Originally Posted by

**Harold14370**
Show us how you summed up the individual cubes., and perhaps we can see where you are going wrong.

Thanks harold, it is very kind of you to listen to my questions.

As per post 44,

**A** is standing at the Northpole and the southpole

**B** is 20 units away. All points P on a circle give the same pull to A, which is 100/x^2 *cos phi, but circles near the equator

**E** have more mass than the others.

The angle phi is at A (PAB) and AP = x = cos phi*20, so the net pull by P is (100/20 = ) 5/x

In the Southern hemisphere pull vary from .35 (at E) to .25 at B:

y= 5/x from 10sqrt2 to 20 - Wolfram|Alpha.

It is useless to do thousands of sums to get the exact average, let's concede (to prevent any objection) that all points here (1/2 of the total mass: 2pi*r^2) give the maximum pull (.35*628, P = 220)

In the Northern hemisphere pulls vary from .35 (at E) to 1 (when phi= 76°) and go from 1 to 4.2 (phi > 87°, for a fractional number of points at distance 1 from A):

y= 5/x from 0 to 10sqrt2 - Wolfram|Alpha
Granting that all points give pull 1 (P = 628)

the global pull is 1+.35/2 =

**.6748 P_O ,** P (= 628+220) = 848 /

**P_O (**1256)

*(but doing individual sums we get about [probably less than] 1/2 P_O (.3+.6/2))*
Is there anything wrong in this procedure?

Thanks again