Can a Gyroscope on a Satellite Detect Orbit?

[zanick]

TL;DR Summary

Since orbits are considered geodesic paths, objects following these paths are considered "in freefall" in Einstein physics. Can gyroscopes detect the constant turn of an orbital path, even though it is considered to be inertial, or in "freefall" as its path is dictated by the warping of spacetime?

Einstein said, when describing someone falling off a building, that the Earth accelerating up to meet him/her. Without the Earth getting larger in all directions as the paradox goes, it curvature of space-time which is why you can have the acceleration up without the surface moving up as you follow a geodesic path. Any deviation from that geodesic will requires a force which is what causes you to have weight on a scale on earth.

However, what if we consider an orbiting satellite which is also considered in free fall. It has gyroscopes on board for maintaining or changing orientation. Do they only measure rotational changes with respect to the geodetic path? I mean, if I think about the inertial state, say you are moving slowly through space in a capsule, with no windows, and you came across a large planet I would think that you divert to the direction of the 'planet',(Due to the warping of spacetime) and you would never know it, as you would still in be in "freefall" Even still, you accelerate, still you feel no acceleration. Then you orbit the planet, So, will the original orientation of the capsule will be held in position? Would it be whatever was in line to your direction through spacetime originally with respect to the planet?

In other words, if the top of the capsule was pointed forward through space time, would it also and always point in line with the geodesic path? And now the real question, would any gyros that happen to be spinning on board, notice an orbital path and the approach to the orbital path? If so, would the gyroscope or craft roll back every revolution around the planet, or would it remain aligned with the geodetic path and not vary its position and would the gyro sense this?

I understand that gyros are fixed in space and detect any kind of deviation, but I'm thinking that if the gyroscope follows the geodesic path (The curved path) of the space capsule, that there would be no forces acting on it, except when the orientation would vary from that path.

Thanks for the help.

 

[Dale]

This is what Gravity Probe B tested. On it they built the most perfect gyroscopes that we have ever produced and mounted a telescope to look at some distant stars. They then kept the satellite stable with respect to the gyroscopes and observed that the stars precessed a small but detectable amount. This is called frame dragging and was the last prediction of GR to be confirmed.

Note that this frame dragging effect is absurdly small. Even with these super-sophisticated gyroscopes it was only detected to within something like 20% if I recall correctly.

Also, note that according to GR this effect is due to the spin of the planet and not the orbit of the satellite. A satellite in a similar orbit around a non-spinning planet would not have any frame dragging.

 

[Buzz Bloom]

I find myself confused. I would appreciate any reference that explains HOW the rotation of the planet creates the effect of frame dragging. Is it perhaps just related to the irregularity of the planet's geometry?

 

[Dale]

I find myself confused. I would appreciate any reference that explains HOW the rotation of the planet creates the effect of frame dragging. Is it perhaps just related to the irregularity of the planet's geometry?

Here are the basic Wiki entries on the topic:
https://en.wikipedia.org/wiki/Lense–Thirring_precession
https://en.wikipedia.org/wiki/Frame-dragging

Irregularities are not part of this effect. This effect remains after irregularities are accounted for.

 

[Ibix]

I find myself confused. I would appreciate any reference that explains HOW the rotation of the planet creates the effect of frame dragging. Is it perhaps just related to the irregularity of the planet's geometry?

It just does... The angular momentum of a body is a source term for gravity and it tends to cause free fall paths nearby to orbit in the same sense as the spin. The relevant exact solution is the Kerr black hole, which is an eternal spinning black hole.

 

[zanick]

My original question is whether or not the gyro will be affected when installed inside an object that is following a curved path (the geodesic path) toward a massive object or orbiting a massive object like a planet. In other words, when in freefall, does the gyro always see a straight, inertial path, due to the warping of spacetime, or can it detect this change of direction relative to the object that it is heading toward or orbiting around?

 

[PeterDonis]

My original question is whether or not the gyro will be affected when installed inside an object that is following a curved path (the geodesic path)

Geodesic paths are not curved. They are straight, by definition: geodesics define what "straight" means.

Locally, a gyroscope can't "detect" any "change" along a path, because a gyroscope defines what "unchanged" means. More precisely, a gyroscope is a physical realization of what is called Fermi-Walker transport of vectors along a worldline, and Fermi-Walker transport defines what it means for a vector to be unchanged when transported along a worldline.

What an experiment like Gravity Probe B does is to compare the gyroscope direction with some globally defined direction, such as the direction pointing towards a particular distant star, in order to see if being transported along some particular geodesic worldline changes the relationship between the local version of "unchanged", which, as above, is defined by the gyroscope, and a global version of "unchanged" defined by continuing to point towards the same distant object. In general, such a relationship will change in curved spacetime as an object travels along a geodesic.

For the case of geodesic orbits around an isolated gravitating body, the change in the relationship between gyroscope direction and a global fixed direction such as pointing towards a distant star can be separated, conceptually, into three different pieces:

(1) Thomas precession. This effect is present even in flat spacetime, and can be viewed, heuristically, as a consequence of the fact that Lorentz boosts in different directions do not commute.

(2) Geodetic or de Sitter precession. This effect is due to the mass of the central body, and gets stronger the closer the orbit is.

(3) Lense-Thirring or "frame dragging" precession. This effect is due to the rotation (if any) of the central body, and is what Gravity Probe B was testing for. Note that the Gravity Probe B experiment had to be carefully designed using two satellites, whose orbital parameters were chosen such that the other precession effects would cancel out when the data from the two satellites was combined, so that the Lense-Thirring precession, which is much smaller than the other two, could be detected.

 

[PeterDonis]

Regarding the general topic of Fermi-Walker transport and "precessions", the following Insights articles might be of interest:

https://www.physicsforums.com/insights/precession-in-special-and-general-relativity/

https://www.physicsforums.com/insights/how-to-study-fermi-walker-transport-in-minkowski-spacetime/

https://www.physicsforums.com/insights/fermi-walker-transport-in-schwarzschild-spacetime/

https://www.physicsforums.com/insights/fermi-walker-transport-in-kerr-spacetime/

 

[Dale]

My original question is whether or not the gyro will be affected when installed inside an object that is following a curved path (the geodesic path) toward a massive object or orbiting a massive object like a planet. In other words, when in freefall, does the gyro always see a straight, inertial path, due to the warping of spacetime, or can it detect this change of direction relative to the object that it is heading toward or orbiting around?

Yes, that is the question I answered. Although @PeterDonis’s answer is more complete since I focused only on one aspect.

Did you think I was answering a different question?

 

[Sagittarius A-Star]

(2) Geodetic or de Sitter precession. This effect is due to the mass of the central body, and gets stronger the closer the orbit is.

A nice animation of geodetic precession in curved spacetime can be found at the bottom of:
https://www.einstein-online.info/en/spotlight/geodetic_precession/

 

[zanick]

Geodesic paths are not curved. They are straight, by definition: geodesics define what "straight" means.

Locally, a gyroscope can't "detect" any "change" along a path, because a gyroscope defines what "unchanged" means. More precisely, a gyroscope is a physical realization of what is called Fermi-Walker transport of vectors along a worldline, and Fermi-Walker transport defines what it means for a vector to be unchanged when transported along a worldline.

What an experiment like Gravity Probe B does is to compare the gyroscope direction with some globally defined direction, such as the direction pointing towards a particular distant star, in order to see if being transported along some particular geodesic worldline changes the relationship between the local version of "unchanged", which, as above, is defined by the gyroscope, and a global version of "unchanged" defined by continuing to point towards the same distant object. In general, such a relationship will change in curved spacetime as an object travels along a geodesic.

For the case of geodesic orbits around an isolated gravitating body, the change in the relationship between gyroscope direction and a global fixed direction such as pointing towards a distant star can be separated, conceptually, into three different pieces:

(1) Thomas precession. This effect is present even in flat spacetime, and can be viewed, heuristically, as a consequence of the fact that Lorentz boosts in different directions do not commute.

(2) Geodetic or de Sitter precession. This effect is due to the mass of the central body, and gets stronger the closer the orbit is.

(3) Lense-Thirring or "frame dragging" precession. This effect is due to the rotation (if any) of the central body, and is what Gravity Probe B was testing for. Note that the Gravity Probe B experiment had to be carefully designed using two satellites, whose orbital parameters were chosen such that the other precession effects would cancel out when the data from the two satellites was combined, so that the Lense-Thirring precession, which is much smaller than the other two, could be detected.

yes, "straight" but curved in relation to large, massive bodies. (i misspoke there, so if the path is straight, then there would be no precession of a gyro in an orbit around a planet? Quite simply, the other question I had was if the "spaceship" was pointed in the direction of a geodesic path, would it naturally turn around the planet it comes to orbit around, but not sense any directional change, including the gyro, unless it was purposed to position the spacecraft or a telescope to other surrounding objects or celestial bodies? And this gets to the reason I asked about this in the first place. The ISS, for example, wouldn't need to have an angular velocity that matched the obit period because it is following a straight geodesic path. It would rotate with respect to the planet but would still remain inertial and gyroscopes would not be able to detect any rotation, unless it varied off the geodesic path or orientation. Is this statement correct? Thank you so much for your response!

 

[Sagittarius A-Star]

The ISS, for example, wouldn't need to have an angular velocity that matched the obit period because it is following a straight geodesic path. It would rotate with respect to the planet but would still remain inertial and gyroscopes would not be able to detect any rotation, unless it varied off the geodesic path or orientation. Is this statement correct? Thank you so much for your response!

No.

The module axes of the ISS are oriented parallel to the earth's surface. Like the moon, she always turns the same "under" side to the earth.

via:
https://translate.google.com/?hl=de...stets dieselbe „Unter“-Seite zu.&op=translate

Source (German):
https://de.wikipedia.org/wiki/Internationale_Raumstation#Umlaufbahn

 

[PeterDonis]

"straight" but curved in relation to large, massive bodies.

Meaning, spatially "curved" with respect to a coordinate system in which the massive body was at rest.

so if the path is straight, then there would be no precession of a gyro in an orbit around a planet?

Do you mean spatially straight in a coordinate system in which the massive body is at rest? If so, such a path would not be a geodesic; any geodesic path in the gravitational field of a massive body will be spatially curved.

if the "spaceship" was pointed in the direction of a geodesic path, would it naturally turn around the planet it comes to orbit around, but not sense any directional change, including the gyro, unless it was purposed to position the spacecraft or a telescope to other surrounding objects or celestial bodies?

Your phrasing is still somewhat vague. Let me rephrase more precisely:

Consider a spaceship--for concreteness, let's say the ISS, since you mentioned it. Suppose we find that at some instant, a gyroscope aboard the ISS is pointed exactly "forward"--that is, exactly in the same direction as the tangent to the orbit. What then happens to the direction of that gyroscope as the ISS goes around in its orbit?

Let's build up the answer to this in stages.

First, as I've already said, we need some global reference to compare the gyroscope to, since otherwise we won't have any way of detecting "change" at all, since the gyroscope's direction locally defines what "unchanged" means. In other words, unless we look out the window of the ISS and compare the direction of the gyroscope with some global reference, we can't run the experiment we're discussing at all. So let's suppose that, at the same instant we referred to above, where the gyro is pointing exactly along the orbit, it is also pointing exactly at some distant star, say Canopus, since that's a commonly used navigational reference (because it's bright and visible from most parts of the Earth).

Now notice: it's obviously impossible for the gyro to both continue to point towards Canopus, and to continue to point exactly along the orbit of the ISS. (This is one way of more precisely defining what it means for the orbit to be "spatially curved".) So we have to pick one of the two to be our "baseline" relative to which we define "precession". The one that is picked in physics is the global reference: i.e., the definition of "zero precession" is that the gyroscope continues to point towards Canopus forever as the ISS goes around in its orbit.

Now we know that the precession won't actually be zero, because of the three effects I listed. Unfortunately, the effects don't combine additively, and they depend on the orbital parameters in different ways, so it's not simple to describe their combined effect on the gyroscope direction. I give some rough descriptions in the Insights articles I referenced, but I'll need to check those over before posting further here.

 

[PeterDonis]

I give some rough descriptions in the Insights articles I referenced, but I'll need to check those over before posting further here.

The rough descriptions in my Insights articles were general, not limited to free-fall orbits (and indeed they included regimes, such as $r < 3M$ in Schwarzschild spacetime, in which there are no free fall orbits). If we limit attention to the case of free-fall orbits, things become somewhat simpler.

I'll look at the case of Schwarzschild spacetime (non-rotating mass, no frame dragging) in this post, and then follow up with the case of a rotating mass in a follow-up post. For Schwarzschild spacetime, the general formula in the Insights article for the vorticity is

$$
\Omega = \frac{\omega}{G^2} \left( 1 - \frac{3M}{r} \right)
$$

where, for our purposes here, we can view $\Omega$ as the rate at which a vector that always points tangent to the orbit rotates, relative to a gyroscope, as seen by the orbiting observer. Also, $\omega$ is the angular velocity of the orbit, as seen by an observer at infinity, and $G$ is the time dilation factor of the orbiting observer relative to the observer at infinity. I am using units in which Newton's gravitational constant and the speed of light are both $1$.

To get from $\Omega$ to the precession of a gyroscope relative to a fixed distant object like Canopus, we need to note two things. First, $\Omega$ is a frequency relative to the orbiting observer; but it is easier to work with frequencies relative to infinity. The frequency relative to infinity will be $\Omega G$. Second, relative to infinity, a vector that always points tangent to the orbit rotates in the prograde direction with frequency $\omega$; to get the frequency at which a gyroscope precesses, we have to subtract $\Omega G$ from this. So our formula for the precession frequency relative to infinity is

$$
P = \omega - \Omega G
$$

For a free-fall orbit, we have $\omega = \sqrt{M / r^3}$, and plugging this into the formula for $G$ given in the Insights article, we can see that $G^2 = 1 - 3M / r$. That means that the $G^2$ in the denominator above cancels the $1 - 3M / r$ factor, so we have

$$
\Omega = \omega
$$

This then gives for the gyroscope precession frequency

$$
P = \omega \left( 1 - G \right)
$$

Since we always have $G < 1$, this means we always have $P > 0$, i.e., the gyroscope will precess in a prograde direction. As $\omega$ increases (meaning we are considering orbits closer and closer to the central mass), $G$ will decrease, so $1 - G$ will get closer and closer to $1$ and $P$ will get closer and closer to $\omega$, which means the gyroscope will get closer and closer to precessing just fast enough to always point along the tangent to the orbit. (However, this turns out to happen at $r = 3M$, i.e., at the photon sphere, where the free fall orbital speed equals the speed of light. The only object which we know for sure is compact enough to allow such an orbit to be reached is a black hole.)

 

[PeterDonis]

Now to follow up with the case of Kerr spacetime, where we now add Lense-Thirring precession to the mix. Here the formulas are more complicated. The general formula for the vorticity is:

$$
\Omega = \frac{\omega}{D^2} \left[ 1 - \frac{3M}{r} \left( 1 - a \omega \right) \right] + \frac{M a}{r^3 D^2} \left( 1 - a \omega \right)^2
$$

where now $D$ is the time dilation factor of the orbiting observer relative to infinity, and $a$ is the angular momentum per unit mass of the rotating body (strictly speaking, Kerr spacetime only exactly describes a rotating black hole, but we'll gloss over that here). This, as before, is a general formula, and we want to restrict to the special case of a free-fall orbit. For that, we need the free-fall orbital frequency, which can be found by setting the proper acceleration $A$ to zero (the formula is given in the Insights article). We could obtain an explicit formula for $\omega$ by solving the resulting quadratic in $\omega$, but for this post I'm going to restrict attention to the case where $a \omega << 1$ (which will be true for any satellite orbiting Earth, and in particular for Gravity Probe B), which will allow us to make some useful approximations to simplify the formulas. The first useful formula is

$$
\omega^2 r^2 = \frac{M}{r} \left( 1 - a \omega \right)^2
$$

for a free-fall orbit (this formula is exact, but we'll make an approximation using it in a moment). Notice that if $a = 0$ this reduces to the corresponding formula for Schwarzschild spacetime, as it should. We next write the formula for the time dilation factor of the orbiting observer (or more precisely its square), which is

$$
D^2 = 1 - \frac{2M}{r} \left( 1 - a \omega \right) - \omega^2 r^2 H^2
$$

For orbits about the Earth, we will have $H = 1$ to a very good approximation, so, plugging in our formula for $\omega^2 r^2$ from above, we obtain

$$
D^2 = 1 - \frac{2M}{r} \left( 1 - a \omega \right) - \frac{M}{r} \left( 1 - a \omega \right)^2
$$

We can rewrite this as

$$
D^2 = 1 - \frac{3M}{r} \left( 1 - a \omega \right) + \frac{M}{r} a \omega \left( 1 - a \omega \right)
$$

And now we can see that this formula also reduces to the Schwarzschild formula for the time dilation factor (there we called it $G^2$) if $a = 0$. What's more, we can see that, except for the last term on the RHS, this factor would exactly cancel the factor in the first term of the formula for $\Omega$ above, just as we saw happened in Schwarzschild spacetime. So we are close to having a much simpler formula--but we're not quite there yet.

Let's see where we are by updating our formula for $\Omega$ using the above and simplifying:

$$
\Omega = \omega \frac{ 1 - \frac{3M}{r} \left( 1 - a \omega \right) + a \omega }{ 1 - \frac{3M}{r} \left( 1 - a \omega \right) + \frac{M}{r} a \omega \left( 1 - a \omega \right) }
$$

If we eliminate the term in $\left( a \omega \right)^2$ in the denominator and rewrite the last term in the numerator slightly, we obtain, after some algebra and after also eliminating a term in $M a \omega$ since it will be small compared to terms in $M$ only or $a \omega$ only,

$$
\Omega = \omega \left[ 1 + \frac{r - M}{r - 3M} a \omega \right]
$$

This then gives for the gyroscope precession frequency

$$
P = \omega - \Omega D = \omega \left( 1 - D \left[ 1 + \frac{r - M}{r - 3M} a \omega \right] \right)
$$

You can see that this looks similar to what we saw for the Schwarzschild case, but there are now extra terms in $a \omega$ in the formula above and also contained in $D$. We can no longer say that $P > 0$ always, because for orbits far enough out, the term in $D$ can be greater than $1$ (indeed, in the limit as $r \to \infty$, where $\omega \to 0$, we can see from the vorticity formula that the gyroscope precession must be retrograde since the only term left in $\Omega$ is the positive term in $M a$). Also, the approximations we made blow up as $r \to 3M$ since the denominator of the $a \omega$ term in the formula above goes to zero, so the behavior for orbits close enough to a Kerr black hole must be analyzed more completely.

 

[member 728827]

The rough descriptions in my Insights articles were general, not limited to free-fall orbits (and indeed they included regimes, such as $r < 3M$ in Schwarzschild spacetime, in which there are no free fall orbits). If we limit attention to the case of free-fall orbits, things become somewhat simpler.

I'll look at the case of Schwarzschild spacetime (non-rotating mass, no frame dragging) in this post, and then follow up with the case of a rotating mass in a follow-up post. For Schwarzschild spacetime, the general formula in the Insights article for the vorticity is

$$
\Omega = \frac{\omega}{G^2} \left( 1 - \frac{3M}{r} \right)
$$

where, for our purposes here, we can view $\Omega$ as the rate at which a vector that always points tangent to the orbit rotates, relative to a gyroscope, as seen by the orbiting observer. Also, $\omega$ is the angular velocity of the orbit, as seen by an observer at infinity, and $G$ is the time dilation factor of the orbiting observer relative to the observer at infinity. I am using units in which Newton's gravitational constant and the speed of light are both $1$.

To get from $\Omega$ to the precession of a gyroscope relative to a fixed distant object like Canopus, we need to note two things. First, $\Omega$ is a frequency relative to the orbiting observer; but it is easier to work with frequencies relative to infinity. The frequency relative to infinity will be $\Omega G$. Second, relative to infinity, a vector that always points tangent to the orbit rotates in the prograde direction with frequency $\omega$; to get the frequency at which a gyroscope precesses, we have to subtract $\Omega G$ from this. So our formula for the precession frequency relative to infinity is

$$
P = \omega - \Omega G
$$

For a free-fall orbit, we have $\omega = \sqrt{M / r^3}$, and plugging this into the formula for $G$ given in the Insights article, we can see that $G^2 = 1 - 3M / r$. That means that the $G^2$ in the denominator above cancels the $1 - 3M / r$ factor, so we have

$$
\Omega = \omega
$$

This then gives for the gyroscope precession frequency

$$
P = \omega \left( 1 - G \right)
$$

Since we always have $G < 1$, this means we always have $P > 0$, i.e., the gyroscope will precess in a prograde direction. As $\omega$ increases (meaning we are considering orbits closer and closer to the central mass), $G$ will decrease, so $1 - G$ will get closer and closer to $1$ and $P$ will get closer and closer to $\omega$, which means the gyroscope will get closer and closer to precessing just fast enough to always point along the tangent to the orbit. (However, this turns out to happen at $r = 3M$, i.e., at the photon sphere, where the free fall orbital speed equals the speed of light. The only object which we know for sure is compact enough to allow such an orbit to be reached is a black hole.)

Do the angular momentum of the gyros itself make any difference?

 

[PeterDonis]

Do the angular momentum of the gyros itself make any difference?

Not in the analysis I give in the Insights articles or in this thread. The gyros themselves are treated as test objects, so they have no effect on the spacetime geometry. For the actual gyros used in, for example, Gravity Probe B, this is an extremely good approximation, since the gyros are some 24 orders of magnitude less massive than the Earth and have angular momenta some 30 orders of magnitude smaller than the Earth's.

 

[A.T.]

My original question is whether or not the gyro will be affected when installed inside an object that is following a curved path (the geodesic path) toward a massive object or orbiting a massive object like a planet. In other words, when in freefall, does the gyro always see a straight, inertial path, due to the warping of spacetime, or can it detect this change of direction relative to the object that it is heading toward or orbiting around?

Here is a description of the various effects that cause precession in orbit according to GR:
https://einstein.stanford.edu/SPACETIME/spacetime4.html

In particular the Geodetic Effect occurs even if the central mass is not spinning, and can be visualized in terms of the curved spatial geometry (Flamm's Paraboloid approximated as a cone):

 

[A.T.]

yes, "straight" but curved in relation to large, massive bodies. (i misspoke there, so if the path is straight, then there would be no precession of a gyro in an orbit around a planet? Quite simply, the other question I had was if the "spaceship" was pointed in the direction of a geodesic path, would it naturally turn around the planet it comes to orbit around, but not sense any directional change, including the gyro, unless it was purposed to position the spacecraft or a telescope to other surrounding objects or celestial bodies? And this gets to the reason I asked about this in the first place. The ISS, for example, wouldn't need to have an angular velocity that matched the obit period because it is following a straight geodesic path. It would rotate with respect to the planet but would still remain inertial and gyroscopes would not be able to detect any rotation, unless it varied off the geodesic path or orientation. Is this statement correct? Thank you so much for your response!

It seems you are asking about the spatial orientation of the gyro-axis relative the local tangent of the spatial-path (orbit). This angle will of course change throughout the orbit, because the orbiting object follows a geodesic only in space-time, not in space. The spatial-path is not a geodesic so there is no reason to expect that the gyro keeps a constant angle to the spatial-path.

But that is not what we mean by procession here. Precession is rather the change of gyro-axis relative to the distant stars. This is what the effects mentioned above explain.

 

[Sagittarius A-Star]

The video shows a gyroscope experiment in the ISS. The gyroscope axis rotates relative to the ISS with 4 degrees per minute.

Source (see at 05:38):

 

[pervect]

My original question is whether or not the gyro will be affected when installed inside an object that is following a curved path (the geodesic path) toward a massive object or orbiting a massive object like a planet. In other words, when in freefall, does the gyro always see a straight, inertial path, due to the warping of spacetime, or can it detect this change of direction relative to the object that it is heading toward or orbiting around?

An orbiting gyroscope will, as others have mentioned, have its spin axis precess relative to distant stars (often called fixed stars). As in the Gravity Probe B experiment (already mentioned), if one points the gyroscope spin axis twoards a distant "fixed" star, as the gyroscope orbits a massive body, the spin axis will shift so as not to point towards the distant "fixed" star.

On the technical end, I found some interesting things while reading a bit. A spinning body should obey the papapetrou equations. https://en.wikipedia.org/wiki/Mathisson–Papapetrou–Dixon_equations

An interesting quote:

This is a form of Fermi–Walker transport of the spin tensor along the trajectory – but one preserving orthogonality to the momentum vector $k^{\mu}$ rather than to the tangent vector $V^{\mu}$. Dixon calls this M-transport.

I think this means that one need the momentum vector to be a directly proportional to the velocity vector (which I believe is usually the case, at least approximately) for a spinning body to Fermi-walker transport the spin axis. I had the impressions that gyroscopes Fermi-walker transported their spin axis (and in general, their basis vectors).

Also, apparently it's an approximation (but a good one) to say that a spinning body follows a geodesic. In general the interaction between the spinning body at the curvature of space-time (as given by the Riemann tensor $R^{a}_{bcd}$ will cause a very slight deviation of trajectory of a spinning body from the geodesic followed by an idealized non-spinning "test" body.