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Working notes from a mechanics document on tethered satellites, in a folder of Phil's physics files. The first part checks the equal-mass, aligned limit of a center-of-gravity expression against an Appendix D result, giving r_cog = b. The second part is a draft digression defining center of gravity via vanishing torque, applying it to an asteroid near Earth, and ending with a reader exercise showing Rcog < Rcms. Equations are partly lost in extraction.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
Limiting calculation. Consider
rcog =
= (D.2.21)
r'12 = r12 + b2 + 2r1b cosθ
r'22 = r22 + b2 - 2r2b cosθ (D.2.4)
This is in notation where r1 and r2 are the short ones.
Suppose the two masses are equal AND things are aligned with m1 on the top. Then we know r1 = r2 and we know r'1 = b + r1 and r'2 = b-r1 . Then μ1 = μ2 = 1/2 and the above becomes
rcog =
=
=
=
= (1/)
= (1/)
= (1/)
Now
r'12 + r'22 = (b+r1)2 + (b-r1)2 = 2b2 + 2r12
r'1r'2 = (b+r1) (b-r1) = b2-r12
so then
rcog = (1/) = (1/2)
= (1/2)
= (1/2) b
Agrees with (D.4.5).
**************************** this from the main text tether section. Added eq in App D.
Digresson on Center of Gravity
The phrase "center of gravity" is often used as a synomym for "center of mass" which complicates searching for information about the former concept. Consider a system of masses mi each of which experiences some force Fi. We define FT = ΣiFi, NT(R) = Σi (ri-R)xFi and NT(0) = ΣirixFi, the latter being total system torques with respect to point R and point 0. The center of gravity is a point R such that the total torque on a system measured with respect to point R vanishes, so NT(R) = 0 . This is exactly the same as saying NT(0) = R x FT. Dotting with FT shows that one must have NT(0) FT = 0 as a condition for R to even exist! This notion of "center of gravity" really has nothing to do with gravity, but can be applied to a situation involving gravity.
For a weird-shaped asteroid near the Earth one has NT(0) = 0 (origin at Earth center) since ri x Fi = 0 for each mi of the asteroid. Thus the condition is met and R does exist. Here is a method for computing R, a variation of Symon p 258. One first computes FT = ΣiFi (summed over all points in the asteroid) and one then knows the direction of FT. One draws a line along FT and then translates it parallel to FT so the line contains the center of the Earth. The center of gravity lies on that translated line a distance R from the center of the Earth such that GMEMasteroid/R2 = FT. One then has R and so R is determined. This R solves NT(0) = R x FT because the left side is 0 and R is parallel to FT. The solution R is not unique, and in this example R = 0 also works, but is not very interesting.
For such an asteroid, R always lies closer to the Earth than the center-of-mass point, and that is the case for our tethered satellite shown in Fig 8.7. Unlike the center of mass, the center of gravity changes position within the asteroid as its orientation changes.
Reader Exercise: Show that if the masses are radially aligned as in Fig 8.7, one has
Rcms = and Rcog = . (none)
Show that Rcog < Rcms (this takes a bit of work). These results are not needed below.