Let’s imagine that you have an idea for an experiment to fly on NASA’s “Vomit Comet.” (What’s special about this plane? It flies in parabolic paths (aka freefall) which result in near weightlessness. This means that you can ignore the effects of gravity when plan your experiment.) You want to mimic the orbital motion of the planets but by using electrostatic force rather than gravitational. And, instead of a planet, you will be orbiting a droplet of water that is 0.5mm in radius and has an deficit of 1.5 x 106 electrons. The droplet is to orbit around a small (1cm radius) sphere. If you want the droplet to move with an orbital radius of 14cm and period of one minute, what should the charge be on the central sphere? BTW The density of water is 997 kg/m3.

(a) What is the escape speed on a spherical asteroid whose radius is 274 km and whose gravitational acceleration at the surface is 0.444 m/s2? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 311 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 289.4 km above the surface?

Around 2.5 centuries ago, several physicists of the time came up with the notion of a dark star.  This was a star so dense, with so much gravity, that not even light could escape.  The calculations used Newtonian mechanics.  In class, we calculated the escape speed from the surface of the earth or the distance from the sun, and the mass of the planet or star. Here, the process is partially reversed.  Calculate the dark star radius from the mass of the star and the escape speed.  Answer in kilometers. c = 3*108 m/s M = 3.2*1030 kg G = 2/3 * 10-10 N*m2/kg2

A satellite in geostationary orbit (also called synchronous orbit) appears to remain stationary in the sky as seen from any particular location on the planet.  a.) In the future, there will be need for satellites in synchronous orbit around Mars to aid colonies. At what altitude would such a satellite need to be above the surface of Mars?Assume that the mass of Mars is 6.39 × 10^23 kg, the length of the Martian solar day (i.e., sol) is 24h 39m 35s, the length of the sidereal day is 24h 37m 22s, and the equatorial radius is 3396 km. (Hint: if you haven’t had a physics class before, you can find this by using the fact that the acceleration of an object in circular motion either as v2/r, where v and r are the velocity and radius of the orbit, or as 4Pi 2r/T2 , where T is the period. Use this second equation and Mathematical Insight 4.5 on p. 131 to find r for T=1 day. Make sure to use values for Mars nstead of Earth, as necessary. Alternatively, you can calculate the answer using Newton’s…