### 1st Putnam 1938

Problem A7

Do either (1) or (2)

(1)   S is a thin spherical shell of constant thickness and density with total mass M and center O. P is a point outside S. Prove that the gravitational attraction of S at P is the same as the gravitational attraction of a point mass M at O.

(2)   K is the surface z = xy in Euclidean 3-space. Find all straight lines lying in S. Draw a diagram to illustrate them.

Solution

(1) Let Q be a point on S. The obvious coordinate is the angle θ = angle QOP. The density is ρ = M/4πr2. By symmetry the attraction on P is towards O. Let the distance PO be d and the radius of the sphere be r. Let the gravitational constant be G. The component of the attraction towards O (per unit mass at P) is G ∫0π r dθ 2 π r sin θ ρ (d - r cos θ) (d2 + r2 - 2dr cos θ)-3/2. Note that the factor (d - r cos θ) (d2 + r2 - 2dr cos θ)-1/2 is needed to resolve the force in the direction PO. Writing x = cos θ, this becomes 2G π r2ρ ∫-11 (d - r x)/(d2 + r2 - 2dr x)3/2 dx.

This is not as bad as it looks. It is just the sum of a (1 - x)-1/2 and a (1 - x)-3/2 integral, both of which are straightforward. Moreover, when we come to substitute x = ±1, the factor (d2 + r2 - 2dr x)1/2 becomes just d - r or d + r. So we get after a little simplification 4 π r2ρ/d2 = MG/d2, which is the same result as if all the mass was concentrated at O.

(2) We can write a general line as x = at + b, y = ct + d, z = et + f, for some constants a, b, c, d, e, f and a parameter t which takes all real values. If this lies in z = xy, then et + f = ac t2 + (bc + ad) t + bd for all t. Hence a or c must be zero. If a is 0, then z = by, so the line can be written as x = b, z = by. Similarly, if c = 0, then the line can be written as y = d, z = dx. Conversely, it is easy to see that these two families of lines lie in the surface.