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## How strong is gravity in the center of the earth?

 Quote by Bjarne Janus Thank you I am just a little confused. I understand you answer that 5000 Meter below the surface of the earth, - in a mine, or in the ocean the acc. of gravity will be 9.786 m/s² (adjusted for density it will be 9.797 m/s) correct? But what do you mean with: “a little under that at the surface. (9.793 m/s²)” ? How deep is “a little under the surface”?
I meant that the value of the acceleration would be less than the value of the acceleration at the surface.
 PS! If we also adjust for the speed of the earth’s rotation at equator? Radius 6378km – 5km = 6373km - circumference = 40.058 Km = 40058000Meter / 86400s = 463,63 m/s Do you know the influence of this? Bjarne
This can be found by a=v^2/r.

So the adjustment would be 0.0337 m/s^2
 Janus what was the values you was using to calculate 9.793 m/s² - at the surface at the earth I mean M and r Bjarne

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 Quote by Bjarne Janus what was the values you was using to calculate 9.793 m/s² - at the surface at the earth I mean M and r Bjarne
M: 5.87e24 kg
r: 6.378e6 m.
 How deep inside the earth approximate would gravity begin to decrease ? Nearly halway? Bjarne
 as soon as you get below earth surface, an dthen move even deeper, it starts to decrease. it decreases proportionately to the distance from centre if you consider earth as a perfect sphere and of uniform density.

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 Quote by Janus So, initially, as you begin to move down towards the center of the Earth, the g force will go increase,.....
Right or almost right, and all albout gravity inside the earth in this thread:

 Recognitions: Gold Member So an object released in a tunnel through the centre of the earth (with the obvious idealizations) would execute simple harmonic motion.? Amusing.
 Due to increased in the density of the mantle, and the cores, gravity increases slightly as you descend to a depth of about 2000km, from 9.81 m/s2 at the surface to about 10 m/s2 at 2000km deep. it then increases faster to a maximum of about 10.7 m/s2 at 3000km depth, then tapers roughly linearly to 0 at the very center. While this general form may be followed for similar sized planets, it is important that the planetary body have the same approximate chemical makeup- iron core, silicate mantle/crust, etc. If the composition is different, an different density gradient will follow as depth increases to the core, and a somewhat different curve would follow. Again, the reason for the deviation from the simple mathematical model curve is the varying densities of the core and mantle of the earth. The actual curve, based on data from the CRC 94th ed., shows two peaks rather than one, before falling to zero. In this case, averaging the planet's density gives very inaccurate results. A

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 Quote by epenguin So an object released in a tunnel through the centre of the earth (with the obvious idealizations) would execute simple harmonic motion.? Amusing.
A cheap way to travel,just jump through the tunnel on one side of the planet and in about 25 minutes you are on the opposite side of the planet.I'm off to get my hammer drill.Does anyone want to invest in my new company?

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 Quote by Dadface A cheap way to travel,just jump through the tunnel on one side of the planet and in about 25 minutes you are on the opposite side of the planet.I'm off to get my hammer drill.Does anyone want to invest in my new company?
Whoops the actual time is closer to 42 minutes

My calculator was playing up first time I did it(this is one of my favourite excuses please feel free to use it)
 I'll have to get more data points, and set up the series on a spreadsheet to get the actual time to accelerate based on the actual acceleration values as you approach the core. This seems like it would be significantly less than 21 minutes (to get to the center) you calculated, so it would be less than the orbital time. The oscillation period in a fixed density is actually the orbital period for the altitude you start at. Of course, on earth you either slow from atmosphere or burn up from 7900 meters per second velocity! Up on the moon, however, you could theoretically orbit just above the surface (the highest points really) for practically ever.

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 Quote by jsandow I'll have to get more data points, and set up the series on a spreadsheet to get the actual time to accelerate based on the actual acceleration values as you approach the core. This seems like it would be significantly less than 21 minutes (to get to the center) you calculated, so it would be less than the orbital time. The oscillation period in a fixed density is actually the orbital period for the altitude you start at. Of course, on earth you either slow from atmosphere or burn up from 7900 meters per second velocity! Up on the moon, however, you could theoretically orbit just above the surface (the highest points really) for practically ever.
I agree jsandow, in my calculation I assumed a uniform density.Of course,because of density variations the motion will not be simple harmonic.
 the actual motion would equal a harmonic motion at four points in a single cycle: start, mid planet, the antipodal point where the direction would reverse, and mid planet again. but the curve would be a little different in between the points.

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