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lpetrich
2017-Jun-01, 01:33 AM
We all know about albedo mapping and height mapping, I'm sure, but there is an additional sort of mapping that has been done on various celestial bodies. Gravity mapping. That kind of mapping can peek beneath a celestial body's surface and give clues about its interior.

What's usually mapped is the departure of the acceleration of gravity from some modeled value. The model here is usually a spherically-symmetric model or a rotating-fluid one. That departure is called the "gravity anomaly", and when adjusted for altitude, it is called the "free-air gravity anomaly." One can extend one's model to include an estimate of the surface terrain's gravitational field. This gives a "Bouguer anomaly".

Gravity acceleration is often measured in a unit called the gal, after Galileo: 1 cm/s2, about 10-3 of the Earth's average surface gravity. Often, one sees milligals, 10-3 of a gal or 10-6 of the Earth's average surface gravity.

An alternative is to map the "geoid", a shape with constant gravitational potential. For the Earth, the geoid is usually set to the sea-level surface, though for other celestial bodies, it is usually much more arbitrary.

Now for some maps.

NASA - Improved Measurement of Earth's Gravity Field (https://www.nasa.gov/mission_pages/Grace/multimedia/GravityFieldCombob.html) from the GRACE mission, two low-Earth-orbit satellites that were tracked to determined the Earth's gravity variations with high precision.

Lunar Background Material (http://minerva.union.edu/hollochk/moon_rocks/background.html) (has some gravity maps) and GRAIL Science Reports - Spaceflight101 (http://www.spaceflight101.net/grail-science-reports.html) -- a NASA mission to map the Moon's gravity.

The Gravity Field of Mercury from MESSENGER (http://www.lpi.usra.edu/meetings/lpsc2012/pdf/2189.pdf) -- solution HgM002: good resolution mainly in the north.

NSSDCA Photo Gallery: Venus (https://nssdc.gsfc.nasa.gov/photo_gallery/photogallery-venus.html) -- look for: Gravity Maps of Venus from Magellan (Mercator Projection).

PGDA - Mars Gravity Field: GMM-3 (https://pgda.gsfc.nasa.gov/products/57), also New Mars Gravity Map (https://www.nasa.gov/feature/goddard/2016/mars-gravity-map)

Space Images | Shape and Gravity of Vesta (https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA15604), also Asteroid Vesta: Topography and Gravity Map - Dawn Mission NASA Video Hd - YouTube (https://www.youtube.com/watch?v=stCml0wlIj8)

Mollweide projection of topography and the Bouguer anomaly. : A partially differentiated interior for (1) Ceres deduced from its gravity field and shape : Nature : Nature Research (https://www.nature.com/nature/journal/v537/n7621/fig_tab/nature18955_F1.html)


For Jupiter, Saturn, Uranus, and Neptune, however, a gravity map will look much less interesting, since their gravitational fields are rotationally symmetric and north-south symmetric to within measurement accuracy.

lpetrich
2017-Jun-01, 04:42 AM
Mathematically, a celestial body's gravitational field can be described as a combination of power series in the distance from its center and a sort-of trigonometric series in the angular coordinates. Here's the expression:

\displaystyle{ V = - \frac{GM}{r} \sum_{l=0}^{\infty} \left( \frac{R}{r} \right)^l \sum_{m=0}^{l} (C_{lm} \cos m\phi + S_{lm} \sin m\phi) P_{lm}(\cos\theta) }

where V is the gravitational potential G is the gravitational constant, M is the mass, r is the distance from the coordinate center, θ is the polar angle or colatitude (90d - latitude), φ is the azimuthal angle or longitude, R is the equatorial radius, and the P's are associated Legendre functions. The θ and φ functions may be combined to make "spherical harmonics" Y_{lm}(\theta,\phi).

The C's and S's are gravitational-field coefficients, and they have some special values. The Sl0 are redundant and thus set to zero. The monopole terms C00 = 1 by definition. The dipole terms C10, C11, and S11 are set to zero by making the coordinate origin the celestial body's centroid or center of mass. The Cl0 are often given another name: Cl0 = - Jl.

In the rotationally-symmetric case, J2 has a simple relation to the polar and equatorial moments of inertia C and A: J2*R2 = C - A. From hydrostatic equilibrium, one can get the flattening to lowest order:
\displaystyle{ f = \frac32 J_2 + \frac12 q ,\ q = \frac{R^3 \omega^2}{GM} }
where ω is the angular velocity of rotation. To lowest order, J2 is proportional to q itself: J2 = (1/3)*k2*q where k2 is a "Love number" after a geophysicist with that last name. For a more even distribution of mass, k2 gets larger, as does the relative moment of inertia C/(M*R2). So one can get clues about a celestial body's mass distribution.

One can find exact solutions in some special cases.

Constant density (Maclaurin spheroid): \frac{C}{MR^2} = \frac25 ,\ f = \frac54 q ,\ J_2 = \frac12 q ,\ k_2 = \frac32
Pressure ~ (density)2 (constant size): \frac{C}{MR^2} = \frac{2(\pi^2-6)}{3\pi^2} ,\ f = \frac{15}{2\pi^2} q ,\ J_2 = \frac{15-\pi^2}{3\pi^2} q ,\ k_2 = \frac{15-\pi^2}{\pi^2}
Darwin-Radau approximation: \frac{C}{MR^2} = \frac23 \left( 1 - \frac25 \sqrt{\frac{5q}{2f}-1} \right)

lpetrich
2017-Jun-01, 09:53 AM
Let's see what the numbers look like.

For the constant-size case, C/(MR2) ~ 0.261, f/q ~ 0.760, J2/q ~ 0.173, k2 ~ 0.520
The Darwin-Radau equation for C from J2 (known from elsewhere, D-R):

Constant density: 2/5, 2/5
Constant size: 0.261, 0.263
Earth: 0.3307, 0.3315
Mars: 0.366, 0.375

Now the J's and k2 for the planets.

For the Earth, the largest coefficient is J2 = 1.083*10-3, and it is mostly due to the equatorial bulge. The others are much smaller, around 10-6 or less, and they are generated by smaller-scale irregularities, like mountain ranges and ocean trenches.

For the Moon, J2 is about 2.03*10-4, giving k2 = 80. That means that the Moon is rigid enough to have stayed that much lopsided for most of its existence. Venus and Mercury are similar, with Venus having J2 = 4.5*10-6 and k2 = 220, and Mercury J2 = 5.0*10-5 and k2 = 150.

Mars, however, is more Earthlike. It has J2 = 1.960*10-3 and k2 = 1.28, close to having constant density. It likely has a small core, smaller than the Earth's core in relative size.

Jupiter has J2 = 1.4736*10-2, J4 = -5.87*10-4, J6 = (3.1+-2.0)*10-5. Other coefficients are zero to within 10-6 or less. Which gives k2 = 0.49, very centrally condensed, close to the constant-radius case.

Saturn has J2 = 1.2691*10-2, J4 = -9.35*10-4, J6 = (8.6+-1.0)*10-5. Which gives k2 = 0.32, very centrally condensed.

Uranus has J2 = 3.343*10-3, J4 = -2.9*10-5. Which gives k2 = 0.34, also very centrally condensed.

Neptune has J2 = 3.411*10-3, J4 = -3.5*10-5. Which gives k2 = 0.39, likewise, though a bit less.

lpetrich
2017-Jun-01, 10:37 AM
First, the J's are sometimes called zonal-harmonic coefficients, the zones being latitude ones. The others are azimuthal ones, or more precisely, mixed ones.

There is an interesting curiosity about the outer planets. - J4/J22 has these values for J, S, U, and N: 2.7, 3.5, 2.6, and 3.0.

For constant density, the ratio is 15/7 ~ 2.14, while for constant size, the ratio is \frac{2025 (2\pi^2-21)}{7(\pi^2-15)^2} ~ 2.70

So that ratio also is consistent with the outer planets being very centrally condensed.

antoniseb
2017-Jun-01, 11:15 AM
No one else has responded yet, but I'd like t point out that this is an interesting collection you've put together here. Thanks!

StupendousMan
2017-Jun-01, 03:51 PM
I would like to echo Antoniseb's comment. Thanks very much.
I may use some of this in a course next year!

lpetrich
2017-Jun-01, 11:11 PM
Earth Gravity Model (https://www.nga.mil/ProductsServices/GeodesyandGeophysics/Pages/EarthGravityModel.aspx), Earth Gravitational Model 2008 (EGM2008) (http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/)

They use n where I'd used l, the overall index. But they still use m for the azimuthal index. Here are the models that they have constructed:

WGS 84: n to 180, m to n
EGM96: n to 360, m to n
EGM2008: n to 2159, m to n, and n from 2160 to 2190, m to 2159

Spatial variation for order n has an approximate wavelength of (2*pi*radius)/n, giving resolutions of 220 km, 110 km, and 18 km for each model.

Melting Glaciers Alter Earth's Gravity (http://www.livescience.com/15605-melting-glaciers-alter-earth-gravity.html) -- in Greenland and Antarctica. A very small effect, but enough to be noticeable.

On Mars, one can watch carbon dioxide condense on each pole during its fall and evaporate from it during its spring -- with its gravity.
Three Cool Things We’ve Learned From NASA's Mars Gravity Map (http://news.nationalgeographic.com/2016/03/160324-mars-nasa-gravity-map-space-science/)
NASA's Got the Best Map of Mars Gravity Ever, and Here It Is (http://www.space.com/32362-mars-gravity-map-best-ever-video.html)

grapes
2017-Jun-02, 12:28 AM
Mathematically, a celestial body's gravitational field can be described as a combination of power series in the distance from its center and a sort-of trigonometric series in the angular coordinates. Here's the expression:

\displaystyle{ V = - \frac{GM}{r} \sum_{l=0}^{\infty} \left( \frac{R}{r} \right)^l \sum_{m=0}^{l} (C_{lm} \cos m\phi + S_{lm} \sin m\phi) P_{lm}(\cos\theta) }

where V is the gravitational potential G is the gravitational constant, M is the mass, r is the distance from the coordinate center, θ is the polar angle or colatitude (90d - latitude), φ is the azimuthal angle or longitude, R is the equatorial radius, and the P's are associated Legendre functions. The θ and φ functions may be combined to make "spherical harmonics" Y_{lm}(\theta,\phi).

The Y_{lm}(\theta,\phi) also have negative m values, which represent the components related to the sine functions.


The C's and S's are gravitational-field coefficients, and they have some special values. The Sl0 are redundant and thus set to zero. The monopole terms C00 = 1 by definition. The dipole terms C10, C11, and S11 are set to zero by making the coordinate origin the celestial body's centroid or center of mass. The Cl0 are often given another name: Cl0 = - Jl.

In the rotationally-symmetric case, J2 has a simple relation to the polar and equatorial moments of inertia C and A: J2*R2 = C - A. From hydrostatic equilibrium, one can get the flattening to lowest order:
\displaystyle{ f = \frac32 J_2 + \frac12 q ,\ q = \frac{R^3 \omega^2}{GM} }
where ω is the angular velocity of rotation. To lowest order, J2 is proportional to q itself: J2 = (1/3)*k2*q where k2 is a "Love number" after a geophysicist with that last name. For a more even distribution of mass, k2 gets larger, as does the relative moment of inertia C/(M*R2). So one can get clues about a celestial body's mass distribution.

One can find exact solutions in some special cases.

Constant density (Maclaurin spheroid): \frac{C}{MR^2} = \frac25 ,\ f = \frac54 q ,\ J_2 = \frac12 q ,\ k_2 = \frac32
Pressure ~ (density)2 (constant size): \frac{C}{MR^2} = \frac{2(\pi^2-6)}{3\pi^2} ,\ f = \frac{15}{2\pi^2} q ,\ J_2 = \frac{15-\pi^2}{3\pi^2} q ,\ k_2 = \frac{15-\pi^2}{\pi^2}
Darwin-Radau approximation: \frac{C}{MR^2} = \frac23 \left( 1 - \frac25 \sqrt{\frac{5q}{2f}-1} \right)

lpetrich
2017-Jun-02, 12:29 AM
I forgot to include the Earth's value of k2. It is 0.94.
Thus, the Earth is somewhat centrally condensed, more than Mars, but less than Jupiter.


Gravity spherical harmonic coefficients of Ceres : A partially differentiated interior for (1) Ceres deduced from its gravity field and shape : Nature : Nature Research (http://www.nature.com/nature/journal/v537/n7621/fig_tab/nature18955_ST1.html)
From these, J2 = 2.6499*10-2 and J4 = - 1.7124*10-3.

This gives us k2 = 1.30 and -J4/J22 = 2.4. So Ceres is close to having constant density. J4, it must be noted, is not much bigger than some of the other coefficients, those due to lumpiness. Those ones max out at around 10-4.

lpetrich
2017-Jun-08, 04:06 AM
The Y_{lm}(\theta,\phi) also have negative m values, which represent the components related to the sine functions.
Actually, it's
\displaystyle{ Y_{lm}(\theta, \phi}) = \pm \sqrt{ \frac{2l+1}{4\pi} \frac{(l-m)!}{(l+m)!} } P_{lm}(\cos\theta) e^{im\phi} }
where the +- depends on the normalization convention. Note that m can be negative as well as positive.

Also, the associated Legendre functions are
\displaystyle{ P_{lm}(x) = \pm (1-x^2)^{m/2} \frac{d^m P_l(x)}{dx^m} }
\displaystyle{ P_{l,-m}(x) = \pm \frac{(l-m)!}{(l+m)!} P_{lm}(x) }
where the +-'s are additional normalization conventions.

lpetrich
2017-Jun-08, 04:46 AM
Now for a big table of k2 and j42 = - J4/J22 values.

Constant density: k2 = 1.5, j42 = 2.14
Constant size: k2 = 0.520, j42 = 2.70
Mercury: k2 = 150
Venus: k2 = 220
Earth: k2 = 0.94, j42 = 1.3
Moon: k2 = 80
Mars: k2 = 1.28, j42 = 4.7
Ceres: k2 = 1.30, j42 = 2.4
Jupiter: k2 = 0.49, j42 = 2.7
Saturn: k2 = 0.32, j42 = 3.5
Uranus: k2 = 0.34, j42 = 2.6
Neptune: k2 = 0.39, j42 = 3.0

For Saturn, J2 ought to be 1.6291*10-2 -- I made a typo in transcribing the number.

Mercury, Venus, and the Moon rotate slowly enough for the main influence on their J2 values to be their topography. Likewise, for the Earth and Mars, the main influence on J4 is their topography.

grapes
2017-Jun-08, 08:28 AM
Actually, it's
\displaystyle{ Y_{lm}(\theta, \phi}) = \pm \sqrt{ \frac{2l+1}{4\pi} \frac{(l-m)!}{(l+m)!} } P_{lm}(\cos\theta) e^{im\phi} }
where the +- depends on the normalization convention. Note that m can be negative as well as positive.

No, you've mixed up two different versions of spherical harmonics. Notice that in the OP the functions are real, whereas the functions you describe in this post have imaginary components. Geophysicists tend to avoid imaginary numbers, but not necessarily. :)


Also, the associated Legendre functions are
\displaystyle{ P_{lm}(x) = \pm (1-x^2)^{m/2} \frac{d^m P_l(x)}{dx^m} }
\displaystyle{ P_{l,-m}(x) = \pm \frac{(l-m)!}{(l+m)!} P_{lm}(x) }
where the +-'s are additional normalization conventions.

lpetrich
2017-Jun-17, 08:56 AM
NASA - NSSDCA - Data Collection - Details (https://nssdc.gsfc.nasa.gov/nmc/datasetDisplay.do?id=PSSB-00621) -- DAWN VESTA GRAVITY SCIENCE DERIVED SCIENCE DATA V1.0 (PDS)

It links to Index of /archive/dawn/grav/DWNVGRS_2 (https://sbn.psi.edu/archive/dawn/grav/DWNVGRS_2/) and the coefficients are at Index of /archive/dawn/grav/DWNVGRS_2/DATA/SHADR (https://sbn.psi.edu/archive/dawn/grav/DWNVGRS_2/DATA/SHADR/) along with a file that explains them. The explanation file also has Vesta's mass, spin rate, and north-pole direction.

ETA:
k2 = 1.53
j42 = 1.94

Close to constant density.

lpetrich
2017-Jun-17, 05:33 PM
There's a problem: the gravity data for Vesta is scaled to a reference radius of 260 km, when the actual size and shape of the asteroid is approximately a triaxial ellipsoid with axes 286.3, 278.6, and 223.2 km. I find an equatorial radius of 282.4 km by taking the geometric mean of the first two axes' lengths. I'd earlier adjusted q for that, but not J2. Adjusting both q and J2, I find

k2 = 1.35, j42 = 1.94

So Vesta has a bit of central concentration.

I'll now see how well it fits a Maclaurin-spheroid solution. For that, I'll use classical mechanics - The gravitational potential of ellipsoid - Physics Stack Exchange (https://physics.stackexchange.com/questions/16412/the-gravitational-potential-of-ellipsoid) and Maclaurin Spheroids (https://farside.ph.utexas.edu/teaching/336L/Fluid/node35.html).

The Maclaurin-spheroid solution is
\displaystyle{ \frac{\omega^2}{2\pi G\rho} = \frac{3-2e^2}{e^3} \sqrt{1-e^2}\arcsin e - \frac{3(1-e^2)}{e^2} }
where e is the eccentricity of the spheroid.

For Vesta's rotation rate and average density, I find e = 0.515, giving a flattening of 0.143. Vesta's actual flattening is 0.210, somewhat larger, corresponding to an eccentricity of 0.613.

I'll now estimate J2 and J4. Scaled to my equatorial radius, they are 0.0626 and -0.0076.

For a Maclaurin spheroid, they can be found from the polar and equatorial potential values by taking a series in 1/r:
\displaystyle{ V_{pol} = - \frac32 GM \left( - \frac{r}{a^2 e^2} + \frac{r^2 + a^2 e^2}{a^3 e^3} \arctan \frac{a e}{r} \right) }
\displaystyle{ V_{eqt} = - \frac34 GM \left( \frac{\sqrt{r^2 - a^2 e^2}}{a^2 e^2} - \frac{r^2 - 2 a^2 e^2}{a^3 e^3} \arcsin \frac{a e}{r} \right) }
where a is the equatorial radius. This gives us J2 = (1/5)*e2 and J4 = - (3/35)*e4.

From Vesta's actual shape, e = 0.613, and I find 0.0752 and -0.0121

So Vesta is a little bit centrally condensed. It is flatter than what one would expect from constant density, and it has less nonspherical gravity than what one would expect from its shape and constant density.