Assume you are 100 m away from your car and you want to walk to your car. You
push with your foot on the ground in a direction opposite to the direction you
want to go. The ground pushes back and pushes you in the direction you want to
go. You accelerate in the desired direction. (Strictly
speaking, some parts of your body are pushing one of your legs backward, while
your leg pushes the rest of your body into the forward direction. These
internal forces are equal in magnitude and opposite in direction, and they
cannot, by themselves, accelerate your center of mass. But the reaction force
from the ground due to static friction cancels the internal force that is
directed into the backward direction, and the internal force directed into the
forward direction now accelerates you and does work.)

Once you have reached your car, you get in and drive off. Again it is
the interaction with the ground via frictional forces that makes your motion
possible. If, however, you are an astronaut in a spacesuit floating un-tethered
100 m from your spaceship, how can you move over to your ship?

To have a force acting on you, you push on another object. The object then
pushes back. But here in space there is nothing to push against. You better
carry something with you that you can throw away. As you throw away an object,
you exert a force **F** on it for a time Δt. The
momentum of the object changes by an amount Δ**p **
= **F**Δt. The object pushes back with a force
equal in magnitude and opposite in direction. Your momentum changes by an
amount -Δ**p **= -**F**Δt.
If you and the object were initially at rest, you now have a velocity
**v **=
-**F**Δt/m, where m is your mass. The total
momentum of you and the object together has not changed. The total momentum of
interacting objects, which are not acted on by outside forces, is always
conserved. You and the object have both gained kinetic energy, because the
forces you have exerted on each other have done work. They have converted
energy that was stored in the system in some other form into kinetic energy.

An astronaut's manned maneuvering unit (MMU) is a
one-man, nitrogen-propelled backpack. Using rotational and
translational hand controllers, the crewmember can fly with precision in
or around the orbiter cargo bay. The MMU propellant is gaseous nitrogen
stored under high pressure. The gas has potential energy. When an
astronaut fires the thrusters, nitrogen gas expands through a nozzle
into low-pressure space. Potential energy is converted into ordered
kinetic energy as the net force on the escaping gas due to the pressure
gradient does work. The gas, which had no net momentum in its
compressed state gains momentum in the direction the nozzle is
pointing. The astronaut is gaining the same amount of momentum in the
opposite direction. The gas gains no net momentum in the direction
perpendicular to the direction the nozzle is pointing, even so it also
expands in this direction. There are gas molecules moving in all
possible perpendicular direction, and, as we add their vector momenta,
the perpendicular components cancel out. The nozzle is designed to
minimize the perpendicular components and maximize the components in the
direction the nozzle is pointing.

When the astronaut reaches the ship, the astronaut and the ship can
accelerate together by firing a rocket. The **rocket**
works on the basis of the same physical principle as the astronaut's jetpack.
Gas is ejected and gains momentum in a given direction. The ship gains momentum
in the opposite direction. High-pressure gas is obtained by burning fuel.
After the rocket has been fired, chemical energy, which was stored in the rocket
fuel, has been converted into kinetic energy of the gas and the ship. The total
momentum of the gas and the ship stays constant (zero if the ship is initially
at rest). The center of mass of the system does not accelerate.

If a rocket takes off from a launch pad on earth, the force of gravity is
acting on it. The center of mass of the system (rocket ship + exhaust gases)
accelerates towards the ground. The force with which the exhaust gases push
against the rocket must be greater than the remaining rocket's weight if the
rocket is to lift off.

Link: Brief History of Rockets

For a single stage rocket in free space, the change in speed is given by

v_{f }- v_{i }= v_{e}ln(M_{i}/M_{f}),

where v_{e} is the speed of the exhaust gases with respect to the
rocket body. The increase in speed
is proportional to the exhaust speed and the natural logarithm of the ratio M_{i}/M_{f}.
To reach a high speed, the exhaust speed must be high, and the initial mass of
the rocket must include the mass of a large quantity of fuel.

The ultimate speed a rocket can reach is governed by the amount of fuel it
can carry and by the speed of its exhaust gases. Because both of these
quantities are limited, **multistage rockets**
are used in the exploration of space. A smaller rocket is mounted on the front
of a larger one. When the fuel in the first stage (the large rocket) has
burned, the empty fuel tank and the motor are ejected. The second stage (the
smaller rocket) is already moving rapidly. Since it does not have to carry the
motor and the fuel tank of the first stage, it can now reach a much higher final
velocity by ejecting its exhaust gases.

The first stage of a Saturn V space vehicle consumes fuel at the rate of
1.5*10^{4 }kg/s, with an exhaust speed of 2.6*10^{3 }m/s
relative to the body of the vehicle.

(a) Calculate the thrust produced by these engines.

(b) Find the initial acceleration of the vehicle on the launch pad if its
initial mass is 3*10^{6 }kg.

(Hint: You must include the force of gravity to solve part b.)

Solution:

(a) The rate at which the momentum of the expelled gas changes is

Δp/Δt
= Δ(mv)/Δt = (1.5*10^{4
}kg/s)(2.6*10^{3 }m/s) = 3.9*10^{7 }N in the backward
direction. The rate at which the momentum of the rest of the rocket changes
therefore is 3.9*10^{7 }N in the forward direction, since the total
momentum of the system is constant. The thrust produced is 3.9*10^{7
}N.

(b) F_{tot }= ma. F_{tot }= 3.9*10^{7
}N - mg =
3.9*10^{7 }N - 2.94*10^{7 }N = 9.6*10^{6 }N.

a = 9.6*10^{6
}N/(3*10^{6 }kg) = 3.2 m/s^{2} upward.

In the previous problem the rocket expels exhaust gas at a constant rate with
constant speed relative to the rocket body. This produces a constant force
F_{thrust} in the forward direction on the body of the vehicle.
Does the speed of the vehicle increase at a constant rate?

Solution:

No! F_{tot
}= ma = constant. But in this problem m is
not constant. The mass of the vehicle decreases at a constant rate, as
it expels fuel. The acceleration of the vehicle, i.e. the rate with
the velocity of the vehicle changes, is not constant, but increases with
time.

Link: The motion of a two-stage rocket

When a satellite is launched, big rockets are used to give it its initial kinetic energy. Once it is in space, smaller rockets are used to increase its kinetic energy and change its direction of travel. Assuming that the satellite's initial kinetic energy must equal the sum of its final kinetic energy and the change in its potential energy, we can calculate the initial speed needed to lift it into any desired orbit.

For a satellite to orbit the earth in a circular orbit of radius r_{f} its
velocity v_{f} is determined by mv_{f}^{2}/r_{f }=
GMm/r_{f}^{2},
since the force of gravity provides the centripetal acceleration. (Here m is the
mass of the satellite and M is the mass of the earth.)

The satellite's kinetic energy
therefore is K = (1/2)mv_{f}^{2 }= GMm/2r_{f}.

Its potential energy is
U = -GMm/r_{f}.

The gravitational force is a conservative force,
K + U = constant.

We
therefore have (1/2)mv_{i}^{2} - GMm/r_{i }= GMm/2r_{f} - GMm/r_{f
}= -GMm/2r_{f}.

(Here r_{i} is the radius of the earth.)

v_{i}^{2} = 2GM/r_{i} - GM/r_{f}.

The bigger the radius of the desired orbit, the bigger is the required initial kinetic
energy. When the radius of the desired orbit goes to infinity, the satellite escapes the
gravitational pull of the earth.

If r_{f} goes to infinity then (1/2)mv_{i}^{2
}- GMm/r_{i }= 0,
v_{i}^{2 }= 2GM/r_{i}.

The **escape velocity**
is 11.2 km/s.