Monday, July 30, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Rockets and Gravitation

Notes to a video lecture on

Rocket in Gravitational Field

In the previous lectures we examined the motion of a rocket with no
external forces (like gravity or drag) acting on it and came with the
Rocket Equation, stating 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.

Now let's add gravity as an external force that acts on a rocket.

There are two cases:

(a) when a rocket is launched from a planet to an orbit, gravity acts
against its movement, thus requiring extra effort by an engine to
overcome the gravity;

(b) when a rocket returns back to a planet for soft landing, gravity
acts in the direction of its movement, but we have to decelerate a
rocket using propellant exhausted also in the same direction, so it also
requires extra effort by an engine.

So, no matter how rocket moves in the gravitational field, an engine
should work harder to either launch it to an orbit or to slow it down
for soft landing.

We will consider the launching from the Earth case only.

Let's follow the same logic as in case of a rocket moving in empty space
with no forces involved and add the effect of gravity in the equation
of conservation of momentum.

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


4. When rocket leaves the planet, the force of gravity F=m(t)·g acts against its movement. During time dt it reduces the impulse of a rocket by

dt = m(t)·g·dt.

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) plus impulse of the gravitational force (see item 4

Equalizing these two momentums at time t and t+dt, according to the Law of Conservation of Momentum, we get the following equation:

m(t)·V(t) = −dm(t)·[ve+V(t)] + [m(t)+dm(t)]·[V(t)+dV(t)] + m(t)·g·dt.

We would like to express the dependency between rocket's speed and the
way it exhausts its propellant without mentioning the time parameter.
This can be done by using the following:

m'(t) = dm(t)/dt (by definition of the derivative and differential)

From this:

dt = dm(t)/m'(t)

The rocket equation above can be simplified. After this the equation looks like

0 = −ve·dm(t) + m(t)·dV(t) + dV(t)·dm(t) + m(t)·g·dt

Ignoring an infinitesimal of a higher order dV(t)·dm(t), the resulting equation looks like


Divide both parts by m(t) and take into consideration that dm(t)/m(t) = d[ln(m(t))]. Then the differential equation of a rocket in the gravitational field looks like

dV(t) + g·dt = ve·d[ln(m(t))]

Replacing dt with its equivalent dm(t)/m'(t), we obtain an equivalent equation

dV(t)+g·dm(t)/m'(t) = ve·dm(t)

Expression on the right is positive because

(a) m(t) is a decreasing function,

(b) ln(m(t)), therefore, is also a decreasing function,

(c) differential of a decreasing function d[ln(m(t))] is always negative,

(d) ve is negative since it is a vector
directed against the movement of a rocket, which we consider as moving
to a positive direction of a coordinate axis,

(e) product of two negative values is positive.

As is obvious from this equation, unless dt is less than ve·d[ln(m(t))], the rocket will not move from the launching pad.

We can simplify this launching condition, using the following:

d[ln(m(t))] = [m'(t)/m(t)dt

This allows to express this condition as

g is less than ve·[m'(t)/m(t)]

ve·m'(t) is greater than g·m(t)

Integrating the differential equation of a rocket in the gravitational field 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:


= ve·

= ve·

= −ve·

That is,

ΔV(t) = −ve·ln[m(tbeg)/m(tend)] − g·(tend−tend)

The last equation does not take into consideration that the force of
gravity decreases with height. It's relatively precise only in the
beginning of the rocket's movement. Obviously, taking this factor into
consideration will complicate the calculations.

The next complication is the drag of the atmosphere, which is not that
important in theory, but for practical matters must be taken into

Another important factor of launching is the planet's rotation. If we
launch a rocket to the East, the Earth's rotation helps to achieve
required speed.

All these and other complications make rocket science a rather involved theory.

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