Grade 12 Rocket propulsion Question

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SUMMARY

The discussion centers on calculating the fuel and oxidizer requirements for a rocket designed to achieve a speed of 10,000 m/s with an initial payload of 3 metric tonnes. The initial calculations incorrectly suggested a fuel mass of -15,000 kg, which is not feasible. The correct approach involves using the ideal rocket equation, Δv = vengineln(mi/mf), to determine the necessary fuel mass based on the exhaust velocity of the engine. For an exhaust velocity of 2,000 m/s, the required fuel mass is approximately 442,000 kg.

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Homework Statement


A rocket for use in deep space is to have the capability of boosting a pay load ( plus the rocket frame and engine) of 3 metric tonnes to an achieved speed of 10000m/s with an engine and fuel designed to produce an exhaust velocity of 2000m/s.
a) How much fuel and oxidizer is req'd, b)and if a diff fuel and engine design could give an exhaust velocity of 5000m/s, what amount of fuel and oxzidier would be req'd?


Homework Equations



P=P'

The Attempt at a Solution


m1 = 3000 kg, v1= 0, v1' = 10 000m/s
m2 = ? kg, v2 = 0, v2' = 2000m/s

m1v1 + m2v2 = m1v1' + m2v2'
0 = m1v1' + m2v2'
-m1v1' / v2'= m2

-15 000 kg = m2

How is the mass negative :S
the answer at the back says "442 x 10^3 kg"

so I know I'm off by a lot... Our fizzicks teacher didn't teach us how to solve these problems, but I assumed it was a momentum problem.

Any help would be greatly appreciated. :)
 
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Can someone please assist me anyway possible.

I'm sorry if it's a stupid question.
 
You are assuming that the 15000 kg are ejected all at once at 2000 m/sec. That is not how the rocket works, however. The rocket fuel burns over a period of time during which the rocket + fuel is accelerated in the other direction. So a lot of the rocket fuel is used to accelerate the rocket fuel still on board the rocket. You also have to take gravity into account if the rocket is moving away from the earth.

This is a hard question if they are expecting you to solve it from first principles. I think they are expecting you to use an equation such as the ideal rocket equation:

\Delta v = v_{engine}\ln{m_i/m_f}

where the left side is the final payload speed, m_i is the initial mass, m_f is the final mass (payload), v_engine is the speed of the gases exiting the engine (ln is the natural logarithm). Do you see this equation anywhere in your materials?

m_fe^{\Delta v/v_{engine}} = m_i

Since the fuel mass is mi-mf you can calculate the amount of fuel used.

AM
 

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