Thrust of a Rocket

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Homework Statement



A rocket has an initial mass of m0 = 30,000 Kg, from which 80% is fuel. The fuel is being burned at a rate of R = 200 kg/s and the gas is expelled with a velocity of vrel = 1.8 km/s. The velocity of the rocket is calculated by:

v = vrel* ln [itex]\frac{m0}{m0 - Rt}[/itex] - gt

Calculate:

(a) The thrust of the rocket

(b) The time until the rocket runs out of fuel

(c) The velocity of the rocket when the rocket runs out of fuel, under the condition that the rocket is always travelling perpendicullarly, and the g is always constant. Air resistance is negligble.

Hint: How does Newton's second law look like with a change of mass with respect to time?


Homework Equations



F = m * a

a = [itex]\frac{dv}{dt}[/itex]

[itex]\frac{dm}{dt}[/itex] = m0 - Rt


The Attempt at a Solution



After diffrentiating the given equation for velocity I got:

a = vrel [itex]\frac{R}{m0 - Rt}[/itex] - g

and the mass relative to time is:

m(t) = m0 - Rt

from this I got:

F = ma = R - gm0 + gRt

This is where I got stuck. The question doesn't seem to ask for the thrust at a specific moment in time, yet the equation for force I got at the end still takes time into account. Did I do something wrong during differentiation?
 

Answers and Replies

  • #2
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The question asked for the thrust, and you're giving it the thrust!
So there is no problem.

EDIT:There is one problem though:

[itex]F=v_{rel}R-gm_0+gRt[/itex]
 
  • #3
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from this I got:

F = ma = R - gm0 + gRt

This is where I got stuck. The question doesn't seem to ask for the thrust at a specific moment in time, yet the equation for force I got at the end still takes time into account. Did I do something wrong during differentiation?
You did something in multiplying m*a. Look at the units. *Always* look at the units. Your R has units of kg/s. The other two terms in ##R-gm_0+gRt## have units of kg·m/s2, or newtons. You can't add a kg/s to a newton. That you obtained these mixed up units means you made a mistake.
 

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