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

In summary, the equation for a rocket that uses atmospheric air to provide thrust is: F = dp/dt Where: 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 = Mass of rocket (at time t)
  • #1
Physyx
2
0
TL;DR Summary
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?
 
Physics news on Phys.org
  • #2
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
 
  • #3
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?
 
  • #4
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?
 
  • #5
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
 
Last edited:

1. What is the rocket equation?

The rocket equation is a mathematical representation of the motion of a rocket in space. It takes into account the mass of the rocket, the mass of the propellant, and the velocity of the rocket.

2. How is the rocket equation derived?

The rocket equation is derived from the principle of conservation of momentum. By taking the derivative of momentum with respect to time, we can determine the change in momentum of the rocket as it expels propellant and accelerates.

3. What are the variables in the rocket equation?

The variables in the rocket equation include the initial mass of the rocket, the final mass of the rocket after expelling propellant, the velocity of the rocket, and the velocity of the expelled propellant.

4. How is the rocket equation used in rocket design?

The rocket equation is used in rocket design to determine the amount of propellant needed to achieve a desired velocity or trajectory. It also helps engineers optimize the design of the rocket by considering factors such as thrust, mass, and fuel efficiency.

5. Are there any limitations to the rocket equation?

Yes, the rocket equation assumes a constant rate of propellant expulsion and does not take into account external forces such as air resistance. It also does not account for the effects of gravity or the changing mass of the rocket as it expels propellant. These limitations can be addressed through more complex models and simulations.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
231
  • Introductory Physics Homework Help
Replies
10
Views
666
  • Introductory Physics Homework Help
Replies
1
Views
882
  • Introductory Physics Homework Help
2
Replies
42
Views
3K
Replies
0
Views
1K
Replies
1
Views
161
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
4
Views
2K
  • Classical Physics
Replies
17
Views
2K
Back
Top