I have been researching the genesis of the Clohessy-Wiltshire (C-W, or Hill's) equations. These equations are used to describe relative motions between the "chaser" and "target" spacecraft in a space rendezvous situation. A fairly decent derivation (I presume, as it is over my head - read on) is given in Clohessy - Wiltshire Analysis, Eugene M. Cliff, October 23, 1998

For reference, here are the rectangular coordinates NASA uses for space rendezvous considerations:

enter image description here

The origin of the depicted coordinate system is located at the "target" vehicle (depicted in red). The "chaser" vehicle is not depicted here.

The version of the C-W equations that is consistent with these coordinate axes is:

$$ \ddot{x}-2\omega\dot{z}=F_x $$

$$ \ddot{y}+\omega^2y=F_y $$

$$ \ddot{z}+2\omega\dot{x}-3\omega^2z=F_z$$

Note that omega ($\omega$) is, of course, angular velocity. Also, the $F$ terms represent component forces due to "chaser" spacecraft thruster firings.

When rearranged into the familiar $F=ma$ format, the equations become:

$$F_x=m(\ddot{x}-2\omega\dot{z}) $$

$$ F_y=m(\ddot{y}+\omega^2y) $$

$$ F_z=m(\ddot{z}+2\omega\dot{x}-3\omega^2z)$$

Note that I've "added" the mass term ($m$) back into the equations for clarification (I can do this with impunity, since these equations are only of interest to me when the thrust force components are all zero, implying a "free drift" situation for the "chaser" spacecraft - note that the "target" spacecraft is assumed to be always in free drift here).

Now my question:

Do the "extra" terms (the ones involving angular velocity) come about because you are dealing with objects that are moving in orbits? The forces referenced all act along orthogonal axes, and each of the acceleration terms include a linear acceleration along one of said axes.

But, we still have those "extra" acceleration terms to deal with.

My supposition is that the "extra" acceleration terms arise because there is orbital motion that is driven by a centripetal force (gravity).

Put into simple dirt farmer terms, am I too far off base here?

  • $\begingroup$ The reference frame is accelerating (rotating) so that means $F = ma$ doesn't exactly hold. Those terms arise because of the coupling between the axes in the rotating frame. In short, yes the orbit motion is included in the equations -- why wouldn't it be? Seems odd to be getting into the equations of motion from a dirt farmer perspective! $\endgroup$ Commented Nov 25, 2015 at 6:06
  • $\begingroup$ Well, there's been a lot of water under the bridge since I graduated from engineering school in 1982...I've sort of lapsed back into "dirt farmer" mode over the decades. Your point about the reference frame not being inertial is a good one (one that I missed, admittedly) and MAY be germane to the discussion, but it may the case that the linearizations that led to these three equations may have assumed that said reference frame is, indeed, inertial under the constraints of said linearizations...not sure. $\endgroup$
    – Digger
    Commented Nov 25, 2015 at 20:07
  • $\begingroup$ Gotcha, I can tell you know your way around the math! No, even with linearization those equations are definitely accounting for the rotating frame (otherwise they would indeed collapse to $F = ma$). $\endgroup$ Commented Nov 25, 2015 at 20:13
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    $\begingroup$ You can find a derivation of the equations here. It starts with the restricted 3-body problem and the linearization come from doing a Taylor series expansion on the spacecraft radius vectors. $\endgroup$
    – Dave
    Commented Nov 25, 2015 at 20:47

3 Answers 3


Using restricted 3-body problem seems like an overkill for this problem to me. The equations can be deduced in a simpler way. Without going into much details, it goes like this: since the frame is rotating around the $y$-axis, the target observes chaser's acceleration due to the centrifugal force, $(\omega^2 x, 0, \omega^2 z)$, and the Coriolis force, $(2\omega \dot{z}, 0, -2\omega \dot{x})$. In addition, both the target and the chaser are accelerated by Earth's gravity, but their accelerations are different since their positions are different. If the target is in a circular orbit and the distance to the chaser is small compared to the orbit's radius, the difference in accelerations due to gravity can be approximated using the same $\omega$ as $(-\omega^2 x, -\omega^2 y, 2\omega^2 z)$. Adding these accelerations together, we get $(2\omega \dot{z}, -\omega^2 y, 3\omega^2 z-2\omega \dot{x})$, which is exactly the extra acceleration: $$ \ddot{x} = f_x + 2\omega \dot{z} \\ \ddot{y} = f_y - \omega^2 y \\ \ddot{z} = f_z + 3\omega^2 z - 2\omega \dot{x} $$


OK, I'll take a stab at this and try not to get too deeply buried in the math...

Thanks to the above comments by members Brian Lynch and Dave, I think I can convey a pretty good intuitive feel for what is going on. Really, the credit for this answer goes to them (however, if the answer is deemed lame, then I'll take the "credit")!

This link, provided by Dave, examines what is referred to therein as the "Restricted Three-Body Problem". While it is acknowledged that the general three-body problem has no analytical solution, the case considered in said reference is one wherein one of the masses is so small as to not have any significant gravitational effect on the other two masses, while the other two masses are large and gravitationally significant.

Now, in order to more easily understand and analyze said restricted three-body system, a three-dimensional rectangular coordinate system is proposed such that the origin is located at the center of mass of the two largest masses and whose axes are stationary with respect to said two largest masses, even though said two masses are orbiting each other in inertial space. So, said adopted coordinate system is rotating in inertial space with a certain, constant angular velocity Ω (note that the equations I posted in the OP refer to the angular velocity as "ω"):

Coordinate System

Note in the above diagram that the smallest mass (mentioned above) is designated as Msub3

Much math is then furiously invoked, and some cool equations are arrived at, but the telling phrase in the whole treatment is:


(Sorry for the small font - I don't know how to enlarge this image.)

For those reading along, said phrase occurs just above Equation (8) in said reference. It says essentially what Brian Lynch was trying to tell me in his above comments to my OP.

Next, this link (again, provided by Dave), goes on to show how the C-W equations can be derived from the results of the above treatment (essentially addressing a special case of said "Restricted Three-Body Problem" wherein one of the two larger masses becomes small enough to be gravitationally insignificant - thus qualifying as the case of a "chaser" spacecraft attempting a space rendezvous with a "target" spacecraft whilst both said spacecraft are orbiting the Earth - the very case I was interested in in my OP). Again, much math is invoked and, voila, the C-W equations pop out (for the case of no thrusting taking place).


So, to put it in a nutshell, the "extra" acceleration terms I was curious about in my OP arise because my coordinate system is rotating in inertial space.

  • $\begingroup$ Looks good, the 3-body problem with 2 significant masses is what you need to investigate Lagrange points, whereas like you mention it reduces further with only 1 significant mass for the target/chaser problem. The coupling in the equations is reminiscent of Euler's equations of motion. $\endgroup$ Commented Nov 26, 2015 at 12:23
  • $\begingroup$ On the toolbar above the area where you write answers is an icon of quote symbols. If you copy and paste text in from a source, then select the text and click on that symbol, it turns into a block quote. In the answer it will appear on a yellow background and indented. If you hover over each of the icons in the toolbar, its action, html tag, and keyboard shortcut are shown. Click the help symbol at the right of the bar for more help. $\endgroup$
    – kim holder
    Commented Nov 26, 2015 at 15:31
  • $\begingroup$ Brian, thanks for looking things over (notice I spelled your name correctly here!). Actually, I was thinking that the rotational terms arose more from kinematics considerations than anything else... $\endgroup$
    – Digger
    Commented Nov 26, 2015 at 17:20
  • $\begingroup$ Kim, thanks for the tips. However, my problem was a little different in that I needed to insert an image if I wanted to show the "telling phrase" in the answer I gave above. This was because I could not figure out any other way to display the x and y double-dot terms that said phrase includes. Any insight on how to do this? $\endgroup$
    – Digger
    Commented Nov 26, 2015 at 17:23

Aerospace Engineer at LMCO here - we use Hills equations all the time for Rendezvous and Formation Flying applications. The "$\omega$" in these equations are the angular rate of motion of the target satellite about the earth. We generally use these equations at GEO altitude, so $\omega$ ~ 360 degrees/day (convert to rad/sec of course). We adjust $\omega$ for drifting satellites that may be moving faster or slower.


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