Equations of Motion in Spherical Coordinates
Arland Thompson
Chief Scientist
Advanced Technology Associates
(www.atacolorado.com)
1) Introduction
It is sometimes useful to formulate the equations of motion in spherical coordinates.
For example, when describing the trajectory of a tactical missile over a relatively short
range where Earth curvature is not a factor, it can be useful to put the equations of motion
in spherical form. In addition, it is often times necessary to be able to convert the equations
of motion from spherical form to Cartesian form, as is the case in orbit determination
where angular and angular rate measures must be taken from tracking antenna gimbal
resolvers and range and range rate are measured by Doppler radar. This brief treatise will
develop the equations of motion in spherical coordinates (position, velocity, and
acceleration) along with the conversion to Cartesian coordinates. The conversion from
Cartesian to spherical will also be developed for completeness.
2) Definition of Topocentric Coordinate Frame
Topocentric Horizon Coordinate frame: X, Y plane is the plane tangent to the
Earth’s surface at a defined (location of RADAR etc.) center point. The X basis vector
points south, the Y basis vector points east. The Z basis vector points up along the local
vertical at the tangent center point.
Z



Y
X
 , , 
- Azimuth from south, elevation from local X, Y plane of the topocentric
coordinate frame, and distance from frame center.
3) Derivation of Position, velocity, and acceleration in Spherical coordinates
The following formulas are used to convert spherical coordinates to rectangular
(Cartesian) coordinates:
x   cos  cos 
y   cos  sin  Eqs. 1
z   sin 
Taking the derivatives to get velocity:
x   cos  cos    sin  cos    cos  sin 
y   cos  sin    sin  sin    cos  cos  Eqs. 2
z   sin    cos 
Taking the derivatives once again to get acceleration:
x   cos  cos    sin  cos    cos  sin 
  sin  cos   2 cos  cos   sin  cos 

  cos  sin   
2

cos  cos   cos  sin  
 2  sin  sin 
y   cos  sin    sin  sin    cos  cos 
  sin  sin   2 cos  sin   sin  sin 

  cos  cos   
2

cos  sin   cos  cos 
Eqs. 3

 2  sin  cos 
z   sin   2  cos   2 sin   cos 
4) Derivation of Conversion from Cartesian coordinates to Spherical
For completeness, the conversion from Cartesian coordinates to spherical will also
be presented here.
The conversion from Cartesian position to spherical coordinates is accomplished
using the following equations:
 
x2  y2  z 2
  a tan 
y

x
 z 
  a sin 

Eqs. 4
Note: The quadrant must be correctly resolved on  . There is no quadrant issue on


 .   
2
2
Taking the derivatives to get velocity
 
 

1 2
x  y2  z2
2
xx  yy  zz
x2  y2  z 2
 2 xx  2 yy  2 zz 

1
2
 r 
r
r
xy  yx
xy  yx
xy  yx
x2
 

 2
2
2
  y   x  y 2 Eqs. 5
 y
1  
x 2 1    
 x 
 x


z  z
z  z
2
 

2
2
z
z
2
1   
 1   


Note that there are singularities in the above equations when the position
magnitude, x component, or y component are exactly 0. These conditions are
unlikely to arise in any practical application. However, software implementation of
these equations must deal with such eventualities.
Taking derivatives once again to get acceleration:
 
 

r   r  r 
x
2

2
y
2
  r r

xy  xy  yx  yx   xy  yx 2 xx  2 yy 
x  y 
2
  z 2 
2 1 

D
1   
2     


2

 
2
1
2
2 2
  z 2 
  z  z  z 
  2  1    
 2 
2
  



  


z
1    z  z  z  z   z  z D

 4   2z2
1
2