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Fuel burn rate

by Simka
Tags: burn, fuel, rate
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Simka
#1
Oct29-06, 02:58 PM
P: 1
A rocket, mass m=4000kg, lifts off the ground vertically with acceleration of a=0.2g. The velocity of gass emmiting from it is u=1200m/s (with respect to the rocket). Find engine's reactive force and then fuel burn rate (mu).

This is what i managed to do:
rocket's speed at any moment t (with respect to the ground) is a*t
gass's speed at any moment t (with respect to the ground) is u-a*t


Mass of rocket at any time t is m-mu*t
Mass of fuel that is burned is mu*t
rocket's mass multiplied by it's speed and gass's mass multiplied by its speed are equal, so

(m-mu*t)*a*t=mu*t*(u-a*t)

and i get that m*a*t=mu*t*u

mu=(m*a)/u=6.53 kg/s

but... I was given 4 answers (5.9kg/s 9.8kg/s 39.2kg/s 19.8kg/s) and mine is wrong...

Any ideas anyone?
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Andrew Mason
#2
Oct30-06, 07:01 AM
Sci Advisor
HW Helper
P: 6,677
Quote Quote by Simka
A rocket, mass m=4000kg, lifts off the ground vertically with acceleration of a=0.2g. The velocity of gass emmiting from it is u=1200m/s (with respect to the rocket). Find engine's reactive force and then fuel burn rate (mu).

This is what i managed to do:
rocket's speed at any moment t (with respect to the ground) is a*t
gass's speed at any moment t (with respect to the ground) is u-a*t


Mass of rocket at any time t is m-mu*t
Mass of fuel that is burned is mu*t
rocket's mass multiplied by it's speed and gass's mass multiplied by its speed are equal...
You are forgetting about gravity. Use Newton's third law: The rate of change of momentum of the rocket + force of gravity = rate of change of momentum of the gas being expelled:

[tex]F_{rocket} = mg + ma = - F_{gas} = - vdm/dt[/tex]

AM


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