Rocket Motion: Solving the Rocket Equation for Lift-Off

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SUMMARY

The discussion focuses on solving the rocket equation to determine the lift-off time of a rocket with an initial mass of 70,000 kg, burning fuel at a rate of 250 kg/s, and an exhaust velocity of 2,500 m/s. The correct formulation of the rocket equation is identified as v = 2500 * ln(70000 / (70000 - 250t)) - 9.8t. It is concluded that the rocket will lift off when the velocity becomes positive, specifically after approximately 47.9 seconds. An alternative approach suggests calculating the force produced by the rocket and equating it to the weight to find the lift-off time, yielding a result of about 25 seconds.

PREREQUISITES
  • Understanding of the rocket equation and its components
  • Familiarity with natural logarithms and their application in physics
  • Basic knowledge of force, mass, and acceleration principles
  • Ability to solve equations numerically
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  • Learn about numerical methods for solving nonlinear equations
  • Explore the concepts of thrust and weight balance in rocket dynamics
  • Investigate the impact of varying fuel burn rates on rocket performance
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Aerospace engineers, physics students, and anyone interested in rocket dynamics and propulsion systems will benefit from this discussion.

SlickJ
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This question was posed in class the other day for extra credit:

A rocket with initial mass 70000kg burns fuel at a rate of 250kg/s; it has an exhaust velocity of 2500m/s. If the rocket is at rest, how long after the engines fire will the rocket lift off?

I've been trying to solve it using some variation of the rocket equation:
v=v(exhaust) * ln(M(0)/M(t)) - gt
but to no success.
Any hints or help would be greatly appreciated
 
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There is a problem with the equation you are using
it is r (rate of burning the fuel) for g you used, also it is dimensionally incorrect we can have log of only natural numbers.
If it works that's Ok, otherwise try by equating force by the ejected gass to mass of rocekt at the moment.
 
SlickJ said:
This question was posed in class the other day for extra credit:

A rocket with initial mass 70000kg burns fuel at a rate of 250kg/s; it has an exhaust velocity of 2500m/s. If the rocket is at rest, how long after the engines fire will the rocket lift off?

I've been trying to solve it using some variation of the rocket equation:
v=v(exhaust) * ln(M(0)/M(t)) - gt
but to no success.
Any hints or help would be greatly appreciated

Be careful. The equation of your rocket is:

v=2500\cdot ln\Big(\frac{70000}{70000-250t}\Big)-9.8t

There is an interval of velocities 0<t<47.9 s in which v<0. The rocket will start to lifting off when v>0 or t>47.9 s. (solve numerically the equation v=0, you will obtain t=0 and t=47.9 s).
 
Thanks both of you for your help, very much appreciated.
 
I think you're all making this much too complicated.
Just calculate the force the rocket produces using the equation force = change in momentum / time, with knowledge of the fact that momentum is mass times velocity. The weight of the rocket is given by acceleration due to gravity x (original mass of rocket - (rate at which the rocket loses mass x time). Set the weight of the rocket to equal the its force of propulsion (which you already calculated) and solve for time. At that time, the force of thrust will balance the rocket's weight and immediately afterwards it will start lifting off.
 
Last edited:
I get about 25 seconds.
 

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