In the Hollow Center of a Large Mass
Date: 12/23/95 at 2:46:21 From: Anonymous Subject: physics and calculus In physics lecture I was told that if a large mass, such as the earth, was hollow and one was to be in the hollow part, then you would be weightless. I hypothesised that unless you were in the exact center you would be drawn towards the closest side; but my teacher (who refused to go into detail) said that Newton had proved that my hypothesis was wrong by using three-dimensional integral calculus. I would very much like to see this proof. Please send it to me.
Date: 5/31/96 at 11:17:39 From: Doctor Ceeks Subject: Re: physics and calculus Hi, Your teacher is correct... an object inside a uniform hollow shell would have no net gravitational force on it. Intuitively, the reason is that although the parts of the sphere closest to you would produce a greater force on the object, there is more mass on the other side to counteract. Imagine a cone with vertex the object. By cone, I mean something that roughly looks like two party hats stuck together at their apexes so that their axes form a single line. (Precisely, I mean something that looks like the solutions to x^2+y^2 = az^2, where a is constant.) Such a cone will intersect the sphere in two pieces, one a distance roughly d from the object, the other a distance roughly d' from the object. The gravitational forces will point roughly in opposite directions and equal (with suitable units) m/d^2 and m'/d'^2 where m and m' are the masses of the two pieces a distance d and d', respectively, from the object. The point is that m (with suitable units) is equal to d^2 and m' is equal to d'^2. This argument is not rigorous. To be made rigorous, careful attention must be paid to the orders of magnitude of the errors made in the rough approximations. This can be done using the techniques of Calculus. Since I am not going to include these computations in my answer, this answer must be considered incomplete. However, if you know Calculus, try to analyze the problem using spherical coordinates. -Doctor Ceeks, The Math Forum
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