In this answer I discuss the first results in the 47th Lunar and Planetary Science Conference (2016) conference paper First Gravity Traverse on the Martian Surface from the Curiosity Rover and a more detailed paper published today in Science A surface gravity traverse on Mars indicates low bedrock density at Gale crater with some important results, but unfortunately though the mission and researcher is mostly taxpayer-funded, we have to pay again to read about it. (i.e. paywalled).

Near the bottom of Gale Crater some flavor of acceleration is reported to be about 3.717 m/s^2. I'm assuming that's along the local gravitational vertical rather than a geodetic vertical, but I don't know if it's the total acceleration (gravitational plus centrifugal effects due to planetary rotation) or it has had centrifugal effects subtracted already.

I tried to tell the difference using math, but my results are ambiguous.

I used $GM$ = 4.282837E+13 m^3/s^2 and the equatorial radius $R_0$ = 3396200 meters, and an altitude of -4500 meters (depth of Gale Crater), and got the following:

$$a_G = -\frac{GM}{R^2} \approx -3.7230 \ m/s^2$$

but I am aware of terms like "free air corrections" and realize that I am "out of my depth" (pardon the pun).

At 5.4 degrees south latitude the rotational velocity is about 240.1 m/s, so the centrifugal force associated with sitting on the surface at the bottom of Gale crater is

$$a_C = +\frac{v^2}{R} \approx +0.0170 \ m/s^2$$

If I add the two, I get -3.7060 m/s^2. The problem is that the value shown in the 2016 conference paper -3.717 m/s^2 is roughly half way between -3.7060 and -3.7230 m/s^2.

Question: Do Curiosity's reported measurements of Mars' surface gravity (~3.717 m/s^2) include centrifugal effects? Is the number the component parallel to the geodetic vertical, or is it the magnitude of acceleration no matter what direction it's pointing?

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Do Curiosity's reported measurements of Mars' surface gravity include centrifugal effects?

Of course. That's how surface gravity is defined.

The measurements in the cited paper were made with accelerometers during periods where when the vehicle was at rest to the surface. Accelerometers cannot measure centrifugal effects, any more than they can measure Coriolis or Euler effects. These are fictitious effects.

Then again, accelerometers cannot measure gravitation, either. The easiest explanation for why this is the case comes from general relativity, where gravitation is a fictitious acceleration, just as are the centrifugal and Coriolis accelerations. No fictitious acceleration can be measured by a local experiment. The Newtonian explanation is that gravitation is very close to uniform at the scale of an accelerometer. Whichever explanation one favors, accelerometers cannot measure gravitation.

What is measurable by a local experiment (e.g., an accelerometer) are accelerations due to non-fictitious, non-gravitational forces. In the special case of an object at rest on the surface of a planet, the measurable acceleration is exactly equal but opposite to the unmeasurable accelerations due to gravitation and the centrifugal effect. The measurements in the cited paper were accelerometer readings made when Curiosity was at rest with respect to Mars's surface.

Regarding the specific number, you've ignored Mars' rather lump gravitational field, the effects of Mars' equatorial bulge, and have only superficially addressed that Curiosity is 4.5 below the reference ellipsoid. It's well accepted that accelerometers at rest with respect to the surface of a planet are proxies for measuring gravity. Note well: geophysicists distinguish gravitation from gravity. Gravitation is the acceleration that results from the attraction of masses toward one another while gravity vectorially adds centrifugal effects to gravitation.

  • $\begingroup$ At high precision, "upward" is not simple to define, is it normal to the local surface (over what scale) or radially away from the centre of mass of the planet, or normal to the isosurfaces of the gravitational field (again on what scale) or normal to some approximation to the shape of the planet (like a best-fit ellipsiod or something)? All of these will be slightly different. $\endgroup$ – Steve Linton Feb 1 at 10:36
  • $\begingroup$ @SteveLinton (at)DavidHammen has said some thing about that here: space.stackexchange.com/a/19729/12102 $\endgroup$ – uhoh Feb 1 at 11:52
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    $\begingroup$ @AtmosphericPrisonEscape - Talk to a physicist who understands relativity. Gravitation is a fictitious force, no different from the centrifugal and Coriolis effects. Newtonian physics has a nice rationale for why the centrifugal and Coriolis effects cannot be measured (they're fictitious), but it's rationale for why gravitation cannot be measured (which it cannot) is a bit dubious. Relativity addresses that issue nicely: Gravitation is a fictitious force in general relativity. $\endgroup$ – David Hammen Feb 1 at 16:18
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    $\begingroup$ @uhoh - Gravitometers essentially are accelerometers, very precise one dimensional accelerometers. $\endgroup$ – David Hammen Feb 18 at 11:00
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    $\begingroup$ @uhoh - To be pedantic, they don't even measure gravity. They use the fact that the observable accelerations / forces on a test mass (accelerations if the test mass is a dropping ball, forces if the test mass is constrained by springs) are a proxy for the unmeasurable gravitational and centrifugal accelerations in the special case that the gravimeter case is at rest with respect to the rotating planet. $\endgroup$ – David Hammen Feb 18 at 11:13

I think the problem here is the calculation of gravity is a bit sensitive to nuances you've over-looked. I am confident that the reported result is the 'experienced' force, as it's the sensible way to do it and fits with the values the literature expects. I'd like to show the value is predictable but that's a bit beyond a short Q/A (and I think I'd quickly run into difficulty). However: I'll try to explain why it's hard to predict and, relatedly, why it's a plausible value.

Unfortunately: "The real world's a very complex and you can't answer the question without considering them." - which is the real answer - is unenlightening so I think we're going to have to wade in to some weeds even without the promise of an exact answer at the end.

One possible issue with your calculation is that the gravitational force experienced between a point mass and a sphere is only a good approximation (it's exact which is a nice result) if the point mass is outside the surface of the sphere.

In fact as you go into further down a mine shaft gravity decreases not increases as your equation would predict, which make sense. Imagine you somehow made it to the centre there's as much pulling you in each direction: gravity goes to zero.

To work out a better approximation, we can use a really nice result: as a spherical shell has no gravitational force on any object inside it. The calculus for this is a bit of a write up but google or this might help. The upshot is we can 'ignore' any of the $M$ further out than the point we want to measure.

Using some geometry we get:

$GM' = (4.282837E+13 m^3s^{-2} * ((3396200-4500)/3396200)^3)$

or $4.265E+13 m^3s^{-2}$, about 99.6% of the value you have.

Plugged back in I get $3.70825ms^{-2}$.

Wait! That takes us further from where we want to be?! Well yes but I was building up to my own "depth" of the problem joke, which I think I just ruined...

Now for some guesswork as to what might be the 'fix':

I think the thing that fixes this situation is the oblate spheroid shape of Mars. In your calculation you use the equatorial radius of Mars. This is significant: if you use your formula but use the equatorial radius instead, even with the depth correction we get $3.7352ms^{-2}$ which is too high (even with centripetal acceleration), so we are well within the range of it being important.

I know what your thinking: using the equatorial radius seems legit. Indeed Gale crater is very equatorial. However, the implicit spherical -> point-mass approximation doesn't work. It's not 'wrong' but if your chasing 0.1% accuracy this is important. And in this case the approximation in moving in the direction we expect/want. Indeed, for a given distance from the centre of mass over the equator, squashing the poles makes gravity stronger. The mass at the poles is moving closer and pulling more in line with the overall direction of gravity. Quantifying this is tricky though.

It might help to see your "just use the radius you're at" intuition fails here by considering what would happen if the rover was on the pole. Should the attraction be higher? What if the planet was really flat? Would you expect an extreme attraction?


@DavidHammen's answer says to include all accelerations, and @drjpizzle's answer recommends to look at Mars' shape and consider both the poles and equator.

So here's a complete tally using

$$a_G = -GM \frac{\mathbf{r}}{r^3}$$

From Geopotential_model; The_deviations of Earth's gravitational field from that of a homogeneous sphere:

$$a_{J2x} = J2 \frac{\mathbf{x}}{r^7}(6z^2 - 1.5(x^2+y^2))$$ $$a_{J2y} = J2 \frac{\mathbf{y}}{r^7}(6z^2 - 1.5(x^2+y^2))$$ $$a_{J2z} = J2 \frac{\mathbf{z}}{r^7}(3z^2 - 4.5(x^2+y^2))$$

$$a_C = \mathbf{r_{xy}} \omega^2 $$

            magnitudes shown only (sign indicates generally "up" or "down")
                  at the equator        at the pole
              h = 0     h = -4500 m        h = 0    
GM          -3.71317     -3.72303        -3.75729
J2          -0.01092     -0.01088        +0.02236
centri      +0.01706     +0.01697         0.0

vector sum  -3.70703     -3.71683        -3.73493

Do Curiosity's reported measurements of Mars' surface gravity (~3.717 m/s^2) include centrifugal effects?

Yes they do! At 5.2 degrees south latitude and -4500 meters altitude (bottom of Gale Crater) the acceleration is -3.7168 m/s taking into account GM, J2, and centrifugal effects.

So as Cheap Trick tells us and Meatloaf reiterated more elegantly in Roadie:

Everything works if you let it!

Here's some Python for double-checking my math:

def accelerations(rr):

    x,   y,   z   = rr
    xsq, ysq, zsq = rr**2

    rsq   = (rr**2).sum()
    rabs  = np.sqrt(rsq)
    nr    = rr / rabs
    rxy   = np.sqrt(xsq + ysq)
    rrxy  = rr * np.array([1.0, 1.0, 0.0])
    nxy   = rrxy/rxy
    rm3   = rsq**-1.5
    rm7   = rsq**-3.5

    acc0  = -GM_mars * rr * rm3

    # https://en.wikipedia.org/wiki/Geopotential_model#The_deviations_of_Earth.27s_gravitational_field_from_that_of_a_homogeneous_sphere
    acc2x = x * rm7 * (6*zsq - 1.5*(xsq + ysq))
    acc2y = y * rm7 * (6*zsq - 1.5*(xsq + ysq))
    acc2z = z * rm7 * (3*zsq - 4.5*(xsq + ysq))

    acc2  = J2_mars * np.hstack((acc2x, acc2y, acc2z))

    accc = nxy * omega**2 * rxy

    return acc0, acc2, accc

import numpy as np

halfpi, pi, twopi = [f*np.pi for f in [0.5, 1, 2]]
degs, rads        = 180./pi, pi/180.

R_mars  = 3396200.0
GM_mars = 4.282837E+13   #  m^3/s^2  https://en.wikipedia.org/wiki/Standard_gravitational_parameter
J2_mars = GM_mars * R_mars**2 * 1960.45E-06     # https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

Req = 3396.2 * 1000. # meters https://en.wikipedia.org/wiki/Mars
R   = Req - 4500.    # https://en.wikipedia.org/wiki/Gale_(crater)
Rpo = 3376.2 * 1000. # meters https://en.wikipedia.org/wiki/Mars

sidereal_day  = 1.025957 # https://en.wikipedia.org/wiki/Mars
T             = sidereal_day * 24 * 3600.
omega         = twopi/T
print "omega:   ", omega
print ''

aaacs = accelerations(np.array([Req, 0, 0]))
print "Req: ", Req
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())

print ''
lat = rads * -5.4
aaacs = accelerations(np.array([R*np.cos(lat), 0, R*np.sin(lat)]))
print "R: ", R
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())

print ''

aaacs = accelerations(np.array([0.0001, 0, Rpo]))  # avoid divide-by-zero (lazy!)
print "Rpo: ", Rpo
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())
  • $\begingroup$ I was going to leave @DavidHammen's answer as accepted, but since someone drive-by, silent, anonymous down voter decided to down vote this, I decided to counter that by accepting this in order to make it absolutely clear it answers the OP's (my) question. $\endgroup$ – uhoh Apr 16 at 1:18

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