Clohessy-Wiltshire/Hill's Equations - F=ma?

Question

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:

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?

Answer 1

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: $$ \ddot{x} = f_x + 2\omega \dot{z} \\ \ddot{y} = f_y - \omega^2 y \\ \ddot{z} = f_z + 3\omega^2 z - 2\omega \dot{x} $$

Answer 2

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"):

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

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).

Bingo!

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.

Answer 3

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.