Nodal Precession: Satellites Orbiting Earth Explained

[Sphere]

Hello, the artificial satellites that orbit the Earth with an orbital inclination greater than 0 are all subject to the nodal precession caused by the equatorial bulge of the Earth at the equator.

But what I don't really understand is that at the equator the gravity is a little less strong than at the poles so how the bulge at the equator can attract and "deviate" satellites while this area at lower gravity?

Thank you!

 

[Hornbein]

Hello, the artificial satellites that orbit the Earth with an orbital inclination greater than 0 are all subject to the nodal precession caused by the equatorial bulge of the Earth at the equator.

But what I don't really understand is that at the equator the gravity is a little less strong than at the poles so how the bulge at the equator can attract and "deviate" satellites while this area at lower gravity?

Thank you!

Gravity is close to the same everywhere on the surface of the Earth. As noted by Isaac Newton, if gravity were stronger somewhere then water would flow there. He also deduced that the Earth is oblate.

Now consider a satellite x miles from the center of mass of the Earth. It will feel most gravity when above the Equator because there is more mass between it and the center of Earth, and some of that mass is closer to the satellite. Minimum gravity is felt above the poles.

 

[Drakkith]

First, satellites do not orbit at the equator or at the poles. These are located on the surface of the Earth and do not extend into space. The force of gravity on a person at the equator is less because the person is further away from center of the Earth than a person is when they are at the poles. But a satellite in a circular orbit is located at the same distance from the center of the Earth no matter what inclination they are in, so the force of gravity is the same. (ignoring very small differences due to the non-symmetrical shape of the Earth)

What DOES change is the direction of the gravitational pull. When a satellite is in an orbit around a non-spherical rotating body, the bulges 'pull' on the satellite at various points in its orbit, rotating the orbit around and causing nodal precession. So it's not the strength of gravity that changes, but the direction that the force points. See the following image:

Per wiki: The centrifugal force deforms the body so that it has an equatorial bulge. Because of the bulge of the central body, the gravitational force on a satellite is not directed toward the center of the central body, but is offset toward its equator. Whichever hemisphere of the central body the satellite lies over, it is preferentially pulled slightly toward the equator of the central body. This creates a torque on the satellite. This torque does not reduce the inclination; rather, it causes a torque-induced gyroscopic precession, which causes the orbital nodes to drift with time.

 

[Drakkith]

Now consider a satellite x miles from the center of mass of the Earth. It will feel most gravity when above the Equator because there is more mass and some of that mass is closer to the satellite. Minimum gravity is felt above the poles.

I don't think this is true. While there will be a very small difference because the Earth isn't a perfectly symmetrical shape, this difference is MUCH smaller than the difference between a point at the equator and one at one of the poles on the surface. See my post above.

 

[Sphere]

For a satellite in circular polar orbit around the Earth, can the equatorial bulge also cause a slight acceleration of the satellite as it heads towards the equator and a slight deceleration as it moves away from it towards a pole? (I hope my question is understandable)

 

[Hornbein]

I don't think this is true. While there will be a very small difference because the Earth isn't a perfectly symmetrical shape, this difference is MUCH smaller than the difference between a point at the equator and one at one of the poles on the surface. See my post above.

I'm not sure I understand what you are saying. The force of gravity is greater at the Equator but the net downward force [which is what Earth people usually call gravity] is almost the same everywhere on the Earth's surface. Is that what you mean?

I was wrong about nodal precession. I didn't even know what it was. Well, posting about stuff you don't know is Internet tradition.

 

[Drakkith]

I'm not sure I understand what you are saying. The force of gravity is greater at the Equator but the net downward force [which is what Earth people usually call gravity] is almost the same everywhere on the Earth's surface. Is that what you mean?

I'm saying that a satellite in a circular orbit experiences virtually no difference in gravitational pull anywhere in its orbit, because it is the same distance from the center of the Earth at all times. This is in contrast to someone at the equator or at the pole, both of which are on the surface and thus have to be at different distances from the center of the Earth.

Note that while it is true that over the equator there is more material closer to the satellite, it ignores that the bulge on the opposite side of the Earth is further away from the satellite than it would be if the Earth was a perfect sphere. This means that the distant bulge attracts the satellite with less force. The net effect of the bulges is that they cancel each other out as far as I know.

 

[Drakkith]

For a satellite in circular polar orbit around the Earth, can the equatorial bulge also cause a slight acceleration of the satellite as it heads towards the equator and a slight deceleration as it moves away from it towards a pole? (I hope my question is understandable)

Yes, that is my understanding as well. For a circular orbit, or rather a near-circular orbit since we can't have a circular orbit if we have changing forces on the satellite, I think these effects end up cancelling so that the orbit doesn't really change on long time scales, but I'm not certain.

 

[Hornbein]

I'm saying that a satellite in a circular orbit experiences virtually no difference in gravitational pull anywhere in its orbit, because it is the same distance from the center of the Earth at all times. This is in contrast to someone at the equator or at the pole, both of which are on the surface and thus have to be at different distances from the center of the Earth.

Note that while it is true that over the equator there is more material closer to the satellite, it ignores that the bulge on the opposite side of the Earth is further away from the satellite than it would be if the Earth was a perfect sphere. This means that the distant bulge attracts the satellite with less force. The net effect of the bulges is that they cancel each other out as far as I know.

Isaac Newton was the first to show that the Earth was oblate. He reasoned that the net downward force everywhere on the surface had to be approx. the same everywhere. If not, the Earth would change shape until it was the same. Isaac calculated the oblacity of the Earth but was off by a factor of 2 because of General Relativity.

The net downward force on the surface (often confused with gravity) is the gravitational force minus the centrifugal force. The centrifugal force is zero at the poles and maximal at the equator. That means the gravitational force is greater at the Equator. Not only is the gravitational force greatest at the Equator, that surface is also furthest from the center of the Earth. This is the opposite of what you would expect from the 1/n^2 gravitational rule. This is not a paradox because that rule applies only to spheres.

Satellites also experience centrifugal force but it is always there, even when above a pole.

Nearby masses are more important than distant ones. This is because the gravitational force decreases exponentially with distance. The greater the exponent, the less influence from distant masses and more from nearby masses. If the exponent were -1 yielding a 1/n formula then it might cancel out in the case you mentioned.

In the GR view the net downward acceleration being the same everywhere on Earth is equivalent to clocks running at the same speed everywhere on the surface of Earth. This is why Newton was off by two.

 

[Drakkith]

The centrifugal force is zero at the poles and maximal at the equator. That means the gravitational force is greater at the Equator.

This is incorrect. The gravitational force is indeed lower at the equator than at the poles by about 0.2% due to the increased distance from the center of the Earth. The addition of mass in the bulge (if any) is more than offset by the increased distance you are from most of that mass. You are experiencing both a centrifugal force AND a reduction in gravity when you are at the equator.

 

[Tracy Hall]

I'm saying that a satellite in a circular orbit experiences virtually no difference in gravitational pull anywhere in its orbit, because it is the same distance from the center of the Earth at all times. This is in contrast to someone at the equator or at the pole, both of which are on the surface and thus have to be at different distances from the center of the Earth.

Note that while it is true that over the equator there is more material closer to the satellite, it ignores that the bulge on the opposite side of the Earth is further away from the satellite than it would be if the Earth was a perfect sphere. This means that the distant bulge attracts the satellite with less force. The net effect of the bulges is that they cancel each other out as far as I know.

This cancellation effect only works for entire spherical shells: When you calculate the integral that gives you the gravitational attraction from any point outside of a spherically symmetric shell, it works out to exactly the same attraction as though all of the mass of the shell were concentrated at its center point. (Incidentally, from any point inside the spherical shell, the gravitational attraction to the entire shell cancels out to exactly zero.) Part of the balancing that makes this work is that, while the inverse square law means that mass close by is much more important than mass that is far away, the shape of a spherical shell means that from a point fairly close to its surface, the amount of mass that is far away is much greater than the amount of mass that is nearby, and the relative importance of the one effect happens to perfectly balance out the relative importance of the other effect.

If you have just two masses, however, similarly balanced in distance from the center, but not part of an entire spherical shell, the cancellation no longer works. Consider three masses, A, B, and C, all of the same mass, such that any pair of them experience a mutual attraction of one Newton at a distance of one kilometer. (This works out to masses of about 122.4 million kilograms each.) If B and C are both at a center point, and A is two kilometers away, then A will experience an attractive force of 1/4 + 1/4 = 0.5 Newtons in the direction of B and C. Now move B one kilometer closer to A (new distance: one kilometer) and C one kilometer further from A (new distance: three kilometers). The attraction that A now feels in the direction of B and C is 1 + 1/9 = 1.111… Newtons—more than twice the attraction that it felt when B and C were together! The inverse square law wins out because the amount of mass far away (at C) is not more than the amount of mass close in (at B).

The wonderful cancellation that happens in the spherical shell integral, that allows us to greatly simplify orbital calculations when we make the assumption of spherically symmetric bodies, becomes much more complicated and difficult once you work with more realistic geometries. In the case of an approximate oblate spheroid like the Earth, the net effect works out so that if two satellites are the same distance from the center of the Earth, but one of them is above the equator and the other is above a pole, the one above the equator feels a slightly stronger attraction—and a third satellite at a latitude somewhere in between will feel a net attraction, not to the exact center point of the Earth, but to a point somewhat closer to the near side of the equator.

 

[Drakkith]

The wonderful cancellation that happens in the spherical shell integral, that allows us to greatly simplify orbital calculations when we make the assumption of spherically symmetric bodies, becomes much more complicated and difficult once you work with more realistic geometries. In the case of an approximate oblate spheroid like the Earth, the net effect works out so that if two satellites are the same distance from the center of the Earth, but one of them is above the equator and the other is above a pole, the one above the equator feels a slightly stronger attraction—and a third satellite at a latitude somewhere in between will feel a net attraction, not to the exact center point of the Earth, but to a point somewhat closer to the near side of the equator.

Hmmm. Perhaps. Do you have a reference for this that I could look at? I'd like to know a bit more on the topic if you have one handy.

 

[David_S]

I don't think this is true. While there will be a very small difference because the Earth isn't a perfectly symmetrical shape, this difference is MUCH smaller than the difference between a point at the equator and one at one of the poles on the surface. See my post above.

Several of these posts appear to contradict NASA's orbital studies of variances in Earth's gravity (GRACE), as measured by two satellites traveling in the same orbit "tethered" by a laser to precisely measure the minute dips & drags that map the stronger and weaker surface areas of Earth's gravity from orbit are too insignificant to be a part of this discussion?

 

[Drakkith]

Several of these posts appear to contradict NASA's orbital studies of variances in Earth's gravity (GRACE), as measured by two satellites traveling in the same orbit "tethered" by a laser to precisely measure the minute dips & drags that map the stronger and weaker surface areas of Earth's gravity from orbit are too insignificant to be a part of this discussion?

The two questions are:

1. The force of gravity at the equator vs the poles on the surface.
2. The force of gravity on a satellite in a circular orbit above the poles vs the equator.

Everything I've looked up says that #1 is exactly how I described it above. It's #2 I'm uncertain about.

 

[Tom.G]

Here is a link to the GRACE satellite data that @David_S referred to above:
https://earthobservatory.nasa.gov/images/3666/Earth's-gravity-field

And a link to a video of more recent data from the European Space Agency;
https://www.esa.int/ESA_Multimedia/Videos/2021/03/Best_view_yet_of_global_gravity/(lang)/en

The problem seems to be not as simple as so-far discussed here!

Have Fun!
Tom

 

[Sphere]

But then does the equatorial bulge of the Earth cause the nodal precession simply because part of the mass at the equator is raised and pulls the satellite a little more towards the equator than towards the center of the Earth or because there is a little more mass at the equator than at the poles and this causes a slightly stronger gravitational force on the satellite in orbit when it approaches the equator and deflects it or else a bit of both ?

 

[Drakkith]

But then does the equatorial bulge of the Earth cause the nodal precession simply because part of the mass at the equator is raised and pulls the satellite a little more towards the equator than towards the center of the Earth or because there is a little more mass at the equator than at the poles and this causes a slightly stronger gravitational force on the satellite in orbit when it approaches the equator and deflects it or else a bit of both ?

I think it's a bit of both. But again, I'm not an expert in this area, so I could be mistaken.

 

[strangerep]

Hmmm. Perhaps. Do you have a reference for this that I could look at? I'd like to know a bit more on the topic if you have one handy.

You could start with Wikipedia's Shell Theorem page.

 

[vanhees71]

Hm, isn't the "proof" a bit overcomplicated in this Wikipedia page (although correct, as far as I can see without checking all calculations carefully "with pencil and paper").

I usually argue with the Laplace equation: We have a mass distribution, that is spherically symmetric around the origin of our Cartesian coordinate system and $\rho(\vec{r})=0$ insight the open solid sphere $r<R$. Because of spherical symmetry in spherical coordinates $\Phi=\Phi(r)$ only (since $r=|\vec{r}|$ is the only scalar you can build from the only independent vector, $\vec{r}$, present in the problem). For $r<R$ we have
$$\Delta \Phi=\frac{1}{r} (r \Phi)''=0.$$
Integrating gives as the general solution
$$\Phi(r)=\frac{C_1}{r}+C_2.$$
Since there are no singularities at $r=0$, we have $C_1=0$, and thus $\Phi(r)=\text{const}$. The gravitational field, $\vec{g}=-\vec{\nabla} \Phi=0$. QED.

 

[strangerep]

For $r<R$ we have
$$\Delta \Phi=\frac{1}{r} (r \Phi)''=0.$$

Already at this stage, your equation is ill-defined at $r=0$, so one must exclude an infinitesimal open neighbourhood around $r=0$. Then...

Integrating gives as the general solution
$$\Phi(r)=\frac{C_1}{r}+C_2.$$
Since there are no singularities at $r=0$, we have $C_1=0$, and thus $\Phi(r)=\text{const}$. The gravitational field, $\vec{g}=-\vec{\nabla} \Phi=0$. QED.

Since we have excluded $r=0$ we don't need to set $C_1=0$.

 

[vanhees71]

You forget that spherical coordinates are singular along the polar axis anyway. It's enough that $C_1/r \rightarrow \infty$ for $r \rightarrow 0$ to argue that $C_1=0$.

 

[Sphere]

Are you sure there is more mass at the equator than at the poles due to the Earth's rotation?
Isn't it rather that matter is less dense at the equator than at the poles because of the centrifugal effect of the Earth's rotation, which would explain why the Earth is a flattened spheroid?

 

[Hornbein]

There is more mass UNDER the equator than the poles. The density of the Earth is roughly the same at the poles and equator.