Separation of Variables for a Rocket Launched in a Gravitational Field

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SUMMARY

The discussion focuses on solving the equation of motion for a rocket launched in a gravitational field, represented by the equation m\dot{v} = -\dot{m}v_{ex} - mg. The rocket ejects mass at a constant rate, defined as \dot{m} = -k, leading to the mass function m = m_{o} - kt. The user attempts to isolate \dot{v} and expresses it as \dot{v} = \frac{kv_{ex}}{m_{o}-kt} - g, but struggles with the presence of v_{ex} and the complexity of the denominator. The user seeks assistance with the separation of variables technique to derive v as a function of time.

PREREQUISITES
  • Understanding of calculus, particularly separation of variables.
  • Familiarity with differential equations and their applications in physics.
  • Knowledge of rocket propulsion principles, including mass flow rate and thrust.
  • Basic understanding of gravitational forces and their effects on motion.
NEXT STEPS
  • Study the method of separation of variables in differential equations.
  • Review the derivation of the Tsiolkovsky rocket equation for insights into rocket motion.
  • Explore online resources or textbooks on classical mechanics, focusing on motion under gravity.
  • Practice solving similar problems involving variable mass systems in physics.
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying dynamics and rocket propulsion, as well as educators seeking to enhance their teaching of differential equations in real-world contexts.

mar2194
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Homework Statement



Taking the equation of motion for a rocket launched from rest in a gravitational field g, m\dot{v} = -\dot{m}v_{ex} - mg, and knowing that the rocket ejects mass (fuel) at a constant rate \dot{m} = -k (where k is a positive constant), so that m = m_{o} - kt. Solve the equation m\dot{v} = -\dot{m}v_{ex} - mg for v as a function of t, using separation of variables.

Homework Equations


m\dot{v} = -\dot{m}v_{ex} - mg
\dot{m} = -k
m = m_{o} - kt.

The Attempt at a Solution



So I've been beating my head at this for a while. It is deceptively simple, and there is probably a simple calculus trick I have forgotten that I need to use to solve this. Here's what I've done so far:

1.) Substituted m = m_{o} - kt for both instances of m, giving:

(m_{o} - kt)\dot{v} = -\dot{m}v_{ex} - (m_{o} - kt)g

2.) Substituted \dot{m} = -k, giving:

m_{o}dv-ktdv+kv_{ex} = -m_{o}g + ktg

This is as far as I can get. I can isolate \dot{v} to get this:

\dot{v}=\frac{kv_{ex}}{m_{o}-kt}-g

This doesn't help me much though because I still have v_{ex} on the right side as well as having a less than elegant m_{o} - kt term on the bottom.

Can someone please help me with this separation of variables? Maybe if you can show me the way on this problem you might know of some good online references to practice doing this sort of thing or direct me to a book (I have Stuarts single variable Calculus book from a while ago when I took my 4 semester Calc sequence; time for a refresher maybe?)

Thanks for anyone who can help me.
 
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mar2194 said:

Homework Statement



Taking the equation of motion for a rocket launched from rest in a gravitational field g, m\dot{v} = -\dot{m}v_{ex} - mg, and knowing that the rocket ejects mass (fuel) at a constant rate \dot{m} = -k (where k is a positive constant), so that m = m_{o} - kt. Solve the equation m\dot{v} = -\dot{m}v_{ex} - mg for v as a function of t, using separation of variables.

Homework Equations


m\dot{v} = -\dot{m}v_{ex} - mg
\dot{m} = -k
m = m_{o} - kt.

The Attempt at a Solution



So I've been beating my head at this for a while. It is deceptively simple, and there is probably a simple calculus trick I have forgotten that I need to use to solve this. Here's what I've done so far:

1.) Substituted m = m_{o} - kt for both instances of m, giving:

(m_{o} - kt)\dot{v} = -\dot{m}v_{ex} - (m_{o} - kt)g

2.) Substituted \dot{m} = -k, giving:

m_{o}dv-ktdv+kv_{ex} = -m_{o}g + ktg

This is as far as I can get. I can isolate \dot{v} to get this:

\dot{v}=\frac{kv_{ex}}{m_{o}-kt}-g

This doesn't help me much though because I still have v_{ex} on the right side as well as having a less than elegant m_{o} - kt term on the bottom.

Can someone please help me with this separation of variables? Maybe if you can show me the way on this problem you might know of some good online references to practice doing this sort of thing or direct me to a book (I have Stuarts single variable Calculus book from a while ago when I took my 4 semester Calc sequence; time for a refresher maybe?)

Thanks for anyone who can help me.

What's the relationship between v and v_{ex}, if any?
 
Last edited:

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