
#1
Jan408, 12:46 AM

P: 78

I am looking for any data on the weight of objects as they decend into the earth. I cant seem to find any, which i find surprising, that would be an ideal way to prove gravity being exclusively due to the attraction of mass.
I think i am correct in saying that as you decend the weight should decrease uniformly, as there is more mass above you pulling you back, and less below you pulling you towards the core. And when you get to the centre you would be completely weightless, as gravity cancels out in every direction. That is what Newtons Shell theorem suggests; http://en.wikipedia.org/wiki/Shell_theorem is there any data on weights at different depths inside the earth? surely there must be some, i cant seem to find any. 


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#2
Jan408, 02:35 AM

P: 54

This is only true if you assume that the density of the earth is uniform, which is not the case. The earth is made of several types of rock, all with different densities. You are correct in that the amount of mass pulling you back does increase as you descend to the center of the earth, but because of the earth's variable desnity the change in the amount of mass pulling you back relative to the amount of mass pulling you to the core as you descend will not result in a uniform decrease in weight. 



#3
Jan408, 02:44 AM

PF Gold
P: 5,450

Been there:
http://www.physicsforums.com/showthread.php?t=32573 Unfortunately the last link is broken. That domain is gone. I see if I can dig up the excel sheet with the calculations again. 



#4
Jan408, 03:19 AM

P: 78

Gravity inside the EarthI just find it odd that there is no data on this. There are plenty of deep caves and pipelines about the earth, some that go over thirty miles down, so i would have thought that at least some experiments would have been carried out to see what gravity is doing inside the earth. there must be some experiments that have been conducted. can anyone find any? i certainly cant 



#5
Jan408, 04:59 AM

P: 54

If the earth were uniformly dense, the weight would be given by (GMem/Re^3)r, where Me is the mass of the earth, m is the mass of the object, Re is the radius of the earth and r is the distance between the CoM of the earth and of the object. This would mean that the gravitational force is only dependent on and directly proportional to r, and so weight would be greatest at the surface of the earth. However, the earth is not uniformly dense so this isn't true. The density of the earth increases as the distance to the core decreases. To find the gravitational force at any point within the earth, you take the sphere of the same radius as the distance from that point to the center of the earth and ignore the rest of the mass, and then you use the equation F=GMm/r^2 where M is the mass of this sphere. Because the density is greater nearer the earth's core, this means that enough mass is concentrated there that M is not much smaller than the mass of the earth while the value of r^2 is much smaller. It shouldn't be surprising, then, that at some point within the earth this expression is greater than it is at the earth's surface. 



#6
Jan408, 05:53 AM

P: 78

I suppose that since the drop in gravitational field strength would be negligable over 20 miles it would be a fairly pointless experiment too. but you could still do it at multiple points on the earth and work with the avergaes, to see if the value is as predicted. Maybe i'm going to have do this myself as a small science project, although there certainly wouldn't be anything small about it I wonder what the graph would look like if you took varying density of the earth into account. It would be a smoother transition for a start, past that I'm not sure what exact shape it would take. 



#7
Jan408, 06:55 AM

PF Gold
P: 5,450

I hope I can find that spreadsheat back.
I seem to remember that going down the (light) crust g increases some procent, then in the upper mantle g is near constant. In the lower mantle g goes up a bit sharper towards the coremantle boundary where it's the highest. In the core the decrease is indeed as expected, near linear to zero. 



#9
Jan408, 01:51 PM

P: 745

You wanted data. Well it has been published, the best I have to hand is in this preliminary reference earth model (PREM) table. If you wanted, you could plot it up on excel next to Andre's to see how they match... might be interesting to see.




#10
Jan408, 02:32 PM

PF Gold
P: 5,450

Seems to be reasonable within the error margin that I mentioned in the old thread.




#11
Jan508, 05:16 AM

Mentor
P: 28,795

Actually, this can be worked out easily even if the earth doesn't have a uniform density. All one has to assume is that it only has a radial dependence of density, which isn't that unreasonable. If one goes by that, then one can apply the Gauss's Law equivalence to calculate the gravitational force because one has the radial symmetry to construct a simple gaussian surface.
Zz. 



#12
Jan508, 06:21 AM

P: 54

I suppose if you had the density values at various depths within the earth you could plot them and do a curve fit to derive an equation (to find the radial dependence of density). Then you could use this equation with Gauss's Law to solve for the gravitational force at any point. However, since the values we have for the density of the earth at various depths are not very accurate, our density equation probably wouldn't be very accurate and therefore our results wouldn't be accurate. 



#13
Jan508, 08:06 AM

P: 745

Note that the PREM assumes spherical symmetry (and also isotropy if you're interested in the seismic aspect).
Note also that if you're looking for more physical understanding about how g can be greater somewhere in the middle of the earth than it is at the surface you need to understand a bit more about what g actually is. g is actually defined as the (maximum negative) rate of change of the "gravitational potential" with respect to position. g=grad(gravitational potential scalar). So if you have some amount of gravitational energy in a gravitational field and you change position in the direction of g, the amount of gravitational energy you lose corresponds to the magnitude of g  if you lose a lot of energy for a very small change in position then g is big, conversely if you lose a little gravitational potential energy over a large positional change then g is small. Bringing it back to the earth, this simply means that you lose more gravitational potential energy if you move some small distance at around 2890 km depth (i.e. at the core mantle boundary), than you do if you move the same small distance at the surface of the earth. Note that this fact is related to the fact that the density contrast at the core mantle boundary is greater even than the density contrast at the surface of the earth! Density within the earth is reasonably well constrained by things such as normal modes, birch's law, and moment of inertia. Temperature is a lot more difficult, and the errors in the earth's temperature profile are more significant. But I don't think the errors in density are too bad. 


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