Question about the Rocket Equation

[Johnny5]

Can someone explain the rocket equation to me in as much detail as they are capable of?

 

Thank you

[Klaynos]

the rocket equation as in rocket motion, the one bassed on conservation of momentum?

 

Which I don't know but have to derive every time I use it?

[Johnny5]

the rocket equation as in rocket motion' date=' the one bassed on conservation of momentum?

 

Which I don't know but have to derive every time I use it?[/quote']

 

Yes that one. Thats the one.

[Klaynos]

Posting in two parts first the basics for rocket motion in any situation...

 

ok well due to conservation of momentum:

 

momentum before = momentum after

 

So we have a trolly on frictionless rails, you are sat in the trolly, total mass M, total velocity 0. Air resistance is negligable, change in magnetic fields gravitational fields etc etc... are negligable...

 

P = M0 = 0

 

You then throw a small marble off of the back of the trolly with mass m, and velocity v, the new velocity of the trolly is u. (note velocities are vectors). So back to orignal equation:

 

P = 0 = mv + (M-m)u

 

Re arange to:

 

(M-m)u = mv

 

u= -mv/(M-u)

 

If you take that you through the marble off of the back then that is a negative velocity, giving you a positive velocity on the trolly.

 

Throw one of the back one of the front and you are stationary if they are at the same velocity...#

[Klaynos]

Now then...

 

Taking the situation of a rocket, we have the rocket effectively throwing out fule at a certain rate.

 

dm/dt

 

in time dt (total time = t + dt)

 

for the original starting momentum:

 

P₁ = mv

 

(m total mass of rocket + starting fule, v = starting velocity)

 

so

 

v_fuel=v+(-v_ex)

 

Where v_ex is the exhaust velocity relative to the ship.

 

the ejected mass (-dm)

 

so the momentum of the ejected mass is:

 

(-dm)v_fuel=(-dm)(v-v_ex)

 

in the time interval dt, the rocket and unburned fule is now increaded to dv+v.

And it's mass has decreased to m+dm (where dm is negative). So the rockets momentum is:

 

(m+dm)(v+dv)

 

Thus the total momentum is P₂ of the rocket + ejected fule is:

 

P₂ = (m+dm)(v+dv) + (-dm)(v-v_ex)

 

So using conservation of motion

 

P₁ = P₂

 

mv = (m+dm)(v+dv) + (-dm)(v-v_ex)

 

simplifying this:

 

mdv=-dmv_ex-dmdv

 

dm and dv are both infitesimally small so dmdv can be neglectied. We now devide by dt to produce 2 rate of change componants (rate of change of velocity, and rate of change of mass)

 

m(dv/dt) = -v_ex(dm/dt)

 

dv/dt = acceleration. F=ma.

 

F = -v_ex(dm/dt)

 

Therefore the thrust is preportional both to the relative speed of the ejected fuel and also to the rate at which it is discharged.

 

If you wish to find such things as differences in velocities or mass between two points v and v₀ say you can integrate this, I'm too lazy to prove this so take my word for it the integral is:

 

v-v₀ = -v_exln(m/m₀) = v_exln(m₀/m)

 

using limits v, v₀, and m, m₀...

[Johnny5]

(M-m)u = mv

 

u= -mv/(M-u)

 

Minor error there' date=' that should read...

 

u = -mv/(M-m)

 

 

PS: If you made the error intentionally, that's a good way to check to see if I followed your work. I still have more reading to do.

 

Kind regards

 

I'm just looking at this formula now.

 

u = -mv/(M-m)

 

M is the system mass, which is composed of three parts, the mass of the trolly, your mass, and the mass of the marble m.

 

So clearly, the system mass M exceeds the marble mass.

 

This means that M-m is greater than zero.

 

Taking that inertial mass is strictly a positive quantity, means that m/(M-m) is a positive quantity.

 

Ok so...

 

initially the system is being observed in a frame in which its center of mass isn't moving. So in this frame, initially, the speed of the trolly is absolute zero.

 

After the marble has been thrown, the trolly will move [i']in this frame [/i] opposite the direction which the marble is thrown, by action/reaction principle.

 

Since it will now be moving in this frame, you could introduce a symbol for the nonzero speed which it will then have, and something to indicate direction, but instead, you used the letter u, to denote the velocity of the trolly in this frame. Following so far...

 

Here is the formula again:

 

u = -mv/(M-m)

 

Since u is a vector quantity, and v is a vector quantity, I am going to boldface them, so that I can follow your argument more easily...

 

u = -m v /(M-m)

 

Sine m/(M-m) is positive, multiplication of any quantity by it does not change algebraic sign.

 

I am going to throw in some specific values.

 

Let M=100

Let m = 1

 

Therefore

 

m/(M-m) = 1/(100-1) = 1/99

 

So we now have this...

 

u = -(1/99)v

 

Now, suppose that the LHS is pointing in the i^ direction of the reference frame.

 

In order to have the equation above be a true statement in that reference frame, it absolutely has to be the case that the direction of v is given by

 

[math] - \hat i [/math]

 

You can see why, by writing each vector as its magnitude times its direction.

 

Let us use nonbold to indicate magnitude. THerefore:

 

[math] \mathbf u = u \hat i [/math]

 

[math] \mathbf v = v ? [/math]

 

[math] \mathbf u = -(1/99) \mathbf v [/math]

 

[math] u \hat i = -(1/99) (v?) [/math]

 

The scalar parts of each side of the equation above must be equivalent, and the direction of each side must be equivalent.

 

Now, speed is a scalar quantity. It is a measure of how fast something is moving through a frame. Negative speed is nonsense. Speed is always a positive quantity. So for example, the initial speed of the trolly was zero in the frame, and the final speed of the trolly might be something like 2 meters per second. But the final speed of the trolly could never be -2 meters per second, because as I said, negative speed is meaningless. (this fact comes straight out of the definition of speed, which is this (speed is distance travelled divided by time of travel). There is no such thing as a negative amount of time, else time would run in reverse. And there is no such thing as negative distance travelled, which dovetails with the fact that distance is measured by rulers, and rulers can only measure positive quantities. The point is, that speed u must be positive, and speed v must be positive. So now we can solve for the direction of the marble in the frame, in terms of the direction of the trolly in the frame.

 

[math] u = (1/99) v [/math]

 

[math] \hat i = -? [/math]

 

[math] ? = - \hat i [/math]

 

Where the symbol ? was used to denote the direction of the trolly in the frame. It is now plainly evident that the direction of the trolly in the frame, and the direction of the marble in the frame are antiparallel, that is they are exact opposite directions.

 

And as you said, if you throw one off the back, and one off the front, the trolly will remain stationary in the frame.

 

Ok I followed your first post. Now onto the second.

[Johnny5]

Now then...

 

Taking the situation of a rocket' date=' we have the rocket effectively throwing out fule at a certain rate.

[/quote']

 

This is one means of propulsion, but is it the only one?

 

But ok, we are doing conventional rocketry here.

 

So you have um, the rocket being propelled by ejecting fuel.

 

Now, there are various frames we can analyze the motion in. The first two which come to mind are these...

 

The center of mass frame of the system (Rocket + fuel).

 

The rest frame of the rocket.

 

In order to help me understand this properly, I am going to imagine a Gaussian sphere, that completely encases the rocket. The center of the sphere is permanently the center of inertia of the rocket. So wherever the rocket goes, this Gaussian sphere of fixed radius goes. The sphere is just large enough to encase the whole rocket, but no larger.

 

What is going to happen is this... material is going to exit from that sphere. There will be mass loss formulas (i am reminded of the continuity equation of electrodynamics).

 

So this frame is what I am calling the rest frame of the rocket, because the center of inertia of the rocket is permanently located at the origin of this frame, and furthermore there is a simple Gaussian sphere that the rocket carries with it.

 

Now, the first thing that you say, is that the rocket effectively throws out fuel at a certain rate dm/dt.

 

Because of relativity, issues have arisen about clocks that make time measurements, so what clock is measuring dt becomes an issue. But, the dt must be measured by a clock in the system in which you have the dm.

 

Now there is also the center of mass frame of the system.

 

When a grenade blows up, if you view the motion of the parts in the CM rest frame, the center of mass of the grenade never moves.

 

So before the engines are turned on... the center of mass of the rocket is at rest in the CM rest frame.

 

Now, once fuel is being ejected, there is a force upon the rocket, which causes it to accelerate in the CM frame. So the center of the Gaussian sphere I was previously talking about accelerates in the CM rest frame.

 

And we can stipulate that the CM frame is inertial.

 

So there is also a force on the fuel particles, and they accelerate in the opposite direction in the CM frame. We have to have conservation of momentum in this frame, since it's inertial.

 

Now, in any Gaussian sphere large enough to contain all the fuel particles as well as the rocket over some large time interval Dt in the CM rest frame, the net mass change through the surface of that sphere is zero, because the sphere was chosen to be so large.

 

But, the Gaussian sphere which is attached to the center of inertia of the rocket, does have nonzero dm, at the very moment the particles are ejected out the back of the rocket. So that is the frame in which dm is defined in, and so that is the frame that dt must be defined in, and that frame, the rocket frame, is a non-inertial reference frame.

 

So for an adherent of SR, the clock there isn't going to measure proper time. Luckily, I don't believe SR is correct, so I can neglect this issue.

 

So here is what we are going to have...

 

There is going to be a clock, permanently located at the rocket's center of inertia, and it will measure dt.

 

And there is a Gaussian sphere, whose center is also permanently attached to the rocket's center of inertia, and whose radius is just large enough to encase the whole rocket.

 

Before the propulsion system is turned on:

 

dm=0

 

After the propulsion system is turned on:

 

not (dm=0)

 

Keep in mind that this dm, as defined by klaynos, is referenced to the Gaussian sphere that is traveling with the rocket's center of inertia, and is just large enough to encase the rocket. So the center of the Gaussian sphere is accelerating through the coordinates of the CM rest frame, and the CM rest frame is the inertial reference frame, not the rocket frame. I just want to keep everything straight is all.

 

So...

 

Then you mention this:

 

in time dt (total time = t + dt)

 

I didn't understand this right away' date=' I think mainly because you didn't specify where the clock is, which is measuring dt. But it is inferrable from the fact that it has to be the system dm is defined/measured in.

 

you have added dt to t, and you are calling that "total time"

 

Let me think...

 

I think of 'dt' as an infinitessimal [i']amount of time[/i], and t as the time coordinate of the reference frame in which dm is measured in (since I don't use SR i dont usually worry about a time coordinate anyway, I am usually focused on amounts of time as measured by specific clocks, be they accelerating through the coordinates an IRF(as in this case) or otherwise). I didn't understand this definition of total time, you will have to explain it to me more.

 

Moving on...

 

for the original starting momentum:

 

P1 = mv

 

(m total mass of rocket + starting fule' date=' v = starting velocity)

[/quote']

 

Velocity only makes sense, if you specify the exact reference frame in which you are definining it in.

 

 

I was imagining the rocket initially being at rest in deep space, in some inertial reference frame. But it could just as well have been initially coasting through interstellar space at a constant speed v.

 

So you are letting m denote the mass of the fuel plus the mass of the rocket.

 

Let the rocket be made mainly of silver AG atomic number 47, so the ship has a silver hull.

 

Right now, it seems to make sense to think of the fuel as a liquid propellant.

 

So we could weight that fuel here on earth, and we could weigh the rocket here on earth. Both measurable quantities.

 

I don't want to get confused about anything so let me do this...

 

Clearly the rocketship itself is made of solid material.

 

It is the amount of liquid fuel thats changing within the boundary of the rocket.

 

Ok so you have the initial momentum equal to (Fuel mass + Rocket mass) times the velocity of the rocket in some frame.

 

[math] \vec P_1 = m \vec v [/math]

 

Now you introduce the velocity of the exhaust relative to the ship, which is easy to understand. So I just have to take a moment, to think about how that is measured.

 

Ok here is what I am going to do. Rather than thinking of this rocket as combining oxygen and hydrogen fuel, I am going to make it far simpler, but I think equivalent so far as rocket physics is concerned...

 

There is a water hose nozzle, at the bottom of the ship. You squeeze the handle, and water jets out the nozzle of the hose.

 

 

When you squeeze the nozzle of a water hose, you can feel a force exerted on your hand, your hand is pushed in the direction opposite to which the water is squirting.

 

So of course the water exits the nozzle at some speed v_ex relative to the nozzle.

 

(Incidentally, this is how toy water rockets work, which you can buy at a hobby store. it was a huge money maker in the seventies. You pump in air into a tube filled with water, and set up your rocket, then flick a switch to allow the water to exit, and the rocket soars 50 to 100 feet into the air)

 

 

So let there be a long ruler attached to the rocket ship, and let the waterfuel be ejected roughly parallel to the ruler.

 

We have a means for time measurement, the clock on the ship, and now a means for length measurement, a ruler attached to the ship, so that we can make speed and acceleration measurements for any given water molecule relative to the ship.

 

Now you introduce this formula here:

 

[math] v_{fuel} = v + (-v_{ex}) [/math]

 

I'm not sure if you mean speed or velocity here. It would help if you specified what frames these speeds are in.

 

Wait I think I get it now...

 

In the case where the rocket starts off from rest v in the above formula is equal to zero, in which case you have:

 

[math] v_{fuel} = -v_{ex} [/math]

 

So they are not speeds (since speed is nonnegative), they must be velocities. That means this is a vector equation. I will write it that way then.

 

[math] \vec v_{fuel} = -\vec v_{ex} [/math]

 

These two vectors are antiparallel. The RHS is the velocity of the water being ejected. I guess the LHS is the remaining fuel located inside the ship.

 

I'm not sure about the truth value of this equation

 

[math] v_{fuel} = v + (-v_{ex}) [/math]

 

but continuing...

 

You next make reference to the ejected mass -dm.

 

You could do a better job explaining this.

 

dm is a differential change in mass, a tiny change in mass. Also, I'd like to add that mass isn't supposed to be a vector, its supposed to be a scalar. (I've read that mass is a vector in a few places though, but I have no idea why anyone does that). Also, mass is taken to be positive. In other words, there is no such thing as negative inertial mass.

 

Quite simplistically, we have a formula for it

 

[math] dm = m_2 - m_1 [/math]

 

m2,m1 are mass values at adjacent moments in time. m1 is the earlier value, m2 the latter value in time.

 

So since the rocket is ejecting material from within its boundaries, the amount of matter inside of the Gaussian sphere is decreasing, not increasing, so this means that m1 is greater than m2, which means that (m2-m1) is a negative number.

 

My point then is that dm = - 5 makes sense, but you suddenly put a negative sign in the formula there. That's not very mathematically rigorous.

 

Here is the portion of your argument which I'm focusing on now:

 

vfuel=v+(-vex)

 

Where vex is the exhaust velocity relative to the ship.

 

the ejected mass (-dm)

 

so the momentum of the ejected mass is:

 

(-dm)vfuel=(-dm)(v-vex)

 

in the time interval dt' date=' the rocket and unburned fule is now increaded to dv+v.

And it's mass has decreased to m+dm (where dm is negative). So the rockets momentum is:

 

(m+dm)(v+dv)

 

Thus the total momentum is P2 of the rocket + ejected fule is:

 

P2 = (m+dm)(v+dv) + (-dm)(v-vex)

 

So using conservation of motion

 

P1 = P2

[/quote']

You could just do this..

 

 

Earlier in your argument you stipulate that the following statement is true:

 

[math] v_{fuel} = v - v_{ex} [/math]

 

Now, regardless of what dm even means, the following is true:

 

[math] dm (v_{fuel}) = dm ( v - v_{ex}) [/math]

 

This here I understood perfectly...

 

in the time interval dt' date=' the rocket and unburned fule is now increaded to dv+v.

[/quote']

 

 

In the formula above, v denotes the previous velocity, dv is the tiny change in velocity, so that dv+v is the new velocity one moment in time later. And dt of course is an infinitessimal amount of time.

 

 

Now, the rocket is having its speed increased, in the CM rest frame. Let's maintain that its direction of travel is constant as it accelerates through the coordinates of the CM frame.

 

So we can focus on the speed of the center of inertia of the rocket in the CM rest frame.

 

[math] dv = v2-v1 [/math]

 

Notice I made no reference to vector notation in the formula above, because speed is a scalar not a vector. Speed is a bit easier to work with rather than velocity.

 

The LHS of the formula above is a differential change in the speed of the rocket's center of inertia in the CM frame. v1 denotes the rocket's initial speed in the frame, and v2 denotes the speed of the rocket in the frame one moment in time later.

 

Next you say this:

 

And it's mass has decreased to m+dm (where dm is negative). So the rockets momentum is:

 

 

In amount of time dt' date=' the total mass interior to the Gaussian sphere which is ship-centered decreases (because matter was ejected, no matter is entering obviously, because the ship's hull is impenetrable)

 

You switched dm on me i think. dm is positive, you previously made it that way remember? when you wrote -dm, what you effectively did was to make dm strictly positive, so that -dm would be strictly negative.

 

See this is why mathematical symbolism is so important in physics.

 

Here is all you need:

 

[math'] dm = m2-m1 [/math]

 

dm is the change in mass, m1 is the initial mass, and m2 is the mass one moment in time later. From the equation above we have:

 

[math] m1+dm = m2 [/math]

 

We are clear that m2 is less than m1 already.

 

So in the formula above dm is negative. That's what you mean, i know it is.

 

You next say that the rockets momentum is:

 

(m+dm)(v+dv)

 

I have absolutely no clue where you came up with that.

 

It really would help your argument, if you labeled which moments in time the values in your formulas correspond to. So that simultaneous quantities would be identifiable.

 

In the formula above, i think m denotes the mass interior to the Gaussian surface at some moment in time, and v is the speed speed at the same moment in time, where v is defined in the CM rest frame.

 

At the very next moment in time, the total mass interior to the Gaussian sphere has changed, and so has the speed of the center of the sphere in the CM frame.

 

The new mass is: m+dm

The new speed is: v+dv

 

The magnitude of the old momentum was: mv

The new magnitude is : (new mass)(new speed) = (m+dm)(v+dv)

 

Ok I am with you this far.

 

Ok so, since the CM rest frame is an inertial reference frame, the law of conservation of momentum is true in that frame, as you stated quite nicely

 

momentum before = momentum after

 

In other words the momentum value doesn't change.

 

next you say

Thus the total momentum is P2 of the rocket + ejected fule is:

 

P2 = (m+dm)(v+dv) + (-dm)(v-vex)

 

 

 

Here was the initial momentum' date=' as defined by you:

 

[math'] \vec P_1 = m_1 \vec v_1 [/math]

 

 

Now, the momentum at the very next moment in time is going to be denoted by:

 

[math] \vec P_2 [/math]

 

So that the law of conservation of momentum can be expressed as:

 

[math] \vec P_1 = \vec P_2 [/math]

 

So a blob of mass was emitted in one direction, and the rocket accelerated in the other direction.

 

The final momentum P2, is the sum of the momentum's of the parts, the new momentum of the rocket, and the new momentum of the blob of ejected material.

 

We already know the magnitude of the new momentum of the rocket in the CM frame, that quantity is (m+dm)(v+dv). As for the blob of ejected fuel...

 

We need the momentum of the blob at the moment in time the momentum of the rocket is (m+dm)(v+dv).

 

Let's assume that mass is conserved. Recall:

 

[math] dm = m2-m1 [/math]

 

The LHS is negative, and mass cannot be negative, therefore the LHS cannot be the "mass of the blob" It makes more sense to say that -dm is the mass of the blob, since mass is supposed to be a strictly positive quantity.

 

So now, what we need to know, is the speed of the center of inertia of the blob, in the CM rest frame, at the moment in time that the momentum of the rocket is (m+dm)(v+dv).

 

You have that as being equal to v- v_ex.

 

It is very important to know what v is in that formula right there. v is the speed/velocity of the center of the gaussian sphere one moment in time before the sphere has momentum (m+dm)(v+dv).

 

Ok I think I have it now...

 

Focus on the speed of a particular water molecule in the CM frame, and forget about the rocket completely.

 

Either the water molecule is being carried along with the rocket, or it just got ejected out the back.

 

Before the engines are turned on, the speed of this water molecule in the CM rest frame is identical to the speed of the rocket in the CM frame. They are moving together.

 

Suppose that initially, the rocket is coasting through interstellar space at a constant speed v.

 

Let the trajectory be a straight line.

 

Therefore, the rocket is initially in an inertial reference frame, that is the CM rest frame is an inertial reference frame, just as we previously stated.

 

So before the engines are turned on the rocket has momentum

 

P1 = m v

 

in this inertial reference frame, the one in which the CM of the system is moving with a constant velocity.

 

So the center of mass of the blob (which is about to be ejected out the back of the rocket) has this same velocity v, as the center of inertia of the rocket, at least initially.

 

Now, the speed of the ship increases in this frame.

 

Therefore, the speed of the center of mass of the blob decreases because it is ejected out the back.

 

Now, we are given that the exhaust velocity v_ex of the rocket is known.

 

v is the initial speed of the center of mass of the blob in the inertial frame.

 

 

I left for awhile and now I am back to finish this up.

 

I left off at a place where I had to determine what the speed of the center of inertia of the blob is, in the CM rest inertial frame. And that's the system CM, not the CM frame of either of the parts.

 

Now this part of the problem has to use the exhaust velocity, because one that is easy to measure or know in advance, and two it's quite intuitive. Its the speed of ejected matter in the rest frame of the rocket/Gaussian sphere.

 

Not to mention it's part of the solution here:

 

Rocket science

 

Before the blob is ejected outside of the Gaussian sphere, it's speed is v, which is identical to the speed of the center of inertia of the Gaussian sphere in the CM rest inertial frame.

 

Then the blob is ejected, and the blob is now moving AWAY from the Gaussian sphere/rocket, with speed v_ex.

 

I think the intelligent thing to do now, is write some formulas for coordinate transformations.

 

I am going to be a bit more mathematical now.

 

 

Many times throughout this argument, I have referred to the CM rest inertial frame.

 

As I recall, I stipulated that before the engines of the rocket are turned on, the velocity of the center of mass of the rocket is constant, so as to ensure that this frame is inertial. Furthermore, I had to do this because Klaynos has made the rocket have an arbitrary initial velocity v.

 

He did not specify the initial conditions of the acceleration, but I took for granted that initially, the acceleration of the rocket is zero in the frame.

 

CM inertial frame:

 

I am going to refer to this frame as frame S.

 

Set up a three dimensional rectangular coordinate system, with its origin somewhere in space. Stipulate that this reference frame is an inertial reference frame. That means that all three of Newton's laws are true statements in this frame. Most importantly, the law of inertia is true.

 

Law of inertia: An object will either remain at rest or in motion in a straight line at a constant speed, unless acted upon by an external force.

 

So initially, the rocket's engines are off, and the object is in outer space, in the vacuum, so there is no external force upon it.

 

Now, the trajectory of the center of inertia of the rocket/Gaussian sphere through this frame must obey Newton's first law in the frame.

 

Let the object be coasting at a constant speed v, in the i^ direction. Thus, the initial velocity of the rocket/Gaussian surface is:

 

[math] \vec v = v \hat i [/math]

 

The letter v in the statement above, is the speed of the object in frame S. Speed is a strictly nonnegative quantity.

 

 

Definition: The speed of the center of mass of an object is loosely defined as the distance travelled by the center of mass in some frame, divided by the amount of time of travel as measured by a clock which is stationary (not moving) in the frame.

 

But we can also define speed in terms of the coordinates of the reference frame the object is moving in (or at rest).

 

In the example here, the object is always located on the positive x axis of the frame. So in order for its position in the frame to change, its x coordinate in the frame must change. So the distance it has travelled in the frame, over any two consecutive moments in time, is found by subtracting the lesser X coordinate from the greater.

 

Since the object is moving in the direction of increasing X coordinates, it follows that over any two consecutive moments in time, the lesser X coordinate comes first in temporal order.

 

We can represent a tiny change in X coordinates by:

 

[math] dx [/math]

 

Let the initial X coordinate of the center of mass of the Gaussian sphere/rocket in frame S be denoted by:

 

[math] X_1 [/math]

 

And let the X coordinate of the center of mass of the Gaussian sphere/rocket in frame S be denoted by:

 

[math] X_2 [/math]

 

Thus we can write:

 

[math] dx = X_2 - X_1 [/math]

 

And we obtain the speed of the object in S by dividing both sides by dt. That is:

 

[math] v = \frac{dx}{dt} = \frac{(X_2 - X_1)}{dt} [/math]

 

And we have for the initial velocity (before blob is emitted) of the Gaussian sphere/rocket in frame S...

 

[math] \vec v = v \hat i = \frac{dx}{dt} \hat i= \frac{(X_2 - X_1)}{dt} \hat i [/math]

 

So right now, the blob is happily inside the sphere, and the sphere is coasting down the X axis of inertial reference frame S.

 

Now, there is another frame in which we can view things, and that frame is the rest frame of the rocket. The rest frame of the rocket is attached to the rocket, and the center of inertia of the rocket is permanently located at the origin. So wherever the rocket goes, so does the rest frame of the rocket.

 

I am going to refer to this frame as S`

 

Now, the origin of frame S` is the center of mass of the Gaussian sphere/rocket, and we just found its velocity in frame S.

 

Now I don't want the axes of frame S` spinning wildly in frame S, so let the x,y,z axes of frame S` be parallel to the x,y,z axes of S, and let the basis vectors (i,j,k) have the same direction as S's basis vectors.

 

I am using the Galilean transformations.

 

Now, all I want to do is focus on the speed of the center of mass of the blob of fuel that gets ejected from the rocket.

 

There comes a moment in time when the blob exerts a force on the rocket, and vice versa. The rocket speeds up in frame S, and the blob slows down in frame S. The initial speed of the blob in frame S is v. All I want now is the final speed of the blob in frame S, in terms of the exhaust velocity v_ex.

 

The exhaust velocity only has meaning in frame S`.

 

The initial speed of the blob in frame S` is zero.

 

At the instant the blob is ejected from the rocket, the speed of the blob in frame S` is v_ex.

 

Now, let me introduce a third reference frame S``

 

At the instant in time at which the blob is ejected from the rocket, we can view the motion of the rocket and the blob in S`` where the origin of S`` is permanently located where the center of mass of the blob+rocket is.

 

So in other words, suppose for the sake of clarity, that the blob has a mass which is identical to the mass of the Gaussian sphere/rocket.

 

Let us use the letter M to denote the mass of either object.

 

Now before the blob is ejected from the rocket, the mass of the rocket/blob system is given by:

 

[math] M+M=2M [/math]

 

It will help out a whole lot if I introduce some numbers.

 

Let the initial speed of the (blob+rocket) be 200 meters per second.

 

Therefore, the center of mass of the (blob+rocket) has the following velocity in reference frame S:

 

[math] \vec v = 200 \hat i [/math]

 

Now, let the rocket have a mass of 15,000 kilograms. Since the blob's mass is identical to the rockets mass, the blob also has a mass of 15,000 kilograms.

 

Suppose that at the moment the blob is ejected from the rocket, the center of mass of the (rocket + blob) is located at X coordinate 80 meters, in inertial reference frame S.

 

Since origin and all points of reference frame S` were initially moving in a straight lines at constant speed through S, reference frame S` is also inertial.

 

 

Now, even after the blob is ejected from the rocket, the origin of S`` will continue to move through S in a straight line at a constant speed. The rocket will have speed V1 along the positive x axis of frame S``, and the blob will have speed V2 along the negative x axis of frame S``. If the masses of the blob and rocket are equal then V2=V1.

 

IN the case where the mass of the blob and the mass of the rocket are different, the speeds of each in reference frame S`` can be found using conservation of momentum in frame S``. Since frame S`` is inertial, we know that conservation of momentum is true in S``.

 

let Mb = mass of blob

let Mr = mass of rocket

 

initially both are at rest in the frame so the total momentum of the (rocket+blob) in frame S`` is zero.

 

From conservation of momentum in S``, after the blob is ejected from the rocket, the sum of the individual momenta of the parts must still be zero.

 

The rocket is going to accelerate in the i^ direction of S, which is also the i^ direction of S``. The blob is going to accelerate in the -i^ direction of S, which is also the -i^ direction of S``.

 

let the final speed of the rocket in S`` be

Vr

let the final speed of the blob in S`` be

Vb

 

From conservation of momentum in S`` we have:

 

MbVb = MrVr

 

So the speed of the blob in S`` is:

 

Vb = MrVr/Mb

 

Keep in mind that the speed of the origin of S`` through S is constant, from memory the speed of S`` through S is 200 meters per second.

 

Since I am using the Galilean transformations, it follows that after the blob is emitted from the Gaussian sphere/rocket, the speed of the blob in reference frame S will be given by:

 

200 - Vb

 

And the speed of the rocket in S will be given by:

 

200 + Vr

 

And the speed of the blob relative to the rocket is

 

Vb+Vr

 

which is just the exhaust velocity v_ex.

 

So we can write this:

 

v_ex = Vb+Vr

 

Vb = MrVr/Mb

 

MbVb/Mr = Vr

 

From which it follows that:

 

v_ex = Vb+Vr = vb+MbVb/Mr = vb (1+Mb/Mr)

 

From which it follows that:

 

v_ex = vb (1+Mb/Mr)

 

 

v_ex/ (1+Mb/Mr) = vb

 

 

Now, the speed of the blob in inertial reference frame S we have as:

 

200 - Vb

 

And now we can write the speed of the blob in frame S right after it is ejected from the rocket as:

200 - [v_ex/ (1+Mb/Mr) ]

 

Now, 200 was a specific choice for v in the original argument given by Klaynos, so that we have instead:

 

 

v - [v_ex/ (1+Mb/Mr) ] = speed of blob in S right after emission from rocket.

 

And that is what I have been looking for.

 

So the instant the blob is ejected from the Gaussian sphere/rocket, the momentum of the blob/fuel in frame S has the following magnitude:

Mb v - [ v_ex Mb/ (1+Mb/Mr) ]

 

Which we can also write as follows:

 

Mb v - [v_ex MbMr/ (Mr+Mb) ]

 

That mixture of masses inside there looks like the formula for the reduced mass, I think thats what that is.

 

MbMr/ (Mr+Mb) = reduced mass = m

 

The point of all this work was to find P2.

 

Klaynos has the answer as

 

P2 = (m+dm)(v+dv) + (-dm)(v-vex)

 

I wanted to double check this.

 

The initial momentum in frame S, is given by Klaynos as:

 

P1 = Mv

 

And the final momentum is P2, which is the sum of the momentum of the rocket in frame S, and the momentum of the ejected fuel in frame S.

 

We already had the momentum of the rocket in frame S, which is:

 

(M+dm)(V+dv)

 

and Klaynos has the momentum of the fuel as being:

 

-dm(v- v_ex)

 

Now, since dm is negative, the mass of the ejected fuel is given by (-dm) just as Klaynos has. So all we need to do is multiply by its velocity in the frame, and then we can add it to (M+dm)(V+dv) to have P2.

 

The value I got for speed in the frame after emission was:

 

v - [v_ex / (1+Mb/Mr) ]

 

Let me multiply this by the inertial mass of the ejected blob of fuel, to obtain the magnitude of the momentum of the blob in frame S after emission.

 

We have:

 

Magnitude of momentum of fuel = (-dm) [ v - [v_ex / (1+Mb/Mr) ] ]

 

Therefore:

 

Magnitude of momentum of fuel = (-dm) v - (-dm)[v_ex / (1+Mb/Mr) ]

 

And I also used Mb to denote the mass of the blob therefore:

 

Mb = (-dm)

 

Therefore:

 

Magnitude of momentum of fuel = (-dm) v - Mb[v_ex / (1+Mb/Mr) ]

 

Therefore:

 

Magnitude of momentum of fuel = (-dm) v - MbMr[v_ex / (Mr+Mb) ]

 

Which we can write in terms of the reduced mass as:

 

Magnitude of momentum of fuel = (-dm) v - mv_ex

 

So we finally have P2...

 

P2 = (m+dm)(v+dv)+ (-dm) v - mv_ex

 

Now this is not the formula you got Klaynos, so did I make an error, or was there an error in the formula you gave me?

 

Here was your result for comparison:

 

P2= (m+dm)(v+dv) + (-dm)(v-v_ex)

 

The only way they are equivalent is if:

 

(-dm) = m

 

And that isn't the case.

 

So where is the error located?

 

Kind regards