Acceleration, velocity and position of a rocket

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SUMMARY

The discussion focuses on the equations for acceleration, velocity, and position of a rocket as functions of time, specifically using the thrust (T), initial mass (m), mass flow rate (ṁ), and gravity (g). The user derived the equations for velocity and position through integration but encountered issues with unit consistency and initial conditions. The community clarified that the natural logarithm function (ln) is dimensionless, which resolves the unit discrepancies in the velocity equation. The conversation emphasizes the importance of correctly applying initial conditions and understanding the dimensional analysis of logarithmic functions in physics.

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  • Understanding of Newton's laws of motion
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of rocket propulsion principles, including thrust and mass flow rate
  • Basic grasp of dimensional analysis in physics
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  • Learn about the implications of initial conditions in differential equations
  • Explore the concept of dimensionless quantities in physics
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Aerospace engineers, physics students, and anyone interested in the mathematical modeling of rocket dynamics will benefit from this discussion.

Belginator
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Hi everyone,

Quick question I may just not be thinking right here but I was trying to find the acceleration, velocity, and position of a rocket as a function of time. I started with acceleration:

a =\frac{T}{m-\dot{m}t} - g where T is the Thrust, m is the initial mass, mdot is the mass flow rate, and g is gravity. This equation seems to work out with dimensional analysis and logically it seems to make sense, but maybe I'm wrong there. So from there I integrated wrt to time to get the velocity:

v = -\frac{T}{\dot{m}} ln(m-\dot{m}t) -gt + v_0 Here is where the problem comes in, while I'm pretty sure I did my integration right, the units don't work out properly and velocity doesn't start out at 0 either, unless you set the v_0 term to some value. Finally I tried to get position by integrating v:

s = \frac{T}{\dot{m}^2} ((m-\dot{m}t)ln(m-\dot{m}t) - (m-\dot{m}t)) - 0.5gt^2 + v_0t + s_0 Again the units don't work out properly.

I'm just considering where the rocket goes vertical for now, no horizontal components.

What am I not seeing here? This should be fairly straight forward. Thanks in advance for any help.
 
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Hi Belginator! :smile:
Belginator said:
v = -\frac{T}{\dot{m}} ln(m-\dot{m}t) -gt + v_0 Here is where the problem comes in, while I'm pretty sure I did my integration right, the units don't work out properly and velocity doesn't start out at 0 either, unless you set the v_0 term to some value.

No, the units are fine.

ln has no units (like sin) …

your ln(m-m't) + vo is really ln((m-m't)/(mo)) for some constant mo :wink:
 

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