Great circles - motion over the surface of a rotating sphereThe red dot represents a particle that can slide frictionless over the surface of a perfect sphere. The particle cannot leave the surface. The view on the left represents the motion as seen from an inertial point of view. The view on the right represents the co-rotating point of view, so that the motion relative to the rotating sphere is clearly seen. If you think of the sphere as a planet, with the planet's gravity confining the object to motion along the surface, then the simulation is physically unrealistic in the following way: a rotating planet always has an equatorial bulge. A rotating planet with a perfectly spherical shape would violate the laws of physics; rotation always induces a bulge. (Of course, this applies only for celestial bodies that are large enough to self-gravitate to an equilibrium shape. Very small celestial bodies, such as asteroids, do not have enough self-gravitation to affect their own shape.) In this simulation the sphere is indeed perfectly spherical, and there is no friction, therefore the sphere does not influence the particle's direction of motion over its surface. Whether the sphere is rotating underneath the particle or not makes no difference for the motion of the object; there's no dynamics going on; there is no exchange of momentum, no change of kinetic energy. Evolution of the simulationWhen the simulation is started with the Play button then initially the particle is located on the starting latitude, co-rotating with the sphere (which of course implies that some retaining mechanism is holding on to the particle). When you press the 'Release' button the particle is released (if not released manually the particle will be released automatically after a full rotation has been completed). Once released the particle proceeds to move along the great circle that is tangent to the starting latitude. Moving along a great circle is the equivalent of moving along a straight line. On a sphere the shortest distance between two points is a (section of) a great circle, just as on a flat surface the shortest distance between two points is a straight line. Uncheck the checkbox 'Spheres': then you get an unobstructed view of the trajectories. You can manipulate the image with the mouse, turning and pitchng the view. You can see then that the motion along the great circle is always tangent to the starting latitude. The simulation's controlsThere are four input fields: 'System rotation rate', 'Starting latitude', 'Release/launch velocity' and 'Object angular velocity'.
General remark: if you input a new value for any of the above four input values then one or more of the other inputs will be updated to keep those four input values in an overall consistent state. When you start altering a value in an input field the field turns yellow. As long as the field is still yellow the input process is not ready. The input is finalized by pressing the Enter key on the keyboard, and then the field will no longer be yellow. PurposeThe reason I designed this simulation is that I wanted to create a vivid illustration of what happens when the particle is launched with a velocity in westward direction. Set 'System rotation rate' to zero. Then both when launched to the left and when launched to the right the particle proceeds to the equator, moving along a great circle. Check the checkbox 'Tangent' to get the tangent great circle. Following the tangent great circle is what you expect for the non-rotating setup. Set 'System rotation rate' to 1 again, and input a moderate westward launch, such as -10 or -20. You will see that the object deflects to the left of the tangent circle. Centrifugal effectI refer to the rotation effect that is illustrated with this simulation as centrifugal effect. The comparison is between what happens in the case of a non-rotating sphere and in the case of a rotating sphere. An object is launched with moderate velocity relative to the sphere. In the case of a non-rotating sphere:
Of course, if you launch with extreme velocity things work out a little differently. The checkbox 'Extra' opens a small window with extra settings, including a section with presets. Choose the preset 'Westward launch'. That will set up a starting latitude of 32 degrees, and a release/launch velocity of -47 (minus 47). Then the object will move along the tangent great circle for quite a stretch. With that particular combination of factors the motion relative to the Sphere comes out very similar to the case of a non-rotating sphere. (The release/launch velocity is a percentage. At 32 degrees latitude, a velocity of 47% of the velocity of co-rotating motion is about the speed of sound.) The case with the least motion towards the equator is when the release/launch velocity is -100, then the object is stopped in its tracks. With an even larger release/launch velocity the object once again will proceed to the equator after launch. Line of sightNote that a great circle is also the line of sight. The shortest distance between two points (on a sphere) is a section of a great circle. Reset the view of the simulation and check the checkbox for the tangent circle. The red dot is at the point where the tangent circle is tangent to a latitude line. Something that you are aiming for that is on that tangent circle will be exactly due West in your line of sight. Relevance for ballisticsA projectile that is expelled by a gun is during it's flight effectively in orbit around the Earth. (But since there is a lot of air resistence the motion of a projectile is generally not referred to as orbital motion.) An orbit is planar, with the Earth's center of mass in the orbital plane, so the line of intersection of the orbital plane and the Earth's surface will be a great circle. This means that this motion-along-a-great circle simulation can serve as a first approximation of projectile motion. Is the great circles simulation relevant for ballistics? Well, generally other factors, which in one form or another are all air resistance factors, are much larger than the rotation-of-Earth-effect. The rotation-of-Earth-effect is going to be swamped anyway, so you can spare yourself the trouble. But if you do want to take the rotation-of-Earth-effect into account, for instance if you want to work out whether a bolt from a crossbow will be affected, then this simulation shows that you need know to when to correct for a deviation to the left and when to correct for a deviation to the right; it can be either way. Not applicable in MeteorologyThe rotation-of-Earth-effect that illustrated with this simulation is distinct from the rotation-of-Earth effect that is taken into account in Meteorology. The effect that is relevant for Meteorology is represented with the following applet: inertial oscillation. Method of computationThe trajectory of the object is calculated by separately calculating the x, y, and z-component of the trajectory. The motion is along a great circle, with uniform velocity. So all three components, x,y and z are sine functions, and calculating the motion is simply a matter of evaluating the sine functions. The left panel, the inertial point of view, is the primary panel In the computations. First the motion as depicted in the inertial point of view is calculated. The next step is to transform the trajectory as computed for the left panel to the coordinate system that is co-rotating with the sphere. Download optionsThis simulation has been created with EJS Also available for download: a standalone version of this applet (requiring the Java Runtime Environment but not a browser to run)
Last time this page was modified: January 05 2011 |
