The attitude of probe body or housing or whatever one calls it is precisely controlled and known so that can indeed determine the true precession of the gyros inside. It used telescopes to keep a precise stellar fix and keep itself aligned as much as possible, and I assume measure any deviation that did occur.
I was thinking this morning that the only thing a controlled housing could do was to keep the gyro aligned extended stay hotel boston in some way. This might actually be important. If the gyro and housing were allowed to turn as they pleased, then they will turn together, and no precession extended stay hotel boston will be measured (relative to the housing itself). If it were aligned so that the same side always faces the Earth, then a positive precession on one side of the Earth will be cancelled by the negative precession on the other, and no precession will be measured. However, if it were aligned so that the gyro is always vertical with the Earth's extended stay hotel boston equatorial plane, it will have a positive precession extended stay hotel boston on one side of the earth and a positive precession on the other as well, since it would be turning the same way on both sides in respect to the Earth (clockwise or counter-clockwise). Looking at publius ' statement above, it seems that a stellar fix would indeed produce this, whereas the same side would always be facing the same fixed stars, and the gyro would be kept vertical in respect to the Earth's equatorial plane. So now we're in business.
Okay. So first we need to find the precession caused by the Earth. GR would predict (3GM/c 2 )/d for the amount of the orbit per orbit. My relation says (v/c)*(R/d). It is probably more precisely [(GM/c 2 )*(R equa -R pole )] 1/2 , but they come out the same for uniform fluid bodies through the production of an equatorial bulge. The extra distance travelled per orbit due to the precession is (v/c)(R/d)(2 pi d)=(v/c)(2 pi R), regardless of the distance of orbit. For the sun, this extra distance is (v/c)(2 pi R)=(2 pi R/T)(2 pi/ R)/c=(2 pi R) 2 /Tc=29088.33622 meters, where R=6.96*10 8 meters and T=2.193*10 6 seconds. This is the extra distance Mercury travels per orbit due to the precession caused by the sun. It is also the extra distance Venus travels, and Earth, and Mars, and so on. For Earth, it is found to be 61.819774 meters, where R=6.36*10 6 meters at the poles and T=86164 seconds. This is much smaller than that of the sun, but still measurable. extended stay hotel boston In fact, it should be very noticable for close range orbits since they would orbit very fast, so that the number of orbits would add up quickly and so would the extra distance over time.
So gravity Probe B will orbit at 640 km from Earth. This would give it an orbit of about 7000 km from its center and a velocity of v B =(GM E /d) 1/2 =7546 m/s, where M E =5.9736*10 24 kg. The time per obit is then P B =(2 pi d)/v B =5828.56 sec. The precession per orbit for a satellite orbitting extended stay hotel boston at the equator at this distance would be pr=(v E /c)(R E /d)[(360 degrees per orbit)*(3600 arc-seconds per degree)]=1.8216 arc-seconds per orbit. Not very much, right? But let's check this over time, say, a year. It would be pr=(v E /c)(R/d)(360 degrees per orbit*3600 arc-seconds per degree)][(3.1536*10 7 sec per year)/P]=(1.8216 arc-seconds per orbit)[(3.1536*10 7 sec per year)/P]=9856 arc-seconds per year=2.7378 degrees per year. This should be quite noticable. extended stay hotel boston Are any satellites that orbit the equator extended stay hotel boston noticably this far ahead in their orbit each year, or behind if orbitting in the opposite direction of the Earth? Are any corrections made in this amount in a year's time?
The precession of Gravity Probe B would be greater on its closest side and less on its other, and so is proportional to (v E /c)[R E /(d-R gyro )]-(v E /c)[R E /(d+R gyro )]=(v E /c)(R E )(2R gyro )/(d 2 -R gyro 2 ) 1/2 , which is approximately (v E /c)(R E *2R gyro )/d 2 . This would add an extra spin in this amount to the gyro, but integrated over its circumference, and so is smaller by an amount of 1/2, or 1/3, or 2/ pi , or something of that nature (I haven't done the calculation yet). Let's just call this fractional ratio for integration z for now. Since the probe will be travelling over the poles, the integration is the same for this circumference, so now we have a fraction of z 2 . So the total extra spin added to Gravity Probe B would be z 2 [(v E /c)(R E *2R gyro )/d 2 ](360*3600)(3.1536*10 extended stay hotel boston 7 /P)=z 2 (9856 arc-seconds per year)(2R[gyro[/sub]/d)=z 2 R gyro *(2.816*10 -3 arc-seconds per year per meter). So if, for instance, extended stay hotel boston z=1/2 and R gyro =1 meter, then the gyro would turn an extra (1/2) 2 (1 meter)(2.816*10 -3 )=7.04*10 extended stay hotel boston -4 arc-seconds in a year's time (or less by this amount, extended stay hotel boston depending on which way it is rotating).
This last part needs a lot of work, so don't take it too literally yet, please, but I thought I might mention the possibility. Regardless of whether my formula for the precession is correct or if that for GR stands, this effect on rotation has one very profound effect. It causes bodies to rotate differently in the presence of gravity. If atoms are oriented randomly, some electrons turning one way might radiate faster while those oriented in the other radiate slower. This would be related to the factor of z as well. This might not cancel out over all of the atoms, but if the atoms radiate extended stay hotel boston over time, then the differential between the minimum and maximum times becomes extended stay hotel boston 1/(P-t extra )-1/(P+t extra )=2t extra /(P 2 -t extra 2 ) 1/2 , which comes to approximately 2t/P. I'm not sure if this is the correct differential I should be using here, but one gets the idea. Anyway, from the formulas in the previous paragraph, one can see that this varies with the inverse of the square of the distance from a gravitating extended stay hotel boston object, the same as gravity itself. So atoms, then, might precess more quickly extended stay hotel boston on average in proportion to the gravity present. In other words, they age faster. This goes along with the GR prediction (and apparent verification) of time dilation in a gravitational field (I'm still working on the details).
Looking at publius ' statement above, it seems that a stellar fix would indeed produce this, whereas the same side would always be facing the sun, and the gyro would be kept vertical in respect to the solar system's plane of rotation, and to the Earth's as well. So now we're in business.
And publius 's statement mentioned stellar fix, which I'm pretty sure means to the fixed stars . Also, the solar system plane of rotation and the Earth's plane of rotation are different by over twenty extended stay hotel boston degrees, you can't be vertical to both of them at the same time.
And publius 's statement mentioned stellar fix, which I'm pretty sure means to the fixed stars . Also, the solar system plane of rotation and the Earth's plane of rotation are different by over twenty degrees, you can't be vertical to both of them at the same time.
Well, basically they both mean the same thing in this case, but not quite, you're right, since a fix on the sun would be angled slightly differently above or below the center of the Earth. The further away the star the telescope is centered on, the better, I suppose. In any case, it would work as long as the probe did not turn with its revolution around the Earth (as with tidal lock), but maintained extended stay hotel boston the same orientation to the fixed stars, like you said. As far as the orientation itself, I guess the calculations would be simpler and the effect the greatest if it is vertical to the Earth's plane of rotation and not the solar system's after all. I was thinking about them as being close to the same, for some reason.
Well, basically they both mean the same thing in this case, but not quite, you're right, since a fix on the sun would be angled slightly differently above or below the center of the Earth. The further away the star the telescope is centered on, the better, I suppose. In any case, it would work as long as the probe did not turn with its revolution around the Earth (as with tidal lock), but maintained the same orientation to the fixed stars, like you said. As far as the orientation itself, I guess the calculations would be simpler and the effect the greatest if it is vertical to the Earth's plane of rotation extended stay hotel boston and not the solar system's after all. I was thinking about them as being close to the same, for some reason.
NASA calls that orbital mode stellar inertial . Many times, you do want to be tidally locked with the earth and keep the same side facing, like when you have instruments, extended stay hotel boston cameras, antennas, etc. There's a word for that I which I forget.
NASA has the logs of some of the Apollo flights posted, with transcripts of the whole mission. You'll see references to this stuff, where they have to get their bearings from various sources, and fire the reaction control extended stay hotel boston rockets to get the spacecraft in the desired orbital mode. And they used stellar fixes to calibrate. In the parking orbit, one of the astronauts had to use a little telescope and get a fix on a number of stars. extended stay hotel boston You'll see the transcipts of this.
Oh, yes. That's important, isn't it? In my examples, extended stay hotel boston even though it may seem I said otherwise (and I guess I did ), the calculations, if correct, are actually based on an orientation of the gyro with the fixed stars and positioned vertically with the Earth's equatorial plane. extended stay hotel boston Sorry about that, and thank you. I have edited the original post to reflect that. Surprisingly, however, extended stay hotel boston a vertical orientation with the ecliptic, although off by about a whopping 23 degrees, would only vary by about 8% in the calculations.
Okay. So we have a precession on the rotation of the gyro of z 2 *R gyro *(2.816*10 -3 arc-seconds extended stay hotel boston per year per meter). I should have stated this as z*z'*R gyro *(2.816*10 -3 arc-seconds per year per meter), since it turns out that z for the integration around the gyro and z' for the integration around the orbit are indeed different. z' for the orbit is simple. It is the summation of the cosine of the angle from zero to ninety degrees divided by the number of integrations (the numbe
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