If earth's density wasn't uniform then the gravitational force

[MissMath]

is it true that if Earth's density wasn't uniform then the gravitational force was half of what it is now? can you give me a mathematical calculation of why that would reduce to half of its current value.thanks

 

[Pengwuino]

You have to say exactly what sort of density you're looking at. By simply stating non-uniform, the gravitational force could be anything depending on exactly what the density is. A 1/r density gives a different force than a \frac{1}{r^2} which gives a different force than a\frac{1}{r^3} which gives a different force then...

 

[Vanadium 50]

The Earth's density isn't uniform.

Nonetheless, the density distribution of the Earth does not influence the gravity at the surface. That's determined solely by the mass and radius.

 

[MissMath]

The Earth's density isn't uniform.

Nonetheless, the density distribution of the Earth does not influence the gravity at the surface. That's determined solely by the mass and radius.

oops! i meant if the density was uniform the gravity was half of its value. are u sure? if yes what does this article say?
http://www.csr.utexas.edu/grace/publications/press/032007_discovermag.pdf

 

[A.T.]

The Earth's density isn't uniform.

Nonetheless, the density distribution of the Earth does not influence the gravity at the surface. That's determined solely by the mass and radius.

oops! i meant if the density was uniform the gravity was half of its value. are u sure? if yes what does this article say?
http://www.csr.utexas.edu/grace/publications/press/032007_discovermag.pdf

Vanadium 50 means a density distribution which varies with depth: like that the core shells are more dense that the outer shell. This distribution doesn't affect the gravity outside the Earth which is also symmetrical.

The article refers to the fact that the Earth is not a sphere, and the density on the surface varies from place to place. This affects gravity in certain places.

I didn't read the article , but where does it say gravity was half of its value if the density was uniform.

 

[D H]

oops! i meant if the density was uniform the gravity was half of its value. are u sure? if yes what does this article say?
http://www.csr.utexas.edu/grace/publications/press/032007_discovermag.pdf

The article doesn't say anything like that.

The Earth's density isn't uniform.

Nonetheless, the density distribution of the Earth does not influence the gravity at the surface. That's determined solely by the mass and radius.

That's not true, either. Knowing how Earth's gravity varies with latitude and longitude is very important if you want to, for example, accurately predict the motion of a satellite or find a new oil field.

The above plot shows the difference between Earth gravity and that from a simple ellipsoidal model of the Earth. A Gal, or galileo, is 1 cm/s². An mGal is 0.001 cm/s². Note that the variations in this plot (plus or minus 100 mGal) are small compared to 1 g, 980.665 Gal.

The above is based on a very large (2160x2190) spherical harmonics model of Earth's gravity field, the Earth Gravity Model 2008 (EGM2008). See http://earth-info.nga.mil/GandG/wgs84/gravitymod/index.html.

 

[Vanadium 50]

That's not true, either. Knowing how Earth's gravity varies with latitude and longitude is very important if you want to, for example, accurately predict the motion of a satellite or find a new oil field.

While that's true, let's go back to the OP's question.

is it true that if Earth's density wasn't uniform then the gravitational force was half of what it is now? can you give me a mathematical calculation of why that would reduce to half of its current value.thanks

We're not talking about the tiny non-uniformities that let one find oil fields, we're talking about factors of two. So while I probably should have had a couple of paragraphs of caveats, the approximation of the Earth as a sphere whose density depends only on radius is a pretty good one to answer the OP's question.

 

[D H]

While that's true, let's go back to the OP's question. ... the approximation of the Earth as a sphere whose density depends only on radius is a pretty good one to answer the OP's question.

In light of the original post which posits that a uniform density would alter Earth's gravity by a factor of two, Vanadium is correct. Assume the Earth is spherical and isn't rotating. So long as the Earth's density depends only on radius, how that density varies with distance from the center of the Earth does not affect gravity on the surface by one bit. All that matters is the total mass of the Earth. Here are three models for this hypothetical spherical, non-rotating Earth:

1. A sphere with radius 6371 km and a uniform density of 5.5153 gram per cubic centimeter.
2. An ultra-dense inner sphere with a radius of 100 km and a density of 1.40036 kg per cubic millimeter surrounded by a very light outer shell with a thickness of 6271 km, an outer radius of 6371 km, and a density of 0.1 gram per cubic centimeter.
3. A very light inner sphere with a radius of 6370.99 km and a density of 0.1 grams per cubic centimeter surrounded by an ultra-dense outer shell with a thickness of 10 meters, an outer radius 6371 km, and a density of 1.15003 kg per cubic millimeter.

All three models have the same outer radius (6371 km), the same total mass (5.9742×10²⁴ kg). Despite the huge differences in internal makeup, all three models have the same surface gravity, 9.8217 m/s².