# Gravity below Earth's surface [duplicate]

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.