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An x y coordinate system is shown. A rocket mass m is moving to the right with velocity v. the rocket’s exhaust mass d m sub g is moving to the left with velocity u. The system consists of the rocket and the exhaust.
The rocket accelerates to the right due to the expulsion of some of its fuel mass to the left. Conservation of momentum enables us to determine the resulting change of velocity. The mass m is the instantaneous total mass of the rocket (i.e., mass of rocket body plus mass of fuel at that point in time). (credit: modification of work by NASA/Bill Ingalls)

At the same moment that the total instantaneous rocket mass is m (i.e., m is the mass of the rocket body plus the mass of the fuel at that point in time), we define the rocket’s instantaneous velocity to be v = v i ^ (in the + x -direction); this velocity is measured relative to an inertial reference system (the Earth, for example). Thus, the initial momentum of the system is

p i = m v i ^ .

The rocket’s engines are burning fuel at a constant rate and ejecting the exhaust gases in the − x -direction. During an infinitesimal time interval dt , the engines eject a (positive) infinitesimal mass of gas d m g at velocity u = u i ^ ; note that although the rocket velocity v i ^ is measured with respect to Earth, the exhaust gas velocity is measured with respect to the (moving) rocket. Measured with respect to the Earth, therefore, the exhaust gas has velocity ( v u ) i ^ .

As a consequence of the ejection of the fuel gas, the rocket’s mass decreases by d m g , and its velocity increases by d v i ^ . Therefore, including both the change for the rocket and the change for the exhaust gas, the final momentum of the system is

p f = p rocket + p gas = ( m d m g ) ( v + d v ) i ^ + d m g ( v u ) i ^ .

Since all vectors are in the x -direction, we drop the vector notation. Applying conservation of momentum, we obtain

p i = p f m v = ( m d m g ) ( v + d v ) + d m g ( v u ) m v = m v + m d v d m g v d m g d v + d m g v d m g u m d v = d m g d v + d m g v .

Now, d m g and dv are each very small; thus, their product d m g d v is very, very small, much smaller than the other two terms in this expression. We neglect this term, therefore, and obtain:

m d v = d m g u .

Our next step is to remember that, since d m g represents an increase in the mass of ejected gases, it must also represent a decrease of mass of the rocket:

d m g = d m .

Replacing this, we have

m d v = d m u

or

d v = u d m m .

Integrating from the initial mass m i to the final mass m of the rocket gives us the result we are after:

v i v d v = u m i m 1 m d m v v i = u ln ( m i m )

and thus our final answer is

Δ v = u ln ( m i m ) .

This result is called the rocket equation    . It was originally derived by the Soviet physicist Konstantin Tsiolkovsky in 1897. It gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass from m 0 down to m . As expected, the relationship between Δ v and the change of mass of the rocket is nonlinear.

Problem-solving strategy: rocket propulsion

In rocket problems, the most common questions are finding the change of velocity due to burning some amount of fuel for some amount of time; or to determine the acceleration that results from burning fuel.

  1. To determine the change of velocity, use the rocket equation [link] .
  2. To determine the acceleration, determine the force by using the impulse-momentum theorem, using the rocket equation to determine the change of velocity.
Practice Key Terms 1

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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