m earth = 5.9742 x 10^24 kg r earth = 6.3781 x 10^6 m m moon = 7.36 x 10^22 kg r

Question

m earth = 5.9742 x 10^24 kg r earth = 6.3781 x 10^6 m m moon = 7.36 x 10^22 kg r moon = 1.7374 x 10^6 m d earth to moon = 3.844 x 10^8 m (center to center) G = 6.67428 x 10^-11 N-m2/kg2 A 1400 kg satellite is orbitting the earth in a circular orbit with an altitude of 1400 km. 1) How much energy does it take just to get it to this altitude? 2)How much kinetic energy does it have once it has reached this altitude? 3)What is the ratio of the this change in potential energy to the change in kinetic energy? (i.e. what is (a)/(b)?) 4)What would this ratio be if the final altitude of the satellite were 4

Explanation / Answer

Angular momentum of the earth = I * ?, where I = moment of inertia of the earth as a sphere about its axis of rotation = (2/5)MR^2 and ? = angular velocity = 2?/(24*3600) radian/sec => angular momentum = [(2/5)MR^2] * [2?/(24*3600)] Plugging mass of the earth, M = 5.9742

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