The international space station makes 15.65 revolutions per day in its orbit around the earth. part a assuming a circular orbit, how high is this satellite above the surface of the earth?

380 km The period of an orbit is expressed by the equation p = sqrt((4*pi^2/u)*a^3) where p = period of orbit u = standard gravitational parameter (product of G and M. For Earth, u is 3.986004418Ă—10^14 m^3/s^2. u is much better to use than either G or M since it's known to a higher precision). a = semi-major axis of orbit. Solving for a, gives p = sqrt((4*pi^2/u)*a^3) p^2 = (4*pi^2/u)*a^3 up^2/(4pi^2) = a^3 (up^2/(4pi^2))^(1/3) = a ((p^2)u/(4pi^2))^(1/3) = a Substituting constants ((p^2)3.986004418Ă—10^14 m^3/s^2/(4*3.141592654^2))^(1/3) = a ((p^2)1.009666714x10^13 m^3/s^2)^(1/3) = a There are 86400 seconds in a day, so the orbital period of the space station is 86400/15.65 = 5520.766773 seconds. Substituting the value for the period and solving gives: ((p^2)1.009666714x10^13 m^3/s^2)^(1/3) = a (((5520.766773 s)^2)1.009666714x10^13 m^3/s^2)^(1/3) = a ((30478865.76 s^2)1.009666714x10^13 m^3/s^2)^(1/3) = a (30478865.76 *1.009666714x10^13 m^3)^(1/3) = a (3.077349624x10^20 m^3)^(1/3) = a 6751375.759 m = a 6751.375759 km = a So the semi-major axis is about 6751 kilometers. Now we need to subtract the radius of the earth to get the altitude of the space station. So 6751.375759 km - 6371 km = 380.375 km Rounding to 4 significant figures gives 380 km.