Gravitational Attraction We use triple integration to calculate the gravitational attraction that a solid body V of mass M exerts on a unit point mass placed at the origin. If the solid V is also a point mass, then according to Newton’s law of gravitation, the force it exerts is given by (1) F = GM r, |R |2 where R is the position vector from the origin 0 to the point V , and the unit vector r = R/|R| is its direction. V R O If however the solid body V is not a point mass, we have to use integration. We concen trate on ﬁnding just the k component of the gravitational attraction — all our examples will have the solid body V placed symmetrically so that its pull is all in the k direction anyway. To calculate this force, we divide up the solid V into small pieces having volume ΔV and mass Δm. If the density function is δ(x, y, z), we have for the piece containing the point (x, y, z) (2) Δm ≈ δ(x, y, z)ΔV, Thinking of this small piece as being essentially a point mass at (x, y, z), the force ΔF it exerts on the unit mass at the origin is given by (1), and its k component ΔFz is therefore ΔFz = G Δm r· k , |R|2 z φ which in spherical coordinates becomes, using (2), and the picture, ΔFz = G cos φ δ ΔV δ ΔV = G 2 cos φ . 2 ρ ρ If we sum all the contributions to the force from each of the mass elements Δm and pass to the limit, we get for the k -component of the gravitational force (3) Fz = G �� V cos φ δ dV . ρ2 If the integral is in spherical coordinates, then dV = ρ2 sin φ dρ dφ dθ, and the integral becomes �� (4) Fz = G δ cos φ sin φ dρ dφ dθ . V 1 ρ Δm 2 GRAVITATIONAL ATTRACTION Example 1. Find the gravitational attraction of the upper half of a solid sphere of � radius a centered at the origin, if its density is given by δ = x2 + y 2 . Solution. Since the solid and its density are symmetric about the z-axis, the force will be in the k -direction, and we can use (3) or (4). Since � x2 + y 2 = r = ρ sin φ , the integral is Fz = G � 2π 0 � π/2 0 � a ρ sin2 φ cos φ dρ dφ dθ 0 which evaluates easily to πGa2 /3 . V a Example 2. Let V√ be the solid spherical cap obtained by slicing a solid sphere of radius a 2 by a plane at a distance a from the center of the sphere. Find the gravitational attraction of V on a unit point mass at the center of the sphere. (Take the density to be 1.) a 2 Solution. To take advantage of the symmetry, place the origin at the center of the sphere, and align the axis of the cap along the z-axis (so the ﬂat side of the cap is parallel to the xy-plane). We use spherical coordinates; the main problem is determining the limits of integration. If we ﬁx φ and θ and let ρ vary, we get a ray which enters V at its ﬂat side z = a, or ρ cos φ = a, √ and leaves V on its spherical side, ρ = a 2. The rays which intersect V in this way are those for which 0 ≤ φ ≤ π/4, as one sees from the picture. Thus by (4), Fz = G � 2π 0 � π/4 0 � √ a 2 sin φ cos φ dρ dφ dθ, a/ cos φ which after integrating with respect to ρ (and θ) becomes = 2πG � = 2πGa π/4 a 0 � 3√2 4 �√ 2− 1 � sin φ cos φ dφ cos φ � −1 . Remark. Newton proved that a solid sphere of uniform density and mass M exerts the same force on an external point mass as would a point mass M placed at the center of the sphere. (See Problem 6a in problem section 5C). This does not however generalize to other uniform solids of mass M — it is not true that the gravitational force they exert is the same as that of a point mass M at their center of mass. For if this were so, a unit test mass placed on the axis between two equal point masses M and M ′ ought to be pulled toward the midposition, whereas actually it will be pulled toward the closer of the two masses.