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Why is gravity reduced if you go below earth surface? Asked and answered before but I do not agree with the concepts followed in several answers. The concept of calculating the gravity attraction force using the mass and the distance to its center is limited to point masses and the error in using it will increase with the ratio of size of the mass to the distance to its center. With large close bodies we need integration (not my best suit) but to illustrate try this exercise.

Consider three masses A, B & C of one mass unit each. Locate them in space one unit distance apart from left to right. Calculate the gravity attraction on the mass A due to masses B & C to its right. Force AB = 1x1/1 (one unit mass each and one unit distance apart) Force AC = 1x1/4 (one unit mass each and two units distance apart) Then Forces (AB + AC) = 1.25

Now consider masses B & C as two halves of a combined mass D (2 units with a CG at 1.5 distance units from A). Force AD = 1x2/(1.5squ) = 0.8' An obvious disagreement, but the las calculation is in error. The principal error is that the location of center of combined masses is calculated by averaging distances. While gravity forces are calculated using squared distances. We cannot add them using a combined mass at an arithmetically averaged distance. I think a math teacher can name the discrepancy.

Now rearrange the masses to have A-B two units and B-C one unit distance apart. Repeat the calculation. You get

Force (AB + AC) = 0.36
Force AD = 0.32 The error is reduced the lower the ratio of size of mass D to distance to A.

About going down in a mine shaft towards the center of earth, thanks to the comment on this answer using integration proves the gravity effect is zero inside an even spherical, because it is zero for every element before integration.

The simplified example of masses ABC points to the error in calculating gravity outside a large mass if we use the distance to the geometric center of the mass.

MechEng
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If you do the integral, you find that inside a spherical shell there is no gravitational force in any direction.

It's a bit difficult to explain without using calculus, but I'll try.

Imagine a point inside the spherical shell, anywhere. Now draw a two-ended cone whose apex is at that point, and extend the two cones until they intersect the shell. Both cones have the same axis, but they extend in opposite directions. Look at the areas of the two portions of the shell carved out by the cones. The areas (and therefore the masses) will be directly proportional to the square of the distance from the point to the middle of the area. The force of gravitational attraction is inversely proportional to the square of the distance, so the force due to the two chunks of shell carved out by the cones is equal and opposite: it adds up to zero. Since it doesn't matter what direction the axis of the cones is pointing, all the forces add up to zero inside the shell. When you're outside the shell, of course, it's a different story.

S. McGrew
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There is a relation g'=g(1-(d/R)) where d= depth below Earth surface R= radius of Earth This relation can be used to find gravity below Earth surface. Hope this helps you.

user226833
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    This doesn’t explain why gravity is less inside, which is the OP asked. – G. Smith Apr 09 '19 at 05:13
  • The comment April9 at 4:29 explained by integration that if the earth mass distribution was homogeneous, the earth mass located in a shell farther from center than your location will have an integral gravity effect of zero at your location. The only mass left is the sphere from your location towards the center. This is obviously less than the total earth mass and will have a lower integral gravity. The value is expected to be less the deeper you go below earth surface and the smaller the sphere between your location and earth center. – MechEng Apr 16 '19 at 21:20