Followup to my previous post! I have now found a figure for the moment of inertia of the Earth, and am able to get close to the OLD value for the framedragging effect. But I'll go through it all from the beginning with new numbers.
To get accurate values, I have tried to use data to five or six figures of accuracy, or whatever is available. I use SI units, unless explicitly given otherwise.
I've checked out the following sources.
Geodetic effect
The geodetic effect is 1.5*(GM)
1.5c
-2r
-2.5, where r is the semilatus-rectum of the probe orbit. This is the semimajor axis times (1-e
2), as indicated in a previous post.
The radius of the orbit of GP-B is measured as 7027.4 km (semi-major axis). The eccentricity is e = 0.0014, so the semilatus rectum is 7027.386 km. Adler and Silbergeit use 7028 km with a circular orbit; this was calculated before launch.
The value of GM (gravitational constant times Earth mass) is known with great precision; much more than either G or M individually. The value of G*M is 3.986004418*10
14 m
3s
-2 in IERS.
NIST currently gives G as 6.67428*10
-11, which would give M as 5.9721864*10
24. The NASA page gives Earth mass as 5.9736*10
24; this corresponds to G = 6.6727*10
-11.
To convert from radians per second to milliarcseconds per year, the factor is 6.50908*10
15. That uses a tropical year of 365.24219 days (IERS). The speed of light is 299792458
The only meaningful source of error here is in r, the radius of the orbit. The largest value for the geodetic precession requires r to be small, and the conversion factor to be large. The conversion factor for milliarcseconds/year uses the length of a tropical year, being 365.24219 days; there's no basis for using anything greater.
The calculation is
1.5 * (3.986004418 * 10^{14})^{1.5} * 299792458^{-2} * (7.027386 * 10^6)^{-2.5} * 365.24219 * 86400 * 360 * 3600 * 1000 / 2 / \pi
This gives the geodetic precession as 6603.77 milliarcseconds/year. I can't see any possible way to make this any bigger.
To make matters worse, Adler and Silbergeit also give a correction factor to take account of the Earth's oblate shape. This factor is given as:
1-\frac{9}{8}*J_2*(R/r)^2)
Here R is the radius of the Earth and J
2 is the quadrupole moment.
For the radius of the Earth, Adler and Silbergeit use 6378 km, NASA gives 6378.1 km, and the IERS gives 6378.1366 km.
For J
2, Adler and Silbergeit use 1.083*10
-3, and the IERS gives 1.0826359*10
-3. The accuracy here will not matter much.
This factor
reduces the geodetic prediction, by (1-1.00*10
-3), to give a final prediction of 6597.14 milliarcsec/year
How anybody ever got 6614.4 I don't know.
There is an additional precession due to the Sun; but this is in a different plane, and is going to have more effect on framedragging. It is discussed in Adler and Silberguit as well, with a magnitude of 19 milliarcsec/yec, but mostly perpendicular to geodetic precession.
Frame dragging
As derived above (and Adler and Silbergeit confirm) the magnitude of the effect works out to be GJr^{-3}c^{-2}\omega / 2
The moment of inertia for a solid body is J = kMR^2, where k is a "moment of inertia ratio". This ratio is 0.4 for a uniform sphere, but it will be more if the mass density is greater near the surface, and less if the mass density is greater near the center.
Adler and Silbergeit use k = 1/3.024 = 0.3307. The NASA fact sheet gives k = 0.3308.
Using the value of GM from IERS, GJ would be kGMR
2, which is about 5.3640*10
27, using the NASA value for the radius R of the Earth.
The IERS gives a value for J directly, which is 8.0365*10
37; and with their value of G as 6.6742*10
-11, this gives GJ = 5.3638*10
27.
The length of a sidereal day 23.9345 hours (NASA sheet) so the rotation velocity ω is 7.2921 * 10
-5 rad/sec. In IERS it is 7.292115*10
-5.
The framedragging effect is
GJr^{-3}c^{-2}\omega/2
This works out to
5.3638*10^{27} * (7.0274 * 10^6)^{-3} * 299792458^{-2} * 7.292115*10^{-5} * 365.24219 * 86400 * 360 * 3600 * 1000 / 2 / \pi / 2
which gives 40.81 milliarcsec/year,
As before, there is a correction factor; this time equal to
1+\frac{9}{8}*J_2*(R/r)^2(1-\frac{3}{7}*MR^2/J))
This works out to 1-2.97*10^{-4}, which brings the prediction back down to 40.80 milliarcseconds/year.
This close to the 40.9; but now I don't know how they are obtaining 39.
Cheers -- Sylas