A day on Splorg lasts the equivalent of 31.7 hours. At the equator, Splorg has a circumference of 54,660 km. It takes a 2 kg mass exactly 0.56 s to fall a distance of 2 m at the surface of Splorg. At what radial distance would you need to orbit a satellite so that it always remained at the same location above the Splorigian equator (i.e. in splorgeosynchronous orbit)? At what velocity does the spacecraft need to reach in order for it to escape the gravity of Splorg.

On February​ 15, 2013, Asteroid 2012 DA14 passed within​ 17,200 miles​ [mi] of the surface of the Earth at a relative speed of 7.8 kilometers per second​ [km/s]. This is considerably closer than the orbit of geosynchronous satellites​ (26,200 miles). This is the closest recorded approach of an object this large. The asteroid 2012 DA14 was estimated to have a diameter of 30 meters​ [m] and a specific gravity of 3. If 2012 DA14 had hit the​ Earth, what is the total amount of energy that would have been released​ (i.e., what was the kinetic energy of the​ asteroid)? Express your answer in megatons​ [Mton]. One megaton is the energy released by one million metric tons of TNT explosive. A metric ton equals​ 1,000 kilograms​ [kg], and the explosive energy of TNT is 4,184 joules per gram​ [J/g].

I need help with this hw problem.

A satellite is in a circular orbit at an altitude of 1.5 earth radii, with an inclination of 90°.  Calculate the satellite's orbital period in minutes.  Report your answer to three significant digits (e.g. 123).  For the radius of the earth use RE = 6371 km; for the gravitation acceleration use g = 9.81 m/s2.  What is the orbital period in minutes?