Earth Average Surface Gravity

Happy new year, everybody 🙂

It would be interesting to know the average surface gravity of Earth. I looked over at Wikipedia: Standard Gravity and found:

The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2

If we look at US Standard Atmosphere, we find the same thing. So what’s the problem?

Both sources tell us:

The value of ɡ0 defined above is a nominal midrange value on Earth, originally based on the acceleration of a body in free fall at sea level at a geodetic latitude of 45°.

Nominal midrange? What is that? That’s not an average for the whole Earth. We will have to figure it out by other means.

A National Geospatial Intelligence Agency document tells us how to do it. Appendix B gives us all the math we need. We will however use the latest values for a and b derived from here. Create a new file called with:

# Zoe Phin 2019/12/15

echo '  function __(s, p, n) { printf "%-3s\t%25.*f\n", s, p, n } 
BEGIN { π = atan2(0,-1)

        # Given

    GM = 3.9860046055e14                    # Earth Mass * G
    a  = 6378137.678                        # Semi-major axis
    b  = 6356752.964                        # Semi-minor axis
    ω  = 7.292115e-5                        # Angular Velocity
    G  = 6.67428e-11                        # Gravity "Constant"
    Mₐ = 5.148e18                           # Mass of the Atmosphere    

        # Derived

    Mₑ  = GM / G                            # Mass of the Earth (with Atmosphere)
    Mₒ  = Mₑ − Mₐ                           # Mass of the Earth (w/o Atmosphere)
    GMₒ = G × Mₒ                            # Atmosphere-less Constant

    F  = (a − b) / b                        # Flattening Factor
    f  = 1 / F                              # Inverse Flattening
    e  = √(1 − b²/a²)                       # 1st Eccentricity
    E  = √(a²/b² − 1)                       # 2nd Eccentricity

    Rₚ = a²/b                               # Polar Radius of Curvature
    R₁ = a × (1 − F/3)                      # Mean Radius of the Three Semi-Axes
    R₂ = Rₚ × (1 − (⅔)E² +                 \# Radius of a Sphere of Equal Area 
        (26/45)E⁴ −                         \
        (100/189)E⁶ +                        \
        (7034/14175)E⁸ −                      \
        (220652/467775)E¹⁰ )

    R₃ = ³√(a² × b)                         # Radius of a Sphere of Equal Volume

    t  = tan⁻¹(E)
    q  = ½ × ( (t + 3×t/E²) − 3/E )
    Q  = 3 × ( (1 + 1/E²) × (1 − t/E) )-1

    __( "GM" , 00 , GM )
    GM  = GMₒ               # Warning ! Switching to Atmosphere-less Constant

    m  = (ω² × a ² × b) / GM                # Normal Gravity Formula Constant
    J₂ = (e²/3) × (1−(2×m×E)/(15×q))        # Dynamical  Form  Factor
    U₀ = t*GM/(a×e) + ⅓×ω²×a²               # Normal Gravity Potential
    Gₑ = GM/(a×b) × (1 − m − (m×E×Q)/(6×q)) # Normal Gravity at the Equator
    Gₚ = GM/a² × (1 + m × (E×Q)/(3×q))      # Normal Gravity at the Poles
    k  = (b × Gₚ)/(a × Gₑ) − 1

    Gₘ = Gₑ × (1 + (⅙)e² + (⅓)k +          \# Mean Value of Normal Gravity
        (59/360)e⁴ + (5/18)ke² +            \
        (2371/15120)e⁶ + (259/1080)ke⁴ +     \
        (270229/1814400)e⁸ + (9623/45360)ke⁶) \

    __( "a"  , 03 , a  )
    __( "b"  , 03 , b  )
    __( "ω"  , 11 , ω  )
    __( "G"  , 16 , G  )
    __( "Mₑ" , 00 , Mₑ )
    __( "Mₐ" , 00 , Mₐ )
    __( "Mₒ" , 00 , Mₒ )
    __( "GMₒ", 03 , GMₒ)
    __( "F"  , 16 , F  )
    __( "f"  , 09 , f  )
    __( "e"  , 15 , e  )
    __( "E"  , 15 , E  )
    __( "Rₚ" , 04 , Rₚ )
    __( "R₁" , 04 , R₁ )
    __( "R₂" , 04 , R₂ )
    __( "R₃" , 04 , R₃ )
    __( "q"  , 12 , q  )
    __( "Q"  , 12 , Q  )
    __( "m"  , 14 , m  )
    __( "J₂" , 15 , J₂ )
    __( "U₀" , 04 , U₀ )
    __( "Gₑ" , 10 , Gₑ )
    __( "Gₚ" , 10 , Gₚ )
    __( "k"  , 15 , k  )
    __( "Gₘ" , 10 , Gₘ )
}'| sed 's/#.*//' | sed '/__/s/,/,\\\n/' | sed -r '
        /__/! { y,πωₚₑₐₒₘ₀₁₂₃₄×−,pWpeaom01234*-,;
        s,³√(.*),(\1)^(1/3),g; s,tan⁻¹\(E\),atan2(E\,1),g;
        s,½,1/2,g; s,⅓,1/3,g; s,⅙,1/6,g; s,⅔,2/3,g; s,²,^2,g; 
        s,⁴,^4,g;  s,⁶,^6,g;  s,⁸,^8,g; s,¹⁰,^10,g; s,√,sqrt,g;
        s,\)ke,\)*k*e,g; s,\)([Eke]),\)*\1,g }' | awk -f -

The program is written in unicode to make reading it and following Appendix B much easier. Sorry if your browser doesn’t render the code properly. I just noticed it doesn’t on my smartphone.

Make this file executable and run it:

> chmod +x
> ./

GM 	          398600460550000
a  	              6378137.678
b  	              6356752.964
ω 	            0.00007292115
G  	       0.0000000000667428
Mₑ	5972186671071637111046144
Mₐ	      5148000000000000000
Mₒ	5972181523071637240414208
GMₒ	      398600116958065.625
F  	       0.0033640939204509
f  	            297.256861326
e  	        0.081819240444396
E  	        0.082094488053751
Rₚ	             6399594.3322
R₁	             6370985.4599
R₂	             6371007.8495
R₃	             6371001.4586
q  	           0.000073346392
Q  	           0.002688044572
m  	         0.00344979040432
J₂	        0.001082631230267
U₀	            62636794.1768
Gₑ	             9.7803152684
Gₚ	             9.8321748731
k  	        0.001931854280022
Gₘ	             9.7976331557

Gₘ (mean normal gravity) is the value we’re looking for, and it’s 9.7976331557.

Now if only the actual earth was a model, then we’d already be done. We want the most accurate answer though, and so we need measured data. We will use the EGM2008 standard, as it’s the most popular. We’ll be using this great resource to seek our answer. But first we’ll need to generate a table of latitudes and longitudes:

> for x in `seq -89.5 89.5`; do for y in `seq -179.5 179.5`; do echo $x $y; done; done > latlon.csv

We will upload this file to get the results. Now go there and fill out the form exactly as shown here (note Gm is our atmosphere-less value GMₒ, not GM)

This will take a while …

But Zoe, Why are we interested in Normal Gravity and not Gravity? Because we’re interested in the acceleration towards the surface and not the center of the earth.

To find the average for the whole Earth we will have to go through each latitude,longitude pair and find out their surface area using the formula from here:

dA = a2 cos[ø (1-e2)] dø dl / (1-e2 sin2ø)2

This will generate an error of ~0.00001281 when we add up all the pairs, but that’s alright because we’ll just divide by a surface area 1.00001281 times as large.

Click on “Download results” when the site is finished, and save to EGM2008.csv in the same folder as the following code.

# Zoe Phin 2019/12/17

sed 1,32d EGM2008.csv | awk '
    function r(x) { return x*atan2(0,-1)/180 } {
    e = 0.081819240444396; printf "%.12f %.12f\n", $5/1e5, \
}' | awk '
    { S += $1*$2; } END { printf "%.10f\n", S/510072261022076.375 }'

Run it

> bash


The best estimate for Earth’s average surface gravity using the most popular reference model is: 9.7977074331

Note how little difference there is between this number and the previous mathematically derived value; It’s only 0.0000742774.

No one should use the “Standard Gravity” value of 9.80655 if they are interested in the whole earth. Such use will lead to a geometric error of ~0.0009025 and an arithmetic error of ~0.0088426.

Now these errors might be small, but why not make them smaller? Use my number. Thank you.

Enjoy 🙂 -Zoe

Published by Zoe Phin

33 thoughts on “Earth Average Surface Gravity

  1. Dear Zoe,

    I am not a scientist or even a Beta person.
    So, trying to understand scientific data is hard for me.
    You seem to me as a very clever person because you explain things in a simpler way than others do.
    (not all of it of course because it is just out of my league)
    Would you help me out please.
    I was watching a video the other day where a guy said that Co2 level rising follows on temperature rising. and not the other way around as IPCC wants me to believe; higher Co2 = higher temp.
    When I asked around someone came up with this explanation.
    Is this true?
    Here it is:
    Both claims are true he says.
    “co2 causes the temperature to rise and a higher temperature causes co2 levels to rise because the solubility of the co2 in the oceans declines.
    The oceans warm up slower than the atmosphere because water has a larger warmth capacity than air so there there is a difference in response time between the 2 processes.”

    When I read your stuff on Principia, you seem to tell me that co2 isn’t able to trap the earths back radiation.
    So co2 can’t warm up the atmosphere.
    Do I understand this correctly?
    Thank you so much Zoe for answering.

    Best Regards

    Liked by 1 person

    1. Hi Tim
      My claim is that CO2 can not warm the SURFACE. CO2 can warm up the atmosphere because it has lower heat capacity than O2/N2. At 2 meters above the surface, we’re talking about a heat up of less than 0.02C if CO2 is doubled from 400 to 800ppm. This assumes no biological or other feedbacks. Nothing to worry about, really.

      Thanks for the comment, Tim.


      1. “CO2 can warm up the atmosphere because it has lower heat capacity than O2/N2. ”
        If had 50% CO2 at 1 atm pressure how does warm up {better}.
        For instance, there is no temperature difference ocean surface and ocean air temperature, but with land surface there is a difference. Would be less difference with land surface if had 50% CO2.
        Or if one has 50 C air temperature and in sunlight, the ground probably close to 70 C.
        Or if you have cooler air, say 25 C, then ground surface in sunlight could heat up to say +50 C
        So with more CO2, would there be less difference in the 30% of area of Earth which is land surface?


        1. “no temperature difference ocean surface and ocean air temperature”

          Ocean air temperature? I take you mean the air above the ocean. Lapse rate still exists over the ocean. It’s not as steep as that over land.


        2. I meant;
          no temperature difference ocean surface temperature and ocean surface air temperature.

          Whereas there is difference with ground temperature {if not wet} and surface air temperature.
          Or dry ground can get much hotter than surface air above it, and question was, do you think more CO2 lessens
          this difference {or makes surface air hotter}.


  2. Earth Average Gravity – just a reminder that the measured acceleration is for everyday matter or ordinary matter which is familiar to us. It’s possible that non-normal matter exists (ie. dark matter) which would have a different acceleration when it falls due to Earth’s gravitational field.

    I personally don’t believe in Einstein’s gravity model and have the imagery of radiated gravitons as the force-carrier of attraction. This concept allows strong gravitational interaction to occur between non-normal matter.

    (Sorry to harp on about it, but..) I’m 💯% convinced that non-normal matter exists at the centre of the Earth. It’s far out, I know, but makes logical sense. (At least to me anyway)


      1. “Would you say that the mass of the Earth and the Moon’s is unknown?”
        It seems he saying there is unknown matter deep within earth which gravity effect. Which roughly speaking is what “everyone thinks”- everyone “knows” there is denser material deep within Earth.
        Or normal rocks have fairly low density as there comprised of +40% oxygen.
        So it’s assume Earth has core of Iron and nickel which not oxidized or less oxided AND due to extreme pressure these denser {more mass in volume of area].
        Dark matter is unknown matter or also something which not like matter we know about but same gravity or more gravity than the matter which we know,
        There is really not much difference, other Alan seems to think dark matter is more exotic/different than the matter we familiar with- which could include being invisible/transparent or different in other ways is hard to detect, other than it gravitation effect.
        Or according modeling we are “missing” something like 97% of the mass in this galaxy {and the universe} and that missing gravitational mass is called “dark matter”.

        I don’t have a lot faith in modeling skills, but I agree there is missing mass. But I also assume we generally don’t know jack about our galaxy {or universe}.

        Liked by 1 person

        1. Thank you for that summary. Yes, I propose that Earth’s core can be non-normal matter, which wasn’t considered by Newton or his peers. Einstein continued the theme that all matter is the same.

          The crucial difference that I make is that the non-normal matter core can have a strong gravitational interaction with the core of the Moon. I further propose that this core-interaction is especially strong when on the same plane. This would account for the extra strong Spring tides when the Moon is on the equator.

          The standard explanation is that the Spring tides occur when the Moon is aligned with the Sun. Yet during an eclipse, the tide is not yet fully in.

          The shape of the solar system and the galaxies we observe all tend towards a plane, as if there were an especially strong attraction bringing them there.


  3. Jarle – the crust is where scientists pick up the rocks and drop them to measure their acceleration. It’s where the Cavendish experiment is performed which measures how weak their gravitational interaction is.

    How about “non-ordinary cores allows a strong planar gravitational interaction at long distances for large bodies, equivalent to Modified Gravity theories of Einstein’s General Relativity”


      1. Jarle, thank you for asking. Yes, the hypothesis predicts that earth-tides are increasing slowly yet exponentially
        This can be tested by using satellites in a careful study.
        The other prediction the hypothesis makes is that geothermal heating of the crust is increasing slowly yet exponentially. This can be tested by boreholes 1km to 5km deep and directly measuring the temperature of the rock in real time.

        Both tests are exceedingly easy and inexpensive yet with the potential for humanity altering consequences.


    1. With respect to remaining an overall constant as per the mainstream science narrative based on Sir Isaac Newton’s foundation and Einstein’s Special Relativity.

      Mainstream science doesn’t allow an extra strong gravitational pull on the Earth’s core to account for climate change via increasing tidal strength.


  4. Are you suggesting that the tidal height is increasing monotonously with time? Exponentially even? Seems to imply that there were no tides in the not so distant past?


    1. I’m talking about the solid body tides of the Earth
      There’s plenty of articles which report ocean water tides are increasing in strength. The mainstream put the cause down to atmospheric warming which is increasing sea level. I’m saying that solid Earth bulge increase is pushing the oceans from beneath, thereby increasing ocean tidal strength. I believe increase in gravity across the Earth is being misinterpreted as sea level rise.


      1. Our moon is moving further away from Earth over over period 1 million year a fair amount of distance away from Earth. Shouldn’t it be decreasing rather than increasing?


      2. If I was standing on my bathroom scale and then swallowed a chunk of dark matter, I would expect a linear step response in my weight, either proportional or non-proportional if dark matter acts differently. I would not expect that there would be a time dependent effect. Can you please explain better what is at play that causes time variability?


        1. I’m not sure exactly what the mean. The variability comes in depending on where you are relative to the equatorial regions.

          If you weighed yourself whilst holding dark matter near the equator you would weigh considerably more than if you weighed yourself with the same amount in North America.


        2. “If you weighed yourself whilst holding dark matter near the equator you would weigh considerably more than if you weighed yourself with the same amount in North America.”

          In terms of “general understanding” what you weigh depends how close you are to mass of Earth. And that you “at rest” rather than being in free fall- or you can change your weigh by bouncing on a trampoline.
          And at rest at equator, you weight tiny bit less:
          “Since there is more centrifugal force at the equator to cancel gravity, your overall weight at the equator versus at the poles is even less. The centrifugal force on your body at the equator is 0.034 m/s2 times the mass of your body. The centrifugal force at the poles is zero.”

          When you are in earth orbit, one can say roughly that you are in zero gravity, but the technically correct term is microgravity. Or there are small gravity or gravity like effect in the free fall of orbit.
          Likewise one say when on Earth surface and at rest, there are likewise mircogravity effects from every body in the solar system, and in the galaxy and rest of the universe.
          But the big gravity effect is from Earth. And next bigger gravity effect is from our Moon, next is sun, and next is whatever body is close to Earth. Or if Jupiter is near, it’s Jupiter. Or when Venus or Mars is near Earth.
          But getting back being on Earth, you have centrifugal force, “canceling” gravity in same sense as jumping cancels gravity, and you also gravitation anomalies which become a consideration when talking about “microgravity effects” and one could someone standing close to you, or effects of light breeze.
          Anyways getting back to:
          “If you weighed yourself whilst holding dark matter near the equator you would weigh considerably more than if you weighed yourself with the same amount in North America.”

          Made me think, that to be true, dark matter gravitational effect “must be” somehow immune to velocity effects.
          Or if had chunk of dark matter and I jumped, my jump would produce less “zero gravity effects”. Or jumping with the dark matter “rock” and the rock not “joining me” in my freefall.
          Or different way to look at it, dark matter less or no inertia. Now the denser anything is, it has more gravity than it has inertia- shrink earth to size of pea- same inertia, far more gravitational effect {if you are are near it]

          So, have baseball size object, which without dark matter in it, weighs a lot less than normal baseball, add tiny amount of dark matter, and weighs 10 lb. And this baseball has less inertia than normal baseball, though weighs a lot more. If throw a shot put ball weighing 10 lb, it’s not going to go far, but my 10 lb baseball, will go a considerable less distance- say 1/10th of distance. And it should weigh more on equator than something else weighing 10 lbs.
          Or in terms sfi, this is what one could loosely call inertia-less drive. or it can make star travel easier.
          Though our problem is leaving Earth, and I don’t see how it helps in that regard.


  5. gbaike – when I speak, I’m talking about a world-changing new insight which I present in the form of a hypothesis that is counter to the concept of “mass” laid down 300 years ago.

    Your mind is unable to comprehend something so profound. I don’t wish to be rude but the concepts I’m trying to convey are very deep and philosophical.

    The only person on the planet who may be able to comprehend something that happens only once in a 2000-year event is Noam Chomsky. There’s yet a breath of hope left for a major positive change out of this current crisis of global humanity.


        1. Others might view that as a tad bit of megamania.

          But I attempt to go in direction of glass 1/2 full, and it seems to me, that you think Einstein achieved a lot.

          And seems to me that Einstein did achieve a lot.
          Merely that he predicted the starlight would be bent by the gravity of sun, alone, was fairly impressive.
          Though of course it was more than that.

          Anyhow, a fair number of people will dismiss Einstein’s theories, and seems that you do not.


        2. Stars bending light has not been observed outside of stars’ plasma atmosphere. It’s not a gravitational thing. Einstein was a mathematician and metaphysical thinker at best. We’re supposed to praise him, but I doubt it’s justified.


        3. Both links just state the theory.

          “Starlight that passes NEAR the edge of the Sun on its way to Earth is deflected.”

          Yes, plasma will deflect light.

          Google “dowdye light”. Dowdye makes a good case, and definitely qualified.


        4. –Yes, plasma will deflect light.

          Google “dowdye light”. Dowdye makes a good case, and definitely qualified.–

          Here it’s got diagram and labeled with “plasma rim” which is unfamiliar term for me, but imagine Ed Dowdye
          is referring to the Sun’s chromosphere which thin layer of plasma around the sun- “appears as a rim of red light during a solar eclipse”. And diagram claimed no one measured it at 2 radius from the Sun.

          And apparently: The diameter of the sun is 0.5 degrees which is 30 arcminutes or 1800 arcseconds.
          So it seems 1 degree from middle of sun is 4 radius from the sun.

          “The deflection caused by the Sun’s gravity decreases as the distance from the Sun’s center, mathematically as 1.75 arcseconds divided by the distance in solar radii. I used calibration images taken during totality to help in the analysis. For these, the telescope was pointed about 7° from the sun, where the gravitational deflection was very small.
          This is something that many of the previous eclipse expeditions had attempted, but kept having problems with. The computerized mount with the automatic scripting worked perfectly, giving me the calibration images that I needed (and that 20th-century astronomers had always wished for).”

          I tend to think astronomers are well qualified.
          I am interested in space, but I am not an astronomer.
          I had fair amount trouble just finding comet neowise.


  6. Physicist Predicts How The Universe Might End: With Explosions Like None in Existence
    “At the end of the Universe, long after the last shining stars flicker out, there might be one final set of explosions. Called black dwarf supernovae, these dazzling blasts will herald in the everlasting darkness as the Universe sinks into dormancy, a new study suggests.”
    Linked from

    I guess people don’t think the universe going to collapse- which does seem rather unlikely to me.
    I have tended to think this universe is inside of black hole.
    And had not got to considering what happens if this black hole evaporates.
    So anyhow, these persons imagine our universe gets cold and lonely. Well actually colder and more lonely as most of our universe at moment appears to be a lot empty space and darkness.
    But space isn’t actually cold.
    But anyhow the vastness and lack of something, in the end of universe could have, if you wait long enough, some fireworks near the end.
    Maybe that is somewhere around the time our blackhole evaporates.


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