PHYS-1500 PHYSICAL MODELING
FALL 2006
Putting a Rocket into Orbit about Earth.
The program that models the transfer of a rocket from the surface of Earth into a circular orbit is
one of the most complex that has been treated in this course. These notes outline the physics
involved, and should help to make the operation of the model more understandable. Briefly, the
radius of the orbit into which the rocket is to be inserted must be selected. Then, the program
calculates the motion of the rocket while the rocket’s engine is on. At the appropriate time, the
engine is turned off, and the rocket coasts, in an elliptical orbit, as its altitude increases. When it
reaches the desired altitude, the engine turns on, and injects the rocket into a circular orbit.
The program ignores atmospheric drag, but includes the thrust of the rocket engine and the
decrease in mass as fuel is expended. The rocket starts on a launch pad, tilted at an angle  from
dm
the vertical. In that configuration, the vertical component of thrust is
u cos , and the
dt
dm
dm
horizontal component is
u sin , where
is the rate at which reaction mass is being
dt
dt
dm
expelled, and u is the speed of the reaction mass relative to the rocket. The values of
, u, and
dt
 can be changed, but a typical initial set is provided. In addition, the rocket is pulled downward
GMm
by Earth’s gravitational attraction, which is described by the formula, FG   2 . In this
r
expression, G is the universal gravitational constant, M is the mass of Earth, m is the mass of the
rocket, and r is the distance between the rocket and the center of Earth. While the rocket’s engine
dm
is turned on, the mass of the rocket decreases with time according to the equation m = m0 –
t,
dt
where m0 is the initial mass of the rocket, and t is time.
When the gravitational force and the thrust are combined, the net upward (or radial) force on the
GMm dm
dm
u sin  . From
u cos , and the lateral (or tangential) force is FT 
rocket is FR   2 
dt
r
dt
Newton’s second law, F = ma, the following expressions for the radial and tangential
accelerations can be obtained:
GM (dm/ dt )u cos 
(dm / dt )u sin 
aR   2 
aT 
, and
.
r
m
m
dm
When the engine is turned off,
is set equal to zero. However, our model was based on
dt
energy and angular momentum concepts, so it used more than the accelerations to calculate the
motion. In addition, it used polar coordinates, so extra terms came in when accelerations were
used.
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For motion in the tangential direction, the model calculates angular velocity from angular
L /m
momentum from the equation, L  mr 2 , so   2 , where L/m is the angular momentum per
r
mass. When the rocket engine is off, L/m is a constant so the rocket’s angular velocity is known
as a function of position.
A key part of the program, is the determination of when the rocket engine should be turned off.
That can be found by use of conservation of mechanical energy and conservation of angular
momentum. The figure shows the path of the rocket.
Point Q represents the farthest point from the center of Earth that the rocket will reach if its
engine is turned off and it goes into an elliptical orbit. If the engine is not turned on again at
point Q, the rocket will continue in its elliptical orbit and return close to Earth, as shown. To find
the value of rQ that will result if the engine is turned off, set the mechanical energy and angular
momentum at the rocket’s location equal to the same quantities at point Q. Then,
GMm 1 2 GMm
2
1
 2 mvQ 
and
mv  r  mvQ rQ where v   v sin  .
2 mv 
r
rQ
v is the component of v that is perpendicular to the radius, r, and  is the angle between the
radius and the velocity. Notice that  is not equal to , the angle between the direction of the
thrust and the radius. At Q, the velocity is perpendicular to rQ, so no sine function is needed. To
find rQ, in terms of r and v, eliminate vQ from the two equations and solve for rQ. From the
r
r
second equation we get, vQ  v 
 v sin  . This can be put into the first equation to give
rQ
rQ
an equation that can be solved for rQ.
2
1
2
2

GMm 1  r
GMm
mv 
 2 m v sin   
r
rQ
 rQ

2
or
2
1
2

GM 1 2  r
GM
v 
 2 v  sin   
.
r
rQ

 rQ
2
 1
 1   2GM

 v2   0 .
Rearranging terms yields,  vr sin      2GM    

 rQ 
 rQ   r
This is a quadratic equation for the quantity, 1/ rQ. Its solutions are,
2
2

2  2GM
2GM  (2GM ) 2  4 vr sin   
 v2 
 1
r


  
.
2
2 vr sin  
 rQ 
If this is inverted, and a few numbers are canceled, the results for rQ are,
rQ 
 vr sin   2
.
2
GM


2
GM  (GM ) 2   vr sin   
 v2 
r


The two solutions correspond to the rockets farthest point from Earth, and its nearest point. We
want the farthest point, and the negative sign will give the larger answer. Therefore, the correct
solution is,
 vr sin   2
.
rQ 
2  2GM
2
2
GM  (GM )   vr sin   
v 
 r

The program calculates this quantity after every step, and compares the result to the radius of the
desired orbit. If the solution is equal to or greater than the radius of the desired orbit, R is set
equal to zero, i.e. the engine is turned off. The rocket then travels in an elliptical orbit until its
distance from the center of Earth is equal to rQ. At that distance, its engine turns on. However,
now the thrust is directed in the tangential direction, at an angle of 90° to the radius. The engine
remains on until the speed of the rocket is sufficient to keep it in a circular orbit of the desired
radius. That speed can be found by setting the gravitational force equal to the mass of the rocket
times the acceleration it would have in a circular orbit of the desired radius. Thus,
GMm
v2
GM
GM
.
 m , then,
 v 2 and v 
2
2
r
r
r
r
When the speed reaches that value, the engine shuts off, and the rocket is in a (nearly) circular
orbit.
Finally, in consideration of the welfare of any passengers or equipment aboard the rocket, the
thrust has been made variable. Whenever the acceleration of the rocket exceed 100 m/s2 (about
ten times the acceleration of gravity, or 10 “gees”), the rate at which mass is ejected is cut in
half, thus cutting the thrust in half.
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