrust by expelling part of its mass with high velocity can thereby move due to the conservation of momentum. This is,

$\Delta V = V_e*ln(\frac{m_0}{m_f})$

Where:

$m_0$ = initial total mass = 1970 kg + 4620 kg = 6590 kg

$m_f$ = final total mass

$V_e$ = effective exhaust velocity

$\Delta V$ = change in rocket velocity

Replacing the values we have that,

$(319-95.7) = (727) ln(\frac{6590}{m_f})$

$223.3 = (727) ln(\frac{6590}{m_f})$

$\frac{223.3 }{(727)} = ln(\frac{6590}{m_f})$

$0.3071 = ln(\frac{6590}{m_f})$

$m_f = 4847.45kg$

Since this is the mass of both the ship and the remaining propellant, the remaining propellant mass, $m_f$, is given by

$m_f = (4847.45 - 4620 ) kg$

$m_f = 227.45 kg$