Question

On November 12, 2014, the Philae lander separated from the European Space Agency's Rosetta spacecraft, and made a landing (after bouncing twice) on a comet known as 67P/Churyumov-Gerasimenko. In this problem, we will explore the gravitational environment that Rosetta (orbiting the comet) and Philae (on the comet's surface) experienced. On September 30, 2016, Rosetta crashed (as planned) on the comet. We will use a very simple model of the comet, modeling it as a perfect sphere. This is not at all the case (some have compared the shape of the comet to a duck), but it will get us a first approximation to the situation. For the purposes of this problem, use the following data: mass of Comet 67P is 1 × 1013 kg; radius of Comet 67P is 2.20 km; mass of Philae = 100 kg; mass of Rosetta = 1000 kg; radius of Rosetta's orbit was 22.8 km (a) What is the acceleration due to gravity on the surface of the comet, where Philae is? (b) What is the escape speed at the surface of the comet? This is the speed above which Philae (or any other object) would not have fallen back to the comet's surface, but would have just gone off into space. For reference, Philae's actual rebound speed off the surface on its first bounce was 0.38 m/s. (c) What was the period of Rosetta's circular orbit around the comet? Express this in days. (d) The comet itself (now accompanied by both Rosetta and Philae) is traveling in an elliptical orbit with a period of about 6.5 years, with the Sun at one focus of the ellipse. As it travels, which one of these quantities is conserved (assuming no loss of mass)? translational kinetic energy, sum of translational + rotational kinetic energy, linear momentum, gravitational potential energy (of the comet-Sun system), angular momentum, speed ,velocity (its only one of these choices)

Answer 1

we know that acceleration due to gravity is given by

       g = GM/R²   where G = gravitational constant = 6.67 * 10^-11 Nm²kg^-2     M=mass of comet R= radius of comet

a. acceleration due to gravity on the surface of comet g = 6.67 * 10^-11 * 10¹³ /(2.2 * 10³)²

                                                                               g = 1.37 * 10^-4 m/s²

b. escape velocity is given by

                   

                    

                      = 778 m/s             [approx]

c. period of Rosetta's circular orbit is given by

                     T = 2(pi)        where r = raius of rosetta's orbit.

                     T = 2* 3.14

                     T = 6.28 * 13.3 * 10⁴

                        = 8.35 * 10⁵ s

                        = 2.65 years