Wednesday, July 25, 2018
Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Equation
Notes to a video lecture on http://www.unizor.com
Ideal Rocket Equation
The most important difference between the motion of a rocket and motions
analyzed before in this course is that the rocket propels itself by
throwing back part of its own mass (propellant), thus becoming lighter.
Its mass is variable. Before, in most cases, we were dealing with motion
of objects of some specific mass, not changing during the movement.
In this lecture we will analyze the motion of an ideal rocket that
throws back propellant with constant (relatively to the rocket) speed.
The formula that will be derived was suggested by Russian mathematician
K.E.Tsiolkovsky and is historically named after him, though he was not
the first to derive it.
In particular, we will analyze the dependency between loss of the total
mass of a rocket, throwing propellant backward in the absence of any
external forces, and its gain in speed caused by this process.
The final formula we will derive states that
m(t) is the mass of a rocket (including propellant) at time t and
V(t) is its speed in some inertial reference frame
(related to stars, for example, and positioned in such a way that a
rocket moves along one axis in a positive direction) and
m(t+Δt) is the mass after time interval Δt, during which a rocket was throwing propellant with constant (relatively to a rocket) effective exhaust speed ve and
V(t+Δt) is its speed after time interval Δt in the same inertial reference frame
the maximum increment of the rocket's speed
ΔV=V(t+Δt)−V(t) during this interval of time Δt is
ΔV = −ve·ln[m(t)/m(t+Δt)]
The equation above should be interpreted as the vector equation.
If inertial frame of reference is directed in such a way that the rocket
moves along one axis in positive direction and the exhaust is directed
backwards relative to a rocket's movement, the ve is negative. The mass during this process decreases, so m(t) is greater than m(t+Δt) and the logarithm is positive. This results in the positive ΔV, that is a rocket accelerates.
If the exhaust is directed forward relative to a rocket's movement, the ve is positive, ΔV is negative and a rocket decelerates.
Here is how to derive this formula.
First of all, let's recall that in the absence of external forces the
combined momentum of motion of a system of objects is constant.
This is the Law of Conservation of Momentum.
Let's apply this law to our situation.
Assume, we are comparing the momentum of the system at times t and infinitesimally incremented t+dt.
1. At moment t the momentum of an entire system of a rocket with its propellant was equal to m(t)·V(t).
2. During the next time interval dt the rocket has exhausted m(t)−m(t+dt)=−dm(t) of propellant with constant relatively to a rocket speed ve. Since a rocket moves in some inertial system with speed V(t) and propellant moves relatively to a rocket with constant speed ve, the speed of propellant in the inertial system equals to ve+V(t). This resulted in the momentum of exhausted propellant at moment t+dt to be −dm(t)·[ve+V(t)].
This equation should be interpreted in the vector form. When a rocket
accelerates, velocity vector of its movement and velocity vector of
exhausted propellant are opposite in their directions.
3. A rocket with remaining propellant at moment t+dt has mass m(t+dt)=m(t)+dm(t), velocity V(t+dt)=V(t)+dV(t) and momentum
Now we are ready to apply the Law of Conservation of Momentum.
Item 1 above describes the momentum of the system at time t.
At the moment t+dt the momentum of the system is a combination of the momentum of the exhausted propellant during time dt (see item 2 above) plus the momentum of the remaining rocket mass (see item 3 above).
Equalizing these two momentums at time t and t+dt, which is the consequence of the Law of Conservation of Momentum, we get the following equation:
= −dm(t)·[ve+V(t)] +
This can be simplified. After this the equation looks like
0 = −ve·dm(t) + m(t)·dV(t) +
The last member in this equation dV(t)·dm(t) is an infinitesimal of the higher order that we can remove from this equation, and the resulting equation looks like
m(t)·dV(t) = ve·dm(t)
Divide both parts by m(t) and take into consideration that dm(t)/m(t) = d[ln(m(t))]. Then our equation looks like
dV(t) = ve·d[ln(m(t))]
Integrating this on the interval Δt of time from the beginning of engine's work tbeg to the end of this period tend, we obtain the equation for an increment of the rocket's speed during this interval:
V(tend) − V(tbeg) = ΔV(t) =
So, the increment of a rocket's speed during a period of Δt=tend−tbeg,
when its engine works, exhausting the propellant, equals to a product
of the speed of exhausted propellant times a logarithm of a ratio of the
rocket's mass at the beginning of this period to its mass at the end of
The minus in front of a formula is very important. This is a vector
equation and, if the exhaust is directed back relatively to rocket's
movement (that is, ve is negative), the
increment of a rocket's speed is positive, a rocket accelerates; if,
however, the exhaust is directed towards a tocket movement (that is, ve is positive), the increment of a rocket's speed is negative, a rocket slows down.
The formula above is the Tsiolkovsky's formula and is called the "ideal rocket equation".