Problem of the Week #1

I thought we could spice things up a bit with a more interactive post on The Virtuosi. Starting this week, a new problem of the week will be posted each week. Solutions will be posted the following week. These problems will be a collection of physics and math problems and riddles, and although hopefully challenging enough to be fun and interesting, they should mostly be solvable using concepts from introductory undergraduate physics and math classes. We welcome you to ponder these problems, and send in solutions, or even any ideas you have about how to solve the problems tothe.physics.virtuosi@gmail.comwith “problem of the week” in the subject line. We will keep track of the top Virtuosi problem solvers. Here it goes…*Maximizing Gravity You are given a blob of Play-Doh (with a fixed mass and uniform density) that you can shape however you choose. How can you shape it to maximize the gravitational force at a given point P* on the surface of the Play-Doh? Solution In cylindrical coordinates, where the point P is taken to be the origin, the z-component of the gravitational field due to any point (s, z) felt at the origin, is proportional to $$\frac{1}{s^{2}+z^{2}}\frac{z}{\sqrt{s^{2}+z^{2}}},$$ where the second factor is necessary to take the z-component. If we have azimuthal symmetry, the magnitude of the gravitational field is given by the sum of all the z-components of the forces due to all points in the planet. For a given contour $$\frac{z}{\left(s^{2}+z^{2}\right)^{3/2}}=C,$$ image for some constant C, each point on the interior contributes more to the total gravitational field than each point on the exterior. Thus, the gravitational field is maximized if the surface of the planet takes the shape of one of these contours. Solving for s(z), we get$$s\left(z\right)=\sqrt{\left(\frac{z}{C}\right)^{2/3}-z^{2}}.$$ If the length of the planet in the z-direction is $$z_0,$$ we can replace C in the above expression to get $$s\left(z\right)=\sqrt{\left(z_{0}^{4}z^{2}\right)^{1/3}-z^{2}}.$$ As can be seen by the plot above, this shape looks a lot like a sphere, but slightly smushed toward the point of interest P. We can rigorously compare the field of the maximal gravity solid to that of a sphere with the same volume. The volume of the maximal solid is given by $$V=\pi\int_{0}^{z_{0}}s^{2}\left(z\right)dz=\pi\int_{0}^{z_{0}}\left[\left(z_{0}^{4}z^{2}\right)^{1/3}-z^{2}\right]dz=\frac{4\pi}{15}z_{0}^{3}$$ The volume of a sphere of radius r is of course $$V=\frac{4}{3}\pi r^{3},$$ so in order for the volumes to be the same, the sphere must have a radius of $$r=z_{0}/5^{1/3}.$$ The acceleration due to the maximal solid is proportional to $$a_{max}=\int dVa_{z}=2\pi G\rho\int_{0}^{z_{0}}dz\int_{0}^{s\left(z\right)}sds\frac{z}{\left(s^{2}+z^{2}\right)^{3/2}}=\frac{4\pi G\rho z_{0}}{5},$$ while the acceleration due to the sphere is just $$a_{sphere}=\frac{G\rho\frac{4}{3}\pi r^{3}}{r^{2}}=\frac{4\pi G\rho z_{0}}{3\cdot5^{1/3}}.$$ Thus, we have $$\frac{F_{max}}{F_{sphere}}=\frac{4\pi G\rho z_{0}/5}{4\pi G\rho z_{0}/\left(3\cdot5^{1/3}\right)}=\frac{3}{5^{2/3}}\approx1.026.$$ So the solid that gives the maximum gravitational field at a point is only about 3% better than a sphere. For a more detailed discussion, see Alemi’s solution.

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By Holmes