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Fuel burn rate 
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#1
Oct2906, 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 ua*t Mass of rocket at any time t is mmu*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 (mmu*t)*a*t=mu*t*(ua*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? 


#2
Oct3006, 07:01 AM

Sci Advisor
HW Helper
P: 6,679

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


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