The International Space Station orbits the Earth (5.98 x 1024 kg) in a circle of radius 6.77 x 106 m. What is its orbital period in hours?

Answer 1

A variation of Kepler's third law is:

allowing you to solve for the period of an orbiting object.

T is the period of the ISS, G is the universal gravitational constant which is 6.674 x 10⁻¹¹ m³/(kg s²), M_E is the mass of earth (given in your problem as 5.98 x 10²⁴ kg) and r is the radius of the ISS (radius of the earth plus the altitude of the ISS which is given in your problem as 6.77 x 10⁶ m)

So T = (2π/√(6.674 x 10⁻¹¹•5.98 x 10²⁴))(6.77 x 10⁶)^3/2

T = 5540 seconds and to convert to hours, one hour is 3600 seconds so 5540/3600 = 1.54 hours