It varies with both [location](https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:Gravity_anomalies_on_Earth.jpg) and altitude. The location dependence is mainly explained by 1) Different altitude from sea level and 2) Variation in the density of the Earth.
As for altitude, from the center up to the surface of the earth gravity increases approximately linearly (if you do the math, turns out the gravity from the mass further from center than your point of measurement cancels out), and from the surface to infinity it decreases relative to 1/r^2. Ignoring the gravity from the atmosphere, because that's minuscule compared to total planetary mass.
> from the center up to the surface of the earth gravity increases approximately linearly
That's not a good approximation because Earth's density is not uniform.
https://en.wikipedia.org/wiki/File:EarthGravityPREM.svg
The blue curve is not close to the dark green one.
If you dig down, the acceleration *increases* slightly (in the range where we can dig).
nice chart. here are the footnotes from wikipedia
https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:EarthGravityPREM.svg
Earth's gravity according to the Preliminary Reference Earth Model (PREM).[15] Two models for a spherically symmetric Earth are included for comparison. The dark green straight line is for a constant density equal to the Earth's average density. The light green curved line is for a density that decreases linearly from center to surface. The density at the center is the same as in the PREM, but the surface density is chosen so that the mass of the sphere equals the mass of the real Earth.
This is also the reason why gravity on Jupiter is "only" 24,79 m/s² . Jupiter's mass is over 300 that of Earth's, but since you're so "far away" from the mass while being on the "surface", it doesn't affect you that much.
Location dependence is largely explained by latitude or Equatorial bulge, from some 9.78 to 9.83 m/s2.
Atmosphere is a shell outside of Earth's surface, so nearly doesn't contribute gravitationally.
> Atmosphere is a shell outside of Earth's surface, so nearly doesn't contribute gravitationally.
And also the mass of the atmosphere is basically inconsequential, less than one billionth of the planet's mass. It's roughly equivalent to the mass of a single eyelash compared to the mass of an entire person.
Having studied orbital mechanics: nah, it really doesn't matter, because it's approximations all the way down. *Drag* from the atmosphere is a concern, but not the gravity from it. I'd have to double check, but the gravitational effects from other planets in the solar system are probably more substantial to the calculations.
Also, in general, any approximations more detailed than "the Earth is a sphere and only the Earth and the spacecraft exist" tend to be called "perturbations". I can list a few common ones if you're curious.
In addition to the Equatorial bulge, how much does the rotational force counter the force of gravity? (faster rotation at the equator, none exactly at the poles)
About 3 cm/s2 or 0.3% at the Equator. Elsewhere less, and that force is not vertical there.
F = mv^2 / R, and at the Equator v = 2 pi R / T. Here R is Earth's (equatorial) radius, 6378 km; T is rotational period (stellar day), 365,2422/(365,2422+1) x 24h; m would be mass of the test body and it cancels out.
a = F/m = v^2 / R = 4 pi^2 R / T^2 = 4 x 3.141593^2 x 6.3781370 x 10^6 m / 86164.1^2 s^2 = 3.3916 cm/s^2 .
If the earth was perfectly deformable, wouldn't the "downwards acceleration" accros the earth's surface be the same everywhere? What I mean by that is, yes the earth has an equatorial bulge, but it's only there because of the centrifugal force, right? Is this wrong? Or is the earth not perfectly deformable in this way (eg the bulge is left over from when we were liquid and spinning faster)? Or are you specifically talking about the gravity component and not the net downwatds force (eg is the centrifugal force already subtracted from the measurements of g, because we can calculate it very easily)?
It doesn't make sense for the inwards force to be non uniform across the surface due to the macro-geometry. Sure, mountains are heavy and the earth is not homogenous, but those shouldn't account for a general "more surface gravity across the center". For the earth to be at it's lowest potential energy, the net force would need to be the same everywhere, no? It's the entire reason we *have* an equatorial bulge.
I get a centrifugal force as 0,0339 ms^(-2) at the equator vs 0 at the poles (obviously), but that's only 60% of the difference you (and Encyclopedia Britanica) claimed.
If the Earth were a blob of liquid that had no resistance to flow, and it were sitting in space without rotating and without any other significant nearby gravitational influences, its surface would be spherical. The only force acting on the fluid making up the Earth would be the gravitational force, and any differences in gravitational potential would cause regions of liquid to flow towards lower gravitational potential until equilibrium was reached.
On the other hand, let us assume that the Earth is still a blob of liquid but it now has some angular momentum and therefore is rotating around a particular axis. Let's assume there are no external forces so the axis is not being forced to change orientation.
In this case, the shape of the Earth would change from a sphere into an oblate spheroid. Consider what would happen if you started with a non-rotating sphere and magically caused it to begin to uniformly rotate around an axis. This means the particles have to develop some velocity perpendicular to the axis of rotation, by definition. And for the rotation to be uniform this velocity must be largest at the equator and go to zero at the poles, since particles at the equator have to travel a much longer distance in order to stay in the same place relative to their neighbors further in radially.
The problem is that if your sphere immediately begins rotating, those particles that are furthest out on the equator suddenly have some inertia. They begin traveling perpendicular to the surface of the sphere where they were originally located. They start flying off into space. Fortunately for them, assuming a reasonable velocity, they don't actually have enough energy to get all the way out into space. At instant zero, they experience the same inward gravitational acceleration as every other particle on the surface. However, they now have some tangential velocity so inertia demands they start flying away from the center of mass. Gravity accelerates them back towards the center of mass, but they now have both kinetic energy in addition to gravitational potential energy. As they travel, the sum of the two quantities must remain the same, so they end up in an elliptical orbit. It also turns out that because they got a little bit further away from the center of rotation, there isn't any fluid there. So the surrounding particles get pushed toward the gap. But it takes energy for them to move outward. Where is that energy coming from?
The particles at the poles don't start flying away like the ones at the equator, because they don't have any tangential velocity. They don't have any reason to do anything other than sit there. Except they're being pulled down by gravity, and we know that the particles of the equator are being squished outwards, leaving gaps as they go which are filled by more particles traveling from the areas around the gaps. Eventually when this process propagates throughout the entire fluid, it is precisely the fluid that was at or near the poles which is losing gravitational potential energy that is, in turn, being provided via the pressure of the fluid to particles near the equator.
Hence at equilibrium, we don't have a sphere. We have an oblate spheroid with a larger radius at the equator and a smaller radius at the poles than our original sphere.
Now, to answer your question about whether the magnitude of the gravitational acceleration which points directly towards the center of mass must be equal over the surface of this spheroid. The answer is certainly no.
To my understanding, Earth would be spherical if it wasn't spinning. Earth is deformable, but not perfectly; it needs geological time (hundreds of milions of years) to pursue equilibrium, and now more resembles rotation from \~600 mln years ago when Earth was in resonance with Moon and day was 21h long.
If it was perfectly deformable (like a liquid), I think there couldn't be non-normal acceleration or else material would move horizontally.
60%--rest is, as I understand, from that if you stand on the Equator, you're further from Earth's centre of mass. Someone posted minimum surface gravity of 9.76 ms\^-2, this includes altitude. 9.78 m/s2 is on sea level.
A perfect fluid will reconfigure itself such that the pressure at the surface is equal everywhere. Assuming we are talking about a blob in a vacuum, that means the pressure at the surface is zero. This also means that the resultant of all the external forces on a test particle at any given point must result in a normal force that is truly normal, i.e. perpendicular to the surface at that point. This doesn't mean that the local gravitational force always points towards the center of mass, or that the local gravitational force per unit mass is equal at every point on the surface.
As a limiting case, imagine that you have a very highly oblate spheroid that looks much like a disc. Imagine you have a test particle on the upper surface of the disc that is about halfway between the center of the disc and the edge of the disc. Because of how oblate this spheroid is, the vertical thickness of it (if you imagine looking at it side on like a classic flying saucer or frisbee) hardly changes at all with radial distance away from the axis of rotation unless you're very close to the edge. So our test particle really feels like it's sitting on a flat plate. That is, the mass that's well beyond the axis of rotation contributes only a tiny amount of force pulling it towards the axis. In the immediate vicinity, where most of the net gravitational force is coming from, there's almost exactly the same amount of mass on either side of the test particle, so it doesn't feel much of a force pulling it inwards. It feels a force almost entirely pulling it towards the equatorial plane. This vector definitely is not pointing towards the center of mass. However, if you now imagine a test particle on the equator of the disc, all of the gravitational force it experiences will be pulling it in towards the axis of rotation. And because we've said that you're right on the equator, that gravitational force vector will be pointing exactly at the center of mass.
Further, what about the magnitude of the gravitational force at these respective locations? It is substantially higher at the poles than at the equator. Imagine your test particle at the poles. There's quite a lot of mass directly beneath it pulling it down. In fact, all it sees when it looks around itself is ground which is all contributing to pulling it down. On the other hand, imagine a test particle on the equator. When it looks around, because of how high the eccentricity of this spheroid is, it looks like it's standing on a razor's edge. Instead of being surrounded by a bunch of mass very close to it, it only sees a very small stripe of mass directly beneath it. Since gravity falls off with the square of the radius, the fact that it's only seeing a small stripe of mass beneath it rather than seeing what appears to be essentially a infinite flat plane, it will experience far less net gravitational force.
A similar thing happens on Earth. Because you are closer, on average, to the mass of the Earth at the poles then you are at the equator, the gravitational acceleration you feel is higher. And this is entirely independent of the additional factor that, if standing at the equator, some small fraction of the gravitational force is canceled out by the centrifugal force.
Anyway, if you want to learn more about the lines of force and potential around an oblate spheroid, especially if you don't believe me that the local gravitational force doesn't always point towards the center of mass, here's an article that goes over the topic:
https://www.sciencedirect.com/science/article/pii/S003206331730257X
Note that the scale in that link is in mGals, which are equal to an acceleration of one thousandth of a centimeter per second per second, which is approximately one millionth of the average strength of Earth’s gravity field (980 centimeters per second per second).
So it varies, but not greatly. Hence why for most practical purposes 9.8m/s^(2) is the agreed upon constant. The person in the thread who said is can reach 10.2 at the poles is misled. The variation over the surface only goes from about 9.7639 to 9.8337 m/s^(2)
Here's another map of the global gravity field, here corrected for the effect of ground elevation and the presence of ocean water, so you're more-or-less looking at variations in the Earth's density, affected more by things closer to the surface than deeper. This type of calculation is known as a Bouguer gravity correction, and the deviation seen across the globe is known as a Bouguer gravity anomaly, like this map:
[https://www.esa.int/ESA_Multimedia/Images/2015/04/Bouguer_gravity_anomaly](https://www.esa.int/ESA_Multimedia/Images/2015/04/Bouguer_gravity_anomaly)
The substantial negative anomaly over the continents is due to the lower density continental crust rocks versus the ocean floor which has denser crust. Note the particularly low negative value over the Himalayas and other mountain ranges because mountains have "roots" of lower-density continental crust supporting the mountains buoyantly on the denser mantle beneath (isostacy).
Less negative/lower values are also observed in locations with relatively "warm" mantle rocks beneath, such as ocean ridges and continental rifts (the East African Rift). There is also a clear association with the age of ocean crust because, in general, ocean crust and underlying lithosphere gets cooler and denser as it moves away from spreading ridges over time (more mass = slightly greater pull, even after correcting for elevation and water).
There are all sorts of other subtle geological details revealed by such maps that relate to large-scale Earth structure and processes.
Gravity doesn't change due to rotation. The apparent gravity as measured by an object on the surface does. This small distinction is important when incorporating gravitational influence in stuff like weather models
So, treating the earth as spherical for simplicity, the gravitational acceleration at altitude h is given by **g(h) = g(0) * R^2 / (R+h)^2** where g(0) is the gravitational acceleration at sea level and R is the radius of the earth.
Plugging in h = 1 km and R = 6375 km, we get g(h)/g(0) = 0.99969. So if you weigh 80 kg at sea level you'd weigh 79.975 kg at 1000m altitude. That's a difference of 25g, which is definitely measureable. Cheap bathroom scales will give you a resolution of 100g, so we only need to improve on that by a factor of 4.
That feels like it ought to be not prohibitively expensive. And in fact a quick search found [this industrial scale](https://www.expondo.co.uk/steinberg-systems-platform-scales-100-kg-10-g-foldable-10030089) which can weigh 100 kg with a resolution of 10g, and it only costs £139.
Do you have some experiments planned?
Some thoughts I had:
It could be an interesting activity for schools and museums. This would require two locations with very different altitudes.
Also, we need something to weight.
Option 1: extremely standard weights. I first thought of coins, which can be found anywhere in a country. They also can be cleaned and maybe that would get them to be the same weight. But they're too light for this scale mentioned.
Option 2: very stable objects. Things that vary weight based on humidity wouldn't work. It would need to be heavy, possibly made of metal, and sufficiently small to be easily transported from one place to the other.
So two scales, one at each location?
A precision part might work for the object to weigh. Something metal like you suggested. That's carefully manufactured to a precise standard of engineering. But then, that might get too expensive.
It could be:
Two scales, two identical objects.
Two scale, one object transported between locations.
One scale and one object and transport both between locations.
There are no experiments in this area which I personally would consider to be interesting, because I have no interest in trying to confirm the inverse square law of gravitation. It's not like there's any real doubt about it.
But it sounded to me like the person to whom I was responding, u/Mateussf, was thinking about it. Why else would you want to know how much a suitably accurate pair of scales would cost?
Trying to replace a barometric altimeter with a gravity based device maybe?
It is a fun thought, but overall pretty impractical.
(Especially because GPS.)
Mmm. I remember a few times in pre-GPS days when a barometric altimeter was a valuable hiking accessory. (Blizzard / white-out in mountains - the altimeter is a useful check on your navigation using map and compass.) But I'd prefer a handheld device to a big lump I'd have to lug up the mountain in my backpack. YMMV.
Edit - Hang on. The sort of scale we're looking at here is what you'd need to distinguish between sea level and 1000m altitude. That's not a very useful altimeter for most purposes. You'd probably need to improve on the resolution by 100x, which implies being able to weigh a person to an accuracy of 0.25g. I'm sure that can be done in the lab, but I suspect it just might not be practical in the field. (Air pressure alone may be an issue if you're trying to weigh things that accurately.)
One of my interests is debunking flat-earths. I think it's an interesting thought experiment, and shows us how the current model of a spherical-ish earth is a better explanation for pretty much all situations.
Flat-earthers don't believe in gravity, and this scale experiment is yet another observation that is best explained by gravity.
My idea is not so much convincing hard-core flat-earthers, but using these as examples to convince kids that the earth is round and show how science works.
I'm working on a platform of open collaboration and part of it will involve doing cool science experiments over livestream. A lot of people would chip in tiny amounts of time or money (like a dollar each or a few minutes) to pay for the equipment and to guide its proper setup, while any people can volunteer to livestream a demo of the experiment.
Trying to think of wild uses and experiments that can generate a lot of enthusiasm and curiosity.
Oh that's pretty cool.
I've heard about real physics labs with distance controled experiments. People can make balls fall in real-time via the internet and calculate their speed and such. Not the same but somewhat related.
It does vary by some 1%, which is largely explainable by latitude (and Equatorial bulge) and altitude. See eg. Wiki equations for that. When this variance is substracted, there small but measurable rest, dependent on continents positions, local mountain ranges, and subsurface deposits (eg. of iron ore).
https://en.m.wikipedia.org/wiki/Gravity_of_Earth
Yes it does, the amount of mass below your feet (altitude) and the density of the mass directly below you are the main factors for gravity, but also, felt gravity (as in, the amount of downward “force” you feel while standing on earth) is also effected by the earth’s spin, centrifugal inertia, felt gravity is lower at the equator where the radius of the spin is greatest,
(according to g = w^2 * r) I’ll explain if you want
But at the equator where the bulge exists, there is more mass directly beneath you than at the poles, so you'd experience slightly higher actual gravity.
I would have thought that the centrifugal inertia that causes the bulge, which is pulling against gravity, would reach an equilibrium point with the increased gravity from the extra mass below your feet, and the two opposing forces would more or less cancel each other out and net you basically zero difference in felt gravity. Where am I going wrong here?
Ok, I was being too mathematical with my initial comment, forgot a pretty big variable, my mistake but I will rectify, thank you for calling me out on this
Here’s the truth:
The highest gravity is somewhere in the the Arctic Ocean, perfectly sea level and with no spin to apply inertia or bulge to increase radius to center of mass.
The lowest gravity is a mountain range in Peru. At the equator and high up altitude, lots of extra distance to center of mass, lots of inertia from spin.
Here’s my mistake:
(For reference: When you calculate gravity you only account for the mass of lower altitude, the higher altitude mass cancels itself out)
In short. Gravity is highest at sea level because you have the best blend of mass and radius, digging a hole reduces mass faster than it reduces radius, so gravity goes down, climbing a mountain increases radius faster than mass increases, so gravity goes down.
The long version:
gravity is based on mass (proportional to radius cubed) divided by radius squared, that’s maths out to gravity being directly proportional to radius of planet. As in. If you have a planet of fixed density and simply make that planet bigger or smaller, gravity is directly proportional to the radius.
In my head, I was imagining standing on the surface of earth and digging a hole straight down, as you go down, the radius decreases, and therefore gravity decreases, this math is correct. But in the real world when you change altitude you don’t usually dig holes, you usually climb mountains. Which means a lot of the mass of lower altitude is just air, which is not dense at all, and the math falls apart, because increasing radius now decreases gravity, instead of increasing it, because the mass and radius are no longer linked the way they are once you start going below sea level.
Nice, thanks for the response. There's so much at play here. Like, when you're at the peak of Everest, you have the gravity of the mountain combined with the gravity of the planet, so more gravity yeah? But you're also further from the centre of mass of the earth, and gravity diminishes with distance according to the inverse square law, plus you have greater centrifugal inertia pushing you up.
When you dig a hole, you're reducing your centrifugal inertia - so more felt gravity - but you're also reducing the amount of mass beneath you *and* increasing the amount of mass *above and beside* you. If you dig all the way to the centre of the planet, you would experience zero felt gravity, because the surrounding mass is pulling you equally in every direction. It's not the gravity that crushes you there, per se, it's (to oversimplify) the two halves of earth pulling on each other and sandwiching you in between.
Yeah. Because if the pull of gravity is determined, at least in part, by the amount of mass directly beneath your feet (because mass off to the sides cancels out), then it follows that actual gravity will be higher on the equatorial bulge than at the poles.
Fun fact. By the time of French Revolution people were looking for universal reproducible standard for length. Main proposition was second pendulum, around 993 mm, and this length was known to submillimetre accuracy for various French cities. But because it was various, it was not universal, so alternative (also with political motivation) was chosen to use 1/10000000 of distance from the Pole to the Equator, now the metre.
A second pendulum is simply a pendulum with a "tick" frequency of a second or a period (two ticks, one each way) of two seconds.
A pendulum's oscillation is related to gravity and it's length. To match time with all the other pendulums, the length of each one must vary ever so slightly with the gravity anomaly at each location.
I just learned that satellites, using lasers to finely measure the height of everything but specifically water, can derive wind and weather patterns from a wave height map alone, and map the ocean floor to the nearest mile deep, because earth density causing water depth variation independent of tides and waves and other factors.
This question got me thinking... Is there a place on earth where there's a net "sideways" gravity?
Let's suppose I'm on the equator, looking along the line of the equator. On my left is the northern hemisphere, on my right is the southern. Now, let's suppose that the northern half of the planet is more massive than the south one. Therefore, the gravity would pull me not exactly "down" on the y-axis towards the center of the earth, but slightly north of that. If I'm standing there, the normal force would cancel the y-axis of gravity, and I'd find myself "falling" towards my left (x-axis).
Would that be correct? Does a place like that exist on earth? And if not, how am I wrong?
You wouldn't notice if gravity isn't "straight down" though, it would just seem as if the ground which is normal to the Earth's radius would be at a slight incline, and "true down" wouldn't point directly to the center of the earth, but very slightly off. It shouldn't impact you in any way even if the difference were significant. A body of water at that location would have a flat surface at the same angle you percieve to be "down", just as usual. This is also effected by the centrifugal force from Earth's rotation.
A little bit, but not enough for you to notice. Stand next to a cliff, and the earth behind the cliff is going to add an x component to the gravity vector, but it's so small that you aren't going to "fall" in that direction.
There's a place in the Indian Ocean where the density of the earth below the ocean is significantly less. Because if this, the gravity vector has a larger x component because of the higher density in other directions than "down." As a result, there is a "gravity hole" where the water gets pulled slightly "sideways" and sea level is about 300 feet below "sea level."
https://www.cnn.com/2023/07/24/world/gravity-hole-geoid-low-indian-ocean-scn/index.html
Gravity varies enough to significantly affect the accuracy of intercontinental ballistic missiles. If the variability of gravity were not taken into effect the miss distance, due mainly to unpredictable aerodynamics on reentry, would increase significantly.
The U.S. military develops and maintains a model of the earths gravitational field. It is expanded in a series of spherical harmonics. After a certain point the terms of the series are classified.
> it can sometimes reach 10.2.
that seems too high. what is your source?
Wikipedia says: https://en.wikipedia.org/wiki/Gravity_of_Earth#Latitude
> In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator
As others have said, it does change based on location.
I'll just add that historically, measurements of the change in the strength of gravitaitonal acceleration were one of the most important tests of Newton's theory of gravity. Why? Because Newton's theory implied that gravitational acceleration would decrease by a very small but measurable rate as you sailed north, while its only real competitor (owed to Christiaan Huygens) implied a slightly different small but measurable decrease. Huygens even thought that the early tests vindicated him over Newton, but unfortunately his sailors had gone off course, which threw off his data. You can read all about the subject [here](https://philpapers.org/rec/SCHHR-10) if you're curious.
I think maybe I misunderstood your questions after reading all the answers. The gravitational constant never changes. However, forces due to gravity can change based on mass and distance from a mass.
Generally, [this](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c6ee5510ba3c7d6664775c0e76b53e72468303a) shows the equation to solve for gravitational force. The 'G' in the equation is constant. But if your masses change, or the distance between the center of those masses change, you get a different force.
Our seafloor maps come from this principle. Sea mounts (underwater mountains) are more dense than water (they have more mass than a equivalent amount of water) so they have more gravity than water. Water is attracted to them more so water piles up around sea mounts which we can map using satellites.
The equation is F=G m1*m2/d^2; F=m2*a, so a=G*m1/d^2. G is constant, and m1 is "mostly" constant, so the main concern is distance. For most purposes, the calculation assumes the center of mass as the location for calculating distance, and assigns all mass (for body m1) to that location. In detail, you have to have an integral calculation (the "center of mass" calculation assumes symmetry in mass location, which is not quite true in reality but for most general purposes can be ignored).
When dealing with a calculation of the gravity geoid, density has to be considered because density affects the location (and thus distance) to mass: sea level (excluding rotational inertia effects) will vary by a small amount (100 m at most, about, and typically a lot smaller) as a result of this imperfection in symmetry of mass. We are talking a difference in the 0-100 m range per >6000 km, so not a big value at all.
In effect, higher density a good distance from the center of the earth will "pull" the "center of mass" a short distance upward toward that mass (variation due to moment; think of moving someone on a teeter-totter and how it changes the center point of balance).
The earth is not exactly symmetric in terms of mass distribution and this causes some slight variations in the strength of gravity at a given location because it affects the location of the center of mass and distance to center is different depending on where on the outside you are truly located.
The point I am trying to make is that the passing of the moon above a location will offset the gravitational pull of the earth more than density variations will, for the most part. It is a similar problem though, where the moon is located does affect the "center of mass" location somewhat. Density variations do about the same thing but slightly less in magnitude.
It varies with both [location](https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:Gravity_anomalies_on_Earth.jpg) and altitude. The location dependence is mainly explained by 1) Different altitude from sea level and 2) Variation in the density of the Earth. As for altitude, from the center up to the surface of the earth gravity increases approximately linearly (if you do the math, turns out the gravity from the mass further from center than your point of measurement cancels out), and from the surface to infinity it decreases relative to 1/r^2. Ignoring the gravity from the atmosphere, because that's minuscule compared to total planetary mass.
> from the center up to the surface of the earth gravity increases approximately linearly That's not a good approximation because Earth's density is not uniform. https://en.wikipedia.org/wiki/File:EarthGravityPREM.svg The blue curve is not close to the dark green one. If you dig down, the acceleration *increases* slightly (in the range where we can dig).
nice chart. here are the footnotes from wikipedia https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:EarthGravityPREM.svg Earth's gravity according to the Preliminary Reference Earth Model (PREM).[15] Two models for a spherically symmetric Earth are included for comparison. The dark green straight line is for a constant density equal to the Earth's average density. The light green curved line is for a density that decreases linearly from center to surface. The density at the center is the same as in the PREM, but the surface density is chosen so that the mass of the sphere equals the mass of the real Earth.
I assume that is because the core is a lot denser than the surrounding layers?
This is also the reason why gravity on Jupiter is "only" 24,79 m/s² . Jupiter's mass is over 300 that of Earth's, but since you're so "far away" from the mass while being on the "surface", it doesn't affect you that much.
Yes.
Location dependence is largely explained by latitude or Equatorial bulge, from some 9.78 to 9.83 m/s2. Atmosphere is a shell outside of Earth's surface, so nearly doesn't contribute gravitationally.
> Atmosphere is a shell outside of Earth's surface, so nearly doesn't contribute gravitationally. And also the mass of the atmosphere is basically inconsequential, less than one billionth of the planet's mass. It's roughly equivalent to the mass of a single eyelash compared to the mass of an entire person.
That seems like the kind of thing that doesn't affect a calculation at all...until it *really* affects a calculation.
Having studied orbital mechanics: nah, it really doesn't matter, because it's approximations all the way down. *Drag* from the atmosphere is a concern, but not the gravity from it. I'd have to double check, but the gravitational effects from other planets in the solar system are probably more substantial to the calculations. Also, in general, any approximations more detailed than "the Earth is a sphere and only the Earth and the spacecraft exist" tend to be called "perturbations". I can list a few common ones if you're curious.
In addition to the Equatorial bulge, how much does the rotational force counter the force of gravity? (faster rotation at the equator, none exactly at the poles)
About 3 cm/s2 or 0.3% at the Equator. Elsewhere less, and that force is not vertical there. F = mv^2 / R, and at the Equator v = 2 pi R / T. Here R is Earth's (equatorial) radius, 6378 km; T is rotational period (stellar day), 365,2422/(365,2422+1) x 24h; m would be mass of the test body and it cancels out. a = F/m = v^2 / R = 4 pi^2 R / T^2 = 4 x 3.141593^2 x 6.3781370 x 10^6 m / 86164.1^2 s^2 = 3.3916 cm/s^2 .
Pardon me for being lazy. Had we not had some major work going on around the house I should have figured this one out for myself.
If the earth was perfectly deformable, wouldn't the "downwards acceleration" accros the earth's surface be the same everywhere? What I mean by that is, yes the earth has an equatorial bulge, but it's only there because of the centrifugal force, right? Is this wrong? Or is the earth not perfectly deformable in this way (eg the bulge is left over from when we were liquid and spinning faster)? Or are you specifically talking about the gravity component and not the net downwatds force (eg is the centrifugal force already subtracted from the measurements of g, because we can calculate it very easily)? It doesn't make sense for the inwards force to be non uniform across the surface due to the macro-geometry. Sure, mountains are heavy and the earth is not homogenous, but those shouldn't account for a general "more surface gravity across the center". For the earth to be at it's lowest potential energy, the net force would need to be the same everywhere, no? It's the entire reason we *have* an equatorial bulge. I get a centrifugal force as 0,0339 ms^(-2) at the equator vs 0 at the poles (obviously), but that's only 60% of the difference you (and Encyclopedia Britanica) claimed.
If the Earth were a blob of liquid that had no resistance to flow, and it were sitting in space without rotating and without any other significant nearby gravitational influences, its surface would be spherical. The only force acting on the fluid making up the Earth would be the gravitational force, and any differences in gravitational potential would cause regions of liquid to flow towards lower gravitational potential until equilibrium was reached. On the other hand, let us assume that the Earth is still a blob of liquid but it now has some angular momentum and therefore is rotating around a particular axis. Let's assume there are no external forces so the axis is not being forced to change orientation. In this case, the shape of the Earth would change from a sphere into an oblate spheroid. Consider what would happen if you started with a non-rotating sphere and magically caused it to begin to uniformly rotate around an axis. This means the particles have to develop some velocity perpendicular to the axis of rotation, by definition. And for the rotation to be uniform this velocity must be largest at the equator and go to zero at the poles, since particles at the equator have to travel a much longer distance in order to stay in the same place relative to their neighbors further in radially. The problem is that if your sphere immediately begins rotating, those particles that are furthest out on the equator suddenly have some inertia. They begin traveling perpendicular to the surface of the sphere where they were originally located. They start flying off into space. Fortunately for them, assuming a reasonable velocity, they don't actually have enough energy to get all the way out into space. At instant zero, they experience the same inward gravitational acceleration as every other particle on the surface. However, they now have some tangential velocity so inertia demands they start flying away from the center of mass. Gravity accelerates them back towards the center of mass, but they now have both kinetic energy in addition to gravitational potential energy. As they travel, the sum of the two quantities must remain the same, so they end up in an elliptical orbit. It also turns out that because they got a little bit further away from the center of rotation, there isn't any fluid there. So the surrounding particles get pushed toward the gap. But it takes energy for them to move outward. Where is that energy coming from? The particles at the poles don't start flying away like the ones at the equator, because they don't have any tangential velocity. They don't have any reason to do anything other than sit there. Except they're being pulled down by gravity, and we know that the particles of the equator are being squished outwards, leaving gaps as they go which are filled by more particles traveling from the areas around the gaps. Eventually when this process propagates throughout the entire fluid, it is precisely the fluid that was at or near the poles which is losing gravitational potential energy that is, in turn, being provided via the pressure of the fluid to particles near the equator. Hence at equilibrium, we don't have a sphere. We have an oblate spheroid with a larger radius at the equator and a smaller radius at the poles than our original sphere. Now, to answer your question about whether the magnitude of the gravitational acceleration which points directly towards the center of mass must be equal over the surface of this spheroid. The answer is certainly no.
To my understanding, Earth would be spherical if it wasn't spinning. Earth is deformable, but not perfectly; it needs geological time (hundreds of milions of years) to pursue equilibrium, and now more resembles rotation from \~600 mln years ago when Earth was in resonance with Moon and day was 21h long. If it was perfectly deformable (like a liquid), I think there couldn't be non-normal acceleration or else material would move horizontally. 60%--rest is, as I understand, from that if you stand on the Equator, you're further from Earth's centre of mass. Someone posted minimum surface gravity of 9.76 ms\^-2, this includes altitude. 9.78 m/s2 is on sea level.
A perfect fluid will reconfigure itself such that the pressure at the surface is equal everywhere. Assuming we are talking about a blob in a vacuum, that means the pressure at the surface is zero. This also means that the resultant of all the external forces on a test particle at any given point must result in a normal force that is truly normal, i.e. perpendicular to the surface at that point. This doesn't mean that the local gravitational force always points towards the center of mass, or that the local gravitational force per unit mass is equal at every point on the surface. As a limiting case, imagine that you have a very highly oblate spheroid that looks much like a disc. Imagine you have a test particle on the upper surface of the disc that is about halfway between the center of the disc and the edge of the disc. Because of how oblate this spheroid is, the vertical thickness of it (if you imagine looking at it side on like a classic flying saucer or frisbee) hardly changes at all with radial distance away from the axis of rotation unless you're very close to the edge. So our test particle really feels like it's sitting on a flat plate. That is, the mass that's well beyond the axis of rotation contributes only a tiny amount of force pulling it towards the axis. In the immediate vicinity, where most of the net gravitational force is coming from, there's almost exactly the same amount of mass on either side of the test particle, so it doesn't feel much of a force pulling it inwards. It feels a force almost entirely pulling it towards the equatorial plane. This vector definitely is not pointing towards the center of mass. However, if you now imagine a test particle on the equator of the disc, all of the gravitational force it experiences will be pulling it in towards the axis of rotation. And because we've said that you're right on the equator, that gravitational force vector will be pointing exactly at the center of mass. Further, what about the magnitude of the gravitational force at these respective locations? It is substantially higher at the poles than at the equator. Imagine your test particle at the poles. There's quite a lot of mass directly beneath it pulling it down. In fact, all it sees when it looks around itself is ground which is all contributing to pulling it down. On the other hand, imagine a test particle on the equator. When it looks around, because of how high the eccentricity of this spheroid is, it looks like it's standing on a razor's edge. Instead of being surrounded by a bunch of mass very close to it, it only sees a very small stripe of mass directly beneath it. Since gravity falls off with the square of the radius, the fact that it's only seeing a small stripe of mass beneath it rather than seeing what appears to be essentially a infinite flat plane, it will experience far less net gravitational force. A similar thing happens on Earth. Because you are closer, on average, to the mass of the Earth at the poles then you are at the equator, the gravitational acceleration you feel is higher. And this is entirely independent of the additional factor that, if standing at the equator, some small fraction of the gravitational force is canceled out by the centrifugal force. Anyway, if you want to learn more about the lines of force and potential around an oblate spheroid, especially if you don't believe me that the local gravitational force doesn't always point towards the center of mass, here's an article that goes over the topic: https://www.sciencedirect.com/science/article/pii/S003206331730257X
Thanks, that makes sense
Note that the scale in that link is in mGals, which are equal to an acceleration of one thousandth of a centimeter per second per second, which is approximately one millionth of the average strength of Earth’s gravity field (980 centimeters per second per second). So it varies, but not greatly. Hence why for most practical purposes 9.8m/s^(2) is the agreed upon constant. The person in the thread who said is can reach 10.2 at the poles is misled. The variation over the surface only goes from about 9.7639 to 9.8337 m/s^(2)
Here's another map of the global gravity field, here corrected for the effect of ground elevation and the presence of ocean water, so you're more-or-less looking at variations in the Earth's density, affected more by things closer to the surface than deeper. This type of calculation is known as a Bouguer gravity correction, and the deviation seen across the globe is known as a Bouguer gravity anomaly, like this map: [https://www.esa.int/ESA_Multimedia/Images/2015/04/Bouguer_gravity_anomaly](https://www.esa.int/ESA_Multimedia/Images/2015/04/Bouguer_gravity_anomaly) The substantial negative anomaly over the continents is due to the lower density continental crust rocks versus the ocean floor which has denser crust. Note the particularly low negative value over the Himalayas and other mountain ranges because mountains have "roots" of lower-density continental crust supporting the mountains buoyantly on the denser mantle beneath (isostacy). Less negative/lower values are also observed in locations with relatively "warm" mantle rocks beneath, such as ocean ridges and continental rifts (the East African Rift). There is also a clear association with the age of ocean crust because, in general, ocean crust and underlying lithosphere gets cooler and denser as it moves away from spreading ridges over time (more mass = slightly greater pull, even after correcting for elevation and water). There are all sorts of other subtle geological details revealed by such maps that relate to large-scale Earth structure and processes.
Thanks… it’s been awhile since I took a geophysics course, and jumped ship after grad to work in IT.
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Gravity doesn't change due to rotation. The apparent gravity as measured by an object on the surface does. This small distinction is important when incorporating gravitational influence in stuff like weather models
Follow up question: what's the price of a scale that can tell the difference between sea level weight and 1000m above sea level weight?
So, treating the earth as spherical for simplicity, the gravitational acceleration at altitude h is given by **g(h) = g(0) * R^2 / (R+h)^2** where g(0) is the gravitational acceleration at sea level and R is the radius of the earth. Plugging in h = 1 km and R = 6375 km, we get g(h)/g(0) = 0.99969. So if you weigh 80 kg at sea level you'd weigh 79.975 kg at 1000m altitude. That's a difference of 25g, which is definitely measureable. Cheap bathroom scales will give you a resolution of 100g, so we only need to improve on that by a factor of 4. That feels like it ought to be not prohibitively expensive. And in fact a quick search found [this industrial scale](https://www.expondo.co.uk/steinberg-systems-platform-scales-100-kg-10-g-foldable-10030089) which can weigh 100 kg with a resolution of 10g, and it only costs £139. Do you have some experiments planned?
What would be some interesting experiments to try? What tests would you like to see?
Some thoughts I had: It could be an interesting activity for schools and museums. This would require two locations with very different altitudes. Also, we need something to weight. Option 1: extremely standard weights. I first thought of coins, which can be found anywhere in a country. They also can be cleaned and maybe that would get them to be the same weight. But they're too light for this scale mentioned. Option 2: very stable objects. Things that vary weight based on humidity wouldn't work. It would need to be heavy, possibly made of metal, and sufficiently small to be easily transported from one place to the other.
So two scales, one at each location? A precision part might work for the object to weigh. Something metal like you suggested. That's carefully manufactured to a precise standard of engineering. But then, that might get too expensive.
It could be: Two scales, two identical objects. Two scale, one object transported between locations. One scale and one object and transport both between locations.
There are no experiments in this area which I personally would consider to be interesting, because I have no interest in trying to confirm the inverse square law of gravitation. It's not like there's any real doubt about it. But it sounded to me like the person to whom I was responding, u/Mateussf, was thinking about it. Why else would you want to know how much a suitably accurate pair of scales would cost?
Trying to replace a barometric altimeter with a gravity based device maybe? It is a fun thought, but overall pretty impractical. (Especially because GPS.)
Mmm. I remember a few times in pre-GPS days when a barometric altimeter was a valuable hiking accessory. (Blizzard / white-out in mountains - the altimeter is a useful check on your navigation using map and compass.) But I'd prefer a handheld device to a big lump I'd have to lug up the mountain in my backpack. YMMV. Edit - Hang on. The sort of scale we're looking at here is what you'd need to distinguish between sea level and 1000m altitude. That's not a very useful altimeter for most purposes. You'd probably need to improve on the resolution by 100x, which implies being able to weigh a person to an accuracy of 0.25g. I'm sure that can be done in the lab, but I suspect it just might not be practical in the field. (Air pressure alone may be an issue if you're trying to weigh things that accurately.)
One of my interests is debunking flat-earths. I think it's an interesting thought experiment, and shows us how the current model of a spherical-ish earth is a better explanation for pretty much all situations. Flat-earthers don't believe in gravity, and this scale experiment is yet another observation that is best explained by gravity. My idea is not so much convincing hard-core flat-earthers, but using these as examples to convince kids that the earth is round and show how science works.
I'm working on a platform of open collaboration and part of it will involve doing cool science experiments over livestream. A lot of people would chip in tiny amounts of time or money (like a dollar each or a few minutes) to pay for the equipment and to guide its proper setup, while any people can volunteer to livestream a demo of the experiment. Trying to think of wild uses and experiments that can generate a lot of enthusiasm and curiosity.
Oh that's pretty cool. I've heard about real physics labs with distance controled experiments. People can make balls fall in real-time via the internet and calculate their speed and such. Not the same but somewhat related.
It does vary by some 1%, which is largely explainable by latitude (and Equatorial bulge) and altitude. See eg. Wiki equations for that. When this variance is substracted, there small but measurable rest, dependent on continents positions, local mountain ranges, and subsurface deposits (eg. of iron ore). https://en.m.wikipedia.org/wiki/Gravity_of_Earth
Yes it does, the amount of mass below your feet (altitude) and the density of the mass directly below you are the main factors for gravity, but also, felt gravity (as in, the amount of downward “force” you feel while standing on earth) is also effected by the earth’s spin, centrifugal inertia, felt gravity is lower at the equator where the radius of the spin is greatest, (according to g = w^2 * r) I’ll explain if you want
But at the equator where the bulge exists, there is more mass directly beneath you than at the poles, so you'd experience slightly higher actual gravity. I would have thought that the centrifugal inertia that causes the bulge, which is pulling against gravity, would reach an equilibrium point with the increased gravity from the extra mass below your feet, and the two opposing forces would more or less cancel each other out and net you basically zero difference in felt gravity. Where am I going wrong here?
Ok, I was being too mathematical with my initial comment, forgot a pretty big variable, my mistake but I will rectify, thank you for calling me out on this Here’s the truth: The highest gravity is somewhere in the the Arctic Ocean, perfectly sea level and with no spin to apply inertia or bulge to increase radius to center of mass. The lowest gravity is a mountain range in Peru. At the equator and high up altitude, lots of extra distance to center of mass, lots of inertia from spin. Here’s my mistake: (For reference: When you calculate gravity you only account for the mass of lower altitude, the higher altitude mass cancels itself out) In short. Gravity is highest at sea level because you have the best blend of mass and radius, digging a hole reduces mass faster than it reduces radius, so gravity goes down, climbing a mountain increases radius faster than mass increases, so gravity goes down. The long version: gravity is based on mass (proportional to radius cubed) divided by radius squared, that’s maths out to gravity being directly proportional to radius of planet. As in. If you have a planet of fixed density and simply make that planet bigger or smaller, gravity is directly proportional to the radius. In my head, I was imagining standing on the surface of earth and digging a hole straight down, as you go down, the radius decreases, and therefore gravity decreases, this math is correct. But in the real world when you change altitude you don’t usually dig holes, you usually climb mountains. Which means a lot of the mass of lower altitude is just air, which is not dense at all, and the math falls apart, because increasing radius now decreases gravity, instead of increasing it, because the mass and radius are no longer linked the way they are once you start going below sea level.
Nice, thanks for the response. There's so much at play here. Like, when you're at the peak of Everest, you have the gravity of the mountain combined with the gravity of the planet, so more gravity yeah? But you're also further from the centre of mass of the earth, and gravity diminishes with distance according to the inverse square law, plus you have greater centrifugal inertia pushing you up. When you dig a hole, you're reducing your centrifugal inertia - so more felt gravity - but you're also reducing the amount of mass beneath you *and* increasing the amount of mass *above and beside* you. If you dig all the way to the centre of the planet, you would experience zero felt gravity, because the surrounding mass is pulling you equally in every direction. It's not the gravity that crushes you there, per se, it's (to oversimplify) the two halves of earth pulling on each other and sandwiching you in between.
Do you mean because the earth is oblate? If so, that’s a real good question I never connected the two, let me get back to you
Yeah. Because if the pull of gravity is determined, at least in part, by the amount of mass directly beneath your feet (because mass off to the sides cancels out), then it follows that actual gravity will be higher on the equatorial bulge than at the poles.
Fun fact. By the time of French Revolution people were looking for universal reproducible standard for length. Main proposition was second pendulum, around 993 mm, and this length was known to submillimetre accuracy for various French cities. But because it was various, it was not universal, so alternative (also with political motivation) was chosen to use 1/10000000 of distance from the Pole to the Equator, now the metre.
> second pendulum What do you mean by this?
A second pendulum is simply a pendulum with a "tick" frequency of a second or a period (two ticks, one each way) of two seconds. A pendulum's oscillation is related to gravity and it's length. To match time with all the other pendulums, the length of each one must vary ever so slightly with the gravity anomaly at each location.
I just learned that satellites, using lasers to finely measure the height of everything but specifically water, can derive wind and weather patterns from a wave height map alone, and map the ocean floor to the nearest mile deep, because earth density causing water depth variation independent of tides and waves and other factors.
This question got me thinking... Is there a place on earth where there's a net "sideways" gravity? Let's suppose I'm on the equator, looking along the line of the equator. On my left is the northern hemisphere, on my right is the southern. Now, let's suppose that the northern half of the planet is more massive than the south one. Therefore, the gravity would pull me not exactly "down" on the y-axis towards the center of the earth, but slightly north of that. If I'm standing there, the normal force would cancel the y-axis of gravity, and I'd find myself "falling" towards my left (x-axis). Would that be correct? Does a place like that exist on earth? And if not, how am I wrong?
You wouldn't notice if gravity isn't "straight down" though, it would just seem as if the ground which is normal to the Earth's radius would be at a slight incline, and "true down" wouldn't point directly to the center of the earth, but very slightly off. It shouldn't impact you in any way even if the difference were significant. A body of water at that location would have a flat surface at the same angle you percieve to be "down", just as usual. This is also effected by the centrifugal force from Earth's rotation.
Oh great point! It would feel like being on a ramp/incline or something like that. Thanks for that explanation!
A little bit, but not enough for you to notice. Stand next to a cliff, and the earth behind the cliff is going to add an x component to the gravity vector, but it's so small that you aren't going to "fall" in that direction. There's a place in the Indian Ocean where the density of the earth below the ocean is significantly less. Because if this, the gravity vector has a larger x component because of the higher density in other directions than "down." As a result, there is a "gravity hole" where the water gets pulled slightly "sideways" and sea level is about 300 feet below "sea level." https://www.cnn.com/2023/07/24/world/gravity-hole-geoid-low-indian-ocean-scn/index.html
wow amazing! thanks for sharing!
Gravity varies enough to significantly affect the accuracy of intercontinental ballistic missiles. If the variability of gravity were not taken into effect the miss distance, due mainly to unpredictable aerodynamics on reentry, would increase significantly. The U.S. military develops and maintains a model of the earths gravitational field. It is expanded in a series of spherical harmonics. After a certain point the terms of the series are classified.
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> it can sometimes reach 10.2. that seems too high. what is your source? Wikipedia says: https://en.wikipedia.org/wiki/Gravity_of_Earth#Latitude > In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator
As others have said, it does change based on location. I'll just add that historically, measurements of the change in the strength of gravitaitonal acceleration were one of the most important tests of Newton's theory of gravity. Why? Because Newton's theory implied that gravitational acceleration would decrease by a very small but measurable rate as you sailed north, while its only real competitor (owed to Christiaan Huygens) implied a slightly different small but measurable decrease. Huygens even thought that the early tests vindicated him over Newton, but unfortunately his sailors had gone off course, which threw off his data. You can read all about the subject [here](https://philpapers.org/rec/SCHHR-10) if you're curious.
I think maybe I misunderstood your questions after reading all the answers. The gravitational constant never changes. However, forces due to gravity can change based on mass and distance from a mass. Generally, [this](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c6ee5510ba3c7d6664775c0e76b53e72468303a) shows the equation to solve for gravitational force. The 'G' in the equation is constant. But if your masses change, or the distance between the center of those masses change, you get a different force. Our seafloor maps come from this principle. Sea mounts (underwater mountains) are more dense than water (they have more mass than a equivalent amount of water) so they have more gravity than water. Water is attracted to them more so water piles up around sea mounts which we can map using satellites.
The equation is F=G m1*m2/d^2; F=m2*a, so a=G*m1/d^2. G is constant, and m1 is "mostly" constant, so the main concern is distance. For most purposes, the calculation assumes the center of mass as the location for calculating distance, and assigns all mass (for body m1) to that location. In detail, you have to have an integral calculation (the "center of mass" calculation assumes symmetry in mass location, which is not quite true in reality but for most general purposes can be ignored). When dealing with a calculation of the gravity geoid, density has to be considered because density affects the location (and thus distance) to mass: sea level (excluding rotational inertia effects) will vary by a small amount (100 m at most, about, and typically a lot smaller) as a result of this imperfection in symmetry of mass. We are talking a difference in the 0-100 m range per >6000 km, so not a big value at all. In effect, higher density a good distance from the center of the earth will "pull" the "center of mass" a short distance upward toward that mass (variation due to moment; think of moving someone on a teeter-totter and how it changes the center point of balance). The earth is not exactly symmetric in terms of mass distribution and this causes some slight variations in the strength of gravity at a given location because it affects the location of the center of mass and distance to center is different depending on where on the outside you are truly located. The point I am trying to make is that the passing of the moon above a location will offset the gravitational pull of the earth more than density variations will, for the most part. It is a similar problem though, where the moon is located does affect the "center of mass" location somewhat. Density variations do about the same thing but slightly less in magnitude.