How strong is gravity in the center of the earth?

Janus

I meant that the value of the acceleration would be less than the value of the acceleration at the surface.
This can be found by a=v^2/r.

So the adjustment would be 0.0337 m/s^2

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

Janus

M: 5.87e24 kg
r: 6.378e6 m.

Bjarne

How deep inside the earth approximate would gravity begin to decrease ?
Nearly halway?

Bjarne

Kushal

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.

Andre

Right or almost right, and all albout gravity inside the earth in this thread:

http://www.physicsforums.com/showthread.php?t=207148

epenguin

So an object released in a tunnel through the centre of the earth (with the obvious idealizations) would execute simple harmonic motion.?

Amusing.

jsandow

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

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?

Dadface

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)

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.

Dadface

I agree jsandow, in my calculation I assumed a uniform density.Of course,because of density variations the motion will not be simple harmonic.

jsandow

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.

mgb_phys

It's also just a special case of a very elliptical orbit

Chronos

Interestingly enough, it takes about 42 minutes to free fall through the earth, no matter what angle the hole is pitched. Of course if the inclination angle relative to the center of the earth is too shallow, you can expect some amount of 'rolling' time where friction will slow the journey.

Bob S

The Earth's surface, and especially the oceans, are at a equipotential, including both the gravitational force and the centrifugal force. So the earth is slightly prolate.

An equatorial bulge is a bulge which a planet may have around its equator, distorting it into an oblate spheroid. The Earth has an equatorial bulge of 42.72 km (26.5 miles) due to its rotation: its diameter measured across the equatorial plane (12756.28 km, 7,927 miles) is 42.72 km more than that measured between the poles (12713.56 km, 7,900 miles).

Stoonroon

The simple harmonic motion prediction for radial free-fall through a uniformly dense spherical mass (such as the idealized Earth discussed here) is very well known. Does anyone feel a desire (or need?) to verify this prediction empirically?

Though a high orbit satellite experiment would be the ideal way to do it, a laboratory version could also be done using a modified Cavendish balance.

Wouldn't it be nice to have all the theoretical discussion about this prediction backed up by a physical demonstration?