Rockets and Missiles - Physics for Game Programmers - page 340

Comparing Equations (11.17) and (11.19), we see that the escape velocity is equal to the
circular orbit velocity times the square root of 2. The escape velocity for an object in a 300 km
orbit around the earth is 10,942 m/s . As was the case with the circular orbit velocity, the mass of
the object has no effect on the escape velocity.

Using the Earth's Rotation
We have seen that fairly high velocities are required to maintain an orbit around the earth.
Fortunately for rocket designers, the earth itself helps things out a bit. As you probably know,
the earth rotates about its north-south axis. The rotational velocity at the surface of the earth is
a function of the distance from the north-south axis of the earth. The maximum rotational
velocity occurs at the equator and is equal to about 460 m/s .
The centripetal force experienced by a rocket traveling around the earth is relative to the
center of the earth. If the rocket travels along the equator in the direction of Earth's rotation
(east), it gets a “free” 460 m/s that it can use to achieve an orbit. Maximizing the available
rotational velocity is the reason that most launch facilities are located as close to the equator
as possible.

Payload to Orbit
Many rockets are used as delivery trucks to outer space. Their job is to lift a satellite or other
payload and place it into orbit. The rocket has to generate enough thrust and maintain it long
enough to overcome gravity and drag forces and lift the payload to the required altitude at the
required orbital velocity.
The amount of payload that a single-stage rocket can lift into orbit can be estimated using
the rocket equation. If drag forces are ignored, the ratio of final mass, m f , to initial mass, m 0 ,
is proportional to the ratio of the final velocity, v f , the burn time, t , and the effective exhaust
velocity, v e .

v t +

f

m

−

f

v

(11.20)

= e

e

m

0

The ratio of the final rocket mass to its initial mass is called the mass ratio . The final mass
of the rocket includes all structural elements of the rocket as well as any payload the rocket is
carrying.
The problem for single-stage rockets is that an unrealistically low mass ratio is often
required to achieve orbital velocities. For example, let's look at a rocket powered by a single
F-1 rocket engine. The vacuum exhaust velocity of the F-1 engine is about 3000 m/s . The burn
time is 150 s . Let's assume that the rocket is supposed to reach the 300 km circular orbit velocity
of 7737 m/s . Under these conditions, the mass ratio is equal to the following:

7737 9.8*150
3000

+

m

−

f

= e = 0.0465

(11.21)

m

0

If the original mass of the rocket was 1.0 e + 6 kg , the final mass of the rocket can only be
46500 kg . This value is probably less than the minimum structural mass of the rocket, meaning
that the rocket can't possibly reach a 300 km circular orbit.