How to model a rocket equation from the derivative of momentum?

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SUMMARY

This discussion focuses on modeling the rocket equation using the derivative of momentum (dp/dt) in conjunction with Newton's 3rd Law and the gravitational force of Earth. The key equation presented is F - (G * m_e * m_r / r²) = v * dm/dt + ma, where F represents the force created by fuel, G is the gravitational constant, m_e is the mass of Earth, m_r is the mass of the rocket, and dm/dt is the instantaneous rate of change of mass. Participants clarify that gravity does not need to be accounted for in the 3rd Law and emphasize the importance of relating the rate of change of momentum of ejected gases to the thrust force required to overcome gravity.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with the concept of momentum and its derivative
  • Knowledge of gravitational forces, specifically the gravitational constant (G)
  • Basic principles of rocket propulsion and mass flow rates
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Aerospace engineers, physics students, and anyone interested in rocket propulsion and the application of Newton's Laws in aerospace dynamics.

Physyx
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TL;DR
Using Newton’s 3rd Law, gravitational force, and derivative of momentum to model a rocket going into space while losing mass.
I am using the derivative of momentum (dp/dt) with Newton’s 3rd Law with the gravitational force of Earth.

F - [Force of gravity on rocket] = dp/dt
F - (G * m_e * m_r / r2 ) = v * dm/dt + ma

F = Force created by fuel (at time t)
G = Gravitational Constant
m_e = Mass of Earth
m_r = Mass of rocket (at time t)
r = Distance between Earth and rocket (at time t)
v = Velocity of rocket relative to Earth (at time t)
dm/dt = Instantaneous rate of change of mass of rocket (at time t)
m = Also mass of rocket (at time t)
a = Instantaneous acceleration of rocket (at time t, equal to dv/dt)

Is my equation correct for a standard rocket? Would dm/dt be negative as the rocket is losing mass over time? Is v relative to the Earth or the expelled gases and why?
 
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Hi Physyx. Welcome to PF!

You do not need to account for gravity in the 3rd law. You just need to relate the rate of change of momentum of the ejected rocket gases to the force on the rocket vehicle and make sure that the force exceeds the force of gravity.

AM
 
So how would I create a separate equation modeling the effect of gravity on the rocket in addition to the force created by the gas consumption?
 
Andrew Mason said:
You do not need to account for gravity. You just need to relate the rate of change of momentum of the ejected rocket gases to the force on the rocket vehicle.
By this reasoning the Apollo lunar module could happily land and take off from earth. I think you need to be more careful here.
This is detailed in many places...has the OP really looked around?
 
Physyx said:
So how would I create a separate equation modeling the effect of gravity on the rocket in addition to the force created by the gas consumption?
The rocket ejects mass at a certain constant speed. This requires a force provided by the rocket : F = dp/dt . Work out what that force is in terms of the rate of change of momentum of the ejected gas and apply the 3rd law to find the thrust force on the rocket vehicle. What does that force have to be to overcome gravity?

AM
 
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