2.5 Burnout Speed and Apogee
The ballistic trajectory calculations of section 2.2 assume that the rocket burned out instantaneously at the ground on take-off. In practice, the burn time for a rocket is appreciably longer. This complicates trajectory calculations, but the problem can be split into two problems.
The trajectory during the burn-time interval (t £ tb) is determined first. At burnout (at time, tb), the rocket takes a ballistic trajectory, beginning at the burnout coordinates, (xb, yb).
The ideal rocket equation gives the burnout speed, where Dv is vb without gravity. The gravity loss in speed is readily calculated using a ballistic equation, and is –g× tb. Then the burnout speed is:
The distance the rocket travels during burn-time due to its propulsive acceleration can be derived from the ideal rocket equation and the definition of position:
This differential equation can be solved using calculus. If
we furthermore assume that y(0) = 0 (launch from surface of
earth) and that the mass of the rocket changes at the constant rate of
propellant usage, 
, then
The height at burnout is then
Apogee is the burnout altitude plus the distance the rocket coasts upward in free-fall after burnout, opposed by gravity, which slows it down until it stops at apogee;
where vb is the burnout speed and g is approximated by the average gravitational constant. For ya much less than the radius of the earth (6378 km), g @ g0. At 32 km, g is in error by –1 %.
Example:
A rocket has a liquid-oxygen (LOX), kerosene engine with an Isp = 200 s. The rocket has a mass ratio of MR = 2.5, a propellant mass, mp = 225 kg and a burn rate,
, of –4.6875 kg/s.
If drag is neglected, how high will the rocket go? Assume g is g0 = 9.81 m/s2.
Solution:
The effective exhaust speed,c = g0× Isp = (9.81 m/s2)× (200 s) = 1.962 km/s
The burn time is
The burnout speed will then be
The burnout altitude is:
Then the apogee includes coasting height, or
This apogee is at the altitude of a minimum low earth orbit. However, to orbit, the rocket must also move horizontally (parallel to the tangent of the earth’s surface) at about 2 km/s, or about 17,500 miles per hour. A MathCAD program, flight1.mcd, is included at the end of the chapter that can perform these calculations.
