Rocket Propulsion, Newton's second law

Jun24-04, 10:31 PM

I was reading a section about rocket propulsion in my general physics text and it came up with the following formula:

$LaTeX Code: m \\frac{dv}{dt} = - v_{ex} \\frac{dm}{dt}$

I don't have much trouble with this formula, but then it went ahead and substituted m dv/dt as the thrust experianced by the rocket. I feel a little hazy about this move because the mass isn't constant. If the net force on the rocket equals the rate of change of momentum, then it seems as if the following is equally valid:

$LaTeX Code: \\begin{align*} F &= \\frac{dp}{dt} = m \\frac{dv}{dt} + v \\frac{dm}{dt} \\\\ &= - v_{ex} \\frac{dm}{dt} + v \\frac{dm}{dt} \\\\ &= \\frac{dm}{dt} ( v - v_{ex} ) \\end{align*}$

Can someone clarify this for me?

**Moe** · Jun25-04, 05:58 AM

The change of mass occurs in the second term. The total momentum of the rocket remains 0. Because of actio=reactio the rocket experiences a change in momentum equal in size but opposite in direction to that of the hot gases ejected from the engine. You have to look at the whole system, rocket and gases. Before launch, the total momentum is 0. After launch, with no external forces, it has to remain 0.

Jun25-04, 12:59 PM

yes, i understand that the momentum of the whole system remains zero momentarily, but i'm still a little puzzeled. for example the momentum of the system at one instant is mv and momentarily later it is still mv because the gas is propelled out but the rocket is propelled forward. however, soon after the gas is propelled out, we no longer consider it as part of the system, now the momentum of our rocket system has apparently changed from what it was a moment ago. if one sets the thrust equal to m dv/dt, it doesn't not appear as if the impulse it provides will equal this apparent change in momentum of the rocket.

**JohnDubYa** · Jun25-04, 01:03 PM

The formula is BS, in my opinion. There is no mass change in the system whatsoever. Mass simply moves from one place to another. Rocket propulsion is simply a collision process run in reverse.

**JohnDubYa** · Jun25-04, 01:06 PM

We have already argued this point. If anyone disagrees with me, check out the following thread:

http://physicsforums.com/showthread.php?t=28879

Jun25-04, 01:41 PM

while you are correct that the mass of the total system never changes, it isn't very practical for the situation at hand. I mean, we are not really concerned about the gas that is propelled back by the rocket except for a moment. we want any equations we formulate to focus their basis on the rocket itself. it's somewhat similar to the centrifugal force. it don't really exist, but that doesn't mean we can't use it as a fictitious forces to keep our problem narrowly focused in a particular frame of reference, etc.

in any case, my problem is only with how we assign what the thrust force is. the book (and other sources) assign it to be m dv/dt. i could concieve how this is true momentarily, but this definition does not appear to give the apparent change in momentum of the rocket by it's impusle.

**Moe** · Jun25-04, 02:15 PM

The total momentum of rocket and gases is
$LaTeX Code: p(t+\\Delta t)= (m-\\Delta m)(v+\\Delta v)+\\Delta m*vsingle-quote$
v' is the velocity of the gases relative to the surface of the earth

is the velocity of the gases relative to the rocket.
For $LaTeX Code: \\Delta t->0$ we therefore get

LaTeX Code: dp/dt = m*(dv/dt) + v_e*(dm/dt)=m*g

Because v is opposite to g, we get (after dividing by m and multiplying by dt)

Jun25-04, 04:59 PM

Thanks Moe, but i'm afraid that's not what i'm confused about. I don't have much qualms about the the equation for acceleration of the rocket. Perhaps my confusion can be enlightened by an example.

Consider a rocket accelerating from rest in outer space. Beginning with the differential equation mentioned in my first post one can derive the velocity of the rocket at a later time. v_ex is relative speed and dm/dt is negative. m0 is the initial mass of rocket with fuel.

$LaTeX Code: m \\frac{dv}{dt} = - v_{ex} \\frac{dm}{dt} \\rightarrow v = v_{ex}\\, \\ln\\left( \\frac{m_0}{m} \\right)$

Consider the thrust to be equal to m dv/dt. Then the impulse is:

$LaTeX Code: \\begin{align*} J &= \\int_0^{\\Delta t} T dt = \\int\\left( m \\frac{dv}{dt} \\right)dt = \\int\\left( -v_{ex} \\frac{dm}{dt} \\right)dt \\\\ &= \\int_{m_0}^m -v_{ex} dm = -v_{ex}(m - m_0) \\end{align*}$

On the other hand the momentum of the rocket has changed as follows:

$LaTeX Code: \\begin{align*} \\Delta p = mv = m v_{ex} \\ln\\left( \\frac{m_0}{m} \\right) \\end{align*}$

Unfortunantly the impulse provided by the thrust (as defined by most resources) on the rocket does not appear (at least on the outset) to equal the change in the momentum of the system. If one considers the thrust to not be m dv/dt but the more complete m dv/dt + v dm/dt, then it does appear to work.

$LaTeX Code: T = m \\frac{dv}{dt} + v \\frac{dm}{dt} = -v_{ex}\\, \\frac{dm}{dt} + v \\frac{dm}{dt}$

$LaTeX Code: \\begin{align*} J &= \\int_0^{\\Delta t} T dt = \\int \\frac{dm}{dt} ( v - v_{ex} ) dt \\\\ &= \\int \\frac{dm}{dt} \\left(v_{ex} \\ln\\left( \\frac{m_0}{m} \\right) - v_{ex}\\right) dt \\\\ &= \\int_{m_0}^m v_{ex} \\left( \\ln\\left( \\frac{m_0}{m}\\right) - 1 \\right)dm \\\\ &= m v_{ex} \\ln\\left( \\frac{m_0}{m} \\right) \\\\ \\end{align*}$

Which does appear to equal the change in momentum of the rocket. I'm not saying that this is neccesarily correct, but so far it makes more sense, especially since the impulse of the thrust on the rocket should give it's change in momentum.

**Moe** · Jun25-04, 06:27 PM

Hmm... you could take the total mass of the ejected gases, and assuming they are ejected at a constant velocity (which they should be, more or less) you can easily calculate the total momentum of the rocket. That should be a lot easier, although this might be incorrect as well.

Jun25-04, 07:05 PM

I'm afraid that the ejected gas is only kicked out with a constant velocity for a short period of time (although the relative velocity is always constant). The ejected gases are fired out with velocity u = v - v_ex. As the rocket begins to pickup velocity, the ejection velocity varies. After an extended period of time, the velocity of gas that was kicked out at the start will not be the same as the velocity of gas kicked out later.

But calculating the momentum is still easy enough. Just take the mass of the rocket at the instant you wish to compute the momentum and multiply it by it's velocity. Thus:

p = mv = mv_ex ln(m0/m)