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.

In atoms, the electron carries a negative elementary or unit charge; the proton carries a positive charge. the two types of charge are equal and opposite. in an atom of matter, an electrical charge occurs whenever the number of protons in the nucleus differs from the number of electrons surrounding that nucleus.