# Tag Info

54

The below figure, taken from Wikipedia shows a model of the free fall acceleration, i.e., 'gravity'. The left-most point corresponds to the center of the Earth; then further right at $6.3\cdot1000$ km you are at the Earth's surface; and then further out you move into space. You can follow the blue line for PREM to get an idea of the average (expected) ...

31

It is the pressure gradient that is proportional to the local gravitational force. When that force is integrated over a distance, the pressure gradient is integrated to accumulate a total pressure. The maximum occurs at the point towards which gravity is directed in a spherical mass, which is the center. True, gravity at that point is zero, but it and ...

23

The speed of rotation of Earth is controlled by its angular momentum. And the conservation of angular momentum is a very serious law of physics (perhaps even stricter than conservation of mass). So in the same way that for the Earth to lose mass, that mass have to go somewhere. For the Earth to lose angular momentum, it'd have to go somewhere. Earth's ...

22

There is an entire field of Geophysics called gravimetry dedicated to measuring the magnitude of the gravitational field. First, we should distinguish between weight (a force) and gravity (an acceleration). Gravity is the acceleration that Earth gives to objects near its surface due to the gravitational force. The acceleration of an object near the ...

19

Earth's radius is about 6400 kilometres. That's 6400000 metres. Let's say that you have a mound 20 metres high, burying an older settlement. Your new "radius" is now 6400020 metres. Let's say that $g = 9.8\ \rm m/s^2$ at 6400, your new gravity will be $g = 9.799939\ \rm m/s^2$. Clearly, this is hardly "lower level of gravity". To make this even less ...

18

I gather, but may be wrong, that the mass of earth at present increases by around 108kg/day. All else being equal, one would expect the earth to have gained a mass, since 75 million years ago, of: $$75^6 \times 365 \times 10^8 kg = 6.496 \times 10^{21} kg$$ First off, that should be $75\times10^6$, not $75^6$. That alone makes your estimate high by a ...

15

Gravitational acceleration does vary. It tends to be strongest at the poles and weakest at the equator, this is due to the oblate shape of the Earth which means that at the Equator you are further from the centre, and gravity decreases the further you are. Also the spinning of the Earth makes gravity seem less at the Equator. The combination of these ...

12

The previous answers do a fine job already. But I'll try to add a simple thought experiment. Imagine three objects floating in space, clumping together by gravity: ### ### #A#|c|#B# ### ### Mass A, the negligible mass c and mass B, equal to A. The center is attracted to A and B and their gravity cancels out. However, A presses against c, because it is ...

11

The link you give seems largely plausible about the cause of the anomaly in that area: a 2007 study found that now-melted glaciers were to blame. The ice that once covered the area during the last ice age has long since melted, but the Earth hasn't entirely snapped back from the burden. Since gravity over an area is proportional to the mass on top of that ...

10

The weight that an object with a constant mass experiences on different parts of the Earth's surface will be slightly different for different locations, largely dependent on the height above sea level and the latitude. This is due to differencies in the gravitational strength on the Earth's surface at different locations. The accelation due to gravity, $g$,...

10

Rings wouldn't decrease the gravity much, but the exact amount would depend on the exact geometry of the rings. One reason is that the gravity of one side of the ring would partially cancel the attraction of the other side. In a similar fashion described by the Shell theorem demonstrated long ago by Newton and proving that if you are inside a shell of mass ...

9

Is there any consensus about the conjecture that gravitational force on Earth may have changed significantly over geological time; No, earth's gravity did not change significantly over time. Yes, earth's mass increases because of meteorites and decreases because of loss of some atmospheric gases to space, but it is extremely negligible. and in ...

9

Pressure at the center of the earth is non-zero. You're correct that there's no gravitational force at the center of the earth, but that doesn't mean pressure is zero - the pressure comes from the many miles of rock sitting above the center of the earth. As an analogy, think of a balloon. The pressure inside is higher than ambient because the elastic skin of ...

8

To add to the previous answer the gravity will vary locally due to the density of the rocks (or other materials) in the area. Gravity meters are used in mineral exploration to find dense rock types associated with certain ore bodies.

8

I had a similar question, and then I learned that at the molecular level it is diffusion what dominates, that means that despite Hydrogen is lighter, it won't rise to the top of the atmosphere. Flotation as we visualize it doesn't really work at the molecular level (because it is overcome by chemical diffusion). Therefore, the Hydrogen in the atmosphere is ...

6

You cannot leave "temperature outside", as temperature is the key factor to know if the balloon would pop or not. Let's set up some assumptions about the problem so we can calculate something: The gas inside the balloon behave according the ideal gas law: $PV=nRT$ That the balloon can hold a differential pressure of 30 mmHg (=4000 Pa) as found in this ...

6

Earth loses more mass as gases (especially Hydrogen and Helium) escape the atmosphere. Also, the numbers you mention as well as the mass of the air that is lost likely has large error bars associated with them. Edit: For example see here or here.

6

Wolfram Alpha reports values between 9.78 and 9.88 m/s^2 depending on the location, which is odd because it should stay within the range 9.76 - 9.83 all over the world. It also seems to be using the ocean floor as a reference elevation if the point falls in water. Using Natural Resources Canada's tool, it reports a gravity of 981933.3 mGal (9.819333 m/s^2) ...

6

This is a companion answer to both Erik's and tfb's answers. tfb's answer assumes a uniform density throughout the Earth. This is highly inaccurate. Not to disparage tfb's answer; I've run across physics PhDs who thought that this was the correct model of gravitation inside the Earth. The problem with this model is that the uniform density model is invalid ...

5

For sure this is not available. To create those maps takes several dozens Earth orbits by GOCE (one orbit ~90 minutes) and possibly many more to bring up the signal to noise Which gravitational features would you expect to vary significantly on any human timescale, so that creating such a scan would make sense? This does not mean however, that no ...

5

First, a distinction between "gravity" and "gravitation" as used in geophysics. "Gravitation" pertains to Newton's law of gravitation. (Nobody uses general relativity to model gravitation for small, lumpy masses such as the Earth, Mars, or the Moon.) "Gravity" pertains to how things appear to fall from the perspective ...

5

To address your original concern, no, the fact that there are buildings underground does NOT mean that the surface of the earth is higher than in the past. What is actually happening is that these buildings are subsiding into the ground. How? Believe it or not, earthworms. Worms were once constantly tunneling through the soil underneath ancient buildings, ...

5

The moon does orbit the sun but it also orbits the earth. But your assertion that the earth's gravity is weaker than the sun's is not universally true. If the sun's gravitational pull were greater than the earth's HERE on the earth, the sun would literally suck everything not attached to the earth's surface into itself. Clearly that's not happening. ...

4

I think this question is less intuitively obvious than it appears. and I like @Universal_learner 's answer. I thought I'd give a different approach to the question. Earth is the only object in the solar system with significant amounts of granite and granite is very important to your question because of it's durability and it's buoyancy. Granite forms ...

4

Citing from Wikipedia: $Y_l^m$ is called a spherical harmonic function of degree $l$ and order $m$. If we take the real part of the spherical harmonics only, there is a nice visual explanation for order and degree. The order $m$ is the zonal wave number, that is how many waves we count walking around the sphere at constant latitude. The degree is a little ...

3

It's Legendre polynomials, not Lagrangian. The zonal harmonics (the $Y^0_l$ terms) depend only on latitude. These zonal harmonics are closely associated with the Legendre polynomials $P_l(x)$, where $x$ is the sine of the geocentric latitude. The tesseral and sectoral harmonics (the $Y^m_l$ terms, where $m\ne 0$) are closely associated with the associated ...

3

Instead of the Nagy prism formula, I suggest you to use the formula quoted in the following paper: B. Banerjee, S.P. Das Gupta (1977): "Gravitational attraction of a rectangular parallelepiped" Geophysics, vol. 42, n. 5, pp. 1053-1055 doi: 10.1190/1.1440766 I wrote, in Fortran, a TC program based on that formula. If you read the paper you better ...

3

According to a slightly old estimate, the total power input from the tides is approx 3.7 TW . Of this about 3.2 TW is from the moon (the rest is from the sun). Most of it goes into moving the oceans around, and some of it into doing the same with the land - but both ultimately end up as heat. The question of how much this power affects the temperature of ...

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