Variable mass F=ma versus Rocket F=ma

In summary, variable mass F=ma is an equation used to calculate force in systems where the mass is changing, while rocket F=ma specifically refers to the case of variable mass in rocket engines. In variable mass F=ma, force increases as mass increases, but in rocket F=ma, force increases as mass decreases due to the conservation of momentum. Both equations have various applications in physics, but a modified version of the equation is needed for rocket motion. Inertia is a key concept in both equations, as it affects the force required to change an object's motion and is overcome by the reaction force in rocket propulsion.
  • #1
walking
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In the book by Tipler & Mosca, the section on F=ma for variable mass derives the following equation:

##\mathbf{F}_{ext}+\frac{dM}{dt} \mathbf{v}_{rel}=M\frac{d\mathbf{v}}{dt}##

where ##\mathbf{F}_ext## is the external force on the system as a whole (ie not just the variable mass sub-system, but the system as a whole), ##\frac{dM}{dt}## is the rate at which the mass of the variable mass sub-system is changing, ##\mathbf{v}_{rel}## is the velocity of the changing mass (whether incoming or outgoing) relative to the variable mass subsystem, ##M## is the mass of the variable mass subsystem at time t, and finally, ##\mathbf{v}## is the velocity of the variable mass subsystem at time t.

Now I may have misunderstood this equation, but if not then I am slightly confused as to the authors's subsequent derivation of the rocket equation. When deriving this latter equation, they seem to interpret the general variable mass equation completely differently, and ##F_{Ext}## no longer represents the total external force on the system for example. Here is their rocket equation:

##M\mathbf{g}-R\mathbf{u}_{ex}=M\frac{d\mathbf{v}}{dt}##,

where ##M## is the rocket's mass at time t, -R is the rate at which fuel-mass is leaving, ##\mathbf{u}_{ex}## is the velocity of the leaving fuel relative to the rocket.

As I said, the authors seem to have changed their interpretation of ##F_{ext}##, because when I use the original variable mass equation on a rocket, I get a different answer. Here is my derivation of the rocket equation based on my understanding (a possibly erroneous one) of the general variable mass equation.

For a rocket-fuel system, ##F_{ext}## in the general equation should be the total gravitational force on the rocket and the fuel. If we let the initial mass of the rocket-fuel be ##M_0##, then ##F_{ext}=M_0\mathbf{g}##. Then deriving the rest of the rocket equation as the authors do, I get

##M_0\mathbf{g}-R\mathbf{u}_{ex}=M\frac{d\mathbf{v}}{dt}##,

which doesn't agree with their equation. The problem, as I highlighted above, seems to be the authors's use of ##F_{ext}##. Here are their derivations of both equations.

General variable mass F=ma:

DSC00320.JPG


DSC00321.JPG


DSC00322.JPG


Rocket equation:

DSC00323.JPG


DSC00325.JPG
 
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  • #2
walking said:
When deriving this latter equation, they seem to interpret the general variable mass equation completely differently, and ##F_{Ext}## no longer represents the total external force on the system for example.

Maybe they just use a little trick: For the current acceleration of the rocket it doesn't matter weather it is already burning for hours or if it has just been started (because exhausted reaction mass doesn't interact with the rocket anymore). Therefore you can always assume M = Mo and neglect the external force acting on the exhaust.

Edit: After reading the sourece again I'm quite sure this is the case and it is not even a trick. The "whole system" that the external force is acting on consists of the rocket (with the mass M) and the reaction mass that is currently exhausted (with the mass ##\Delta M##). Reaction mass that has been exhausted in the past (with the mass ##M_0 - M - \Delta M##) is not included.
 
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1. What is the difference between variable mass F=ma and rocket F=ma?

The main difference between these two concepts is the inclusion of the variable mass term in the rocket equation. In traditional F=ma equations, the mass of the object remains constant. However, in the rocket equation, the mass of the rocket decreases as fuel is burned, resulting in a change in acceleration.

2. How does the variable mass term affect the rocket's acceleration?

The variable mass term in the rocket equation directly affects the acceleration of the rocket. As the mass decreases, the acceleration increases, resulting in a higher velocity and ultimately allowing the rocket to overcome the force of gravity and reach space.

3. Can the variable mass term be ignored in rocket calculations?

No, the variable mass term is a crucial component in accurately calculating the motion of a rocket. Ignoring this term would result in incorrect calculations and could potentially lead to disastrous consequences.

4. What other factors are involved in rocket motion besides F=ma?

In addition to F=ma, there are other factors that play a role in rocket motion, such as air resistance, thrust, and gravity. These factors must all be taken into account in order to accurately predict the trajectory and performance of a rocket.

5. How does the rocket equation apply to real-life space missions?

The rocket equation is a fundamental concept in space travel and is used in the design and operation of rockets for real-life space missions. Engineers and scientists use this equation to calculate the necessary thrust and fuel requirements for a successful launch and trajectory to a desired destination in space.

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