An 800 kg satellite orbits the Earth at an average height of 450.0 km above the surface.

Question

An 800 kg satellite orbits the Earth at an average height of 450.0 km above the surface. (The radius of the Earth is 6.38 x 103 km, while the mass of the Earth is 5.98 x 1024 kg.) Determine:

a) the force of gravitational attraction between them

b) the orbital speed of the satellite around the Earth.

c) the orbital period (in hours) of the satellite (the time it takes to make 1 revolution around Earth).

Answer 1

a) Gravitational force F_g = G M m / r². The distance r is from the center of Earth, so r = 6,380 + 450 km = 6.83 x 10³ km • (1,000 m/1 km) = 6.83 x 10⁶ m. F_g = (6.67 x 10^-11) (5.98 x 10²⁴) (800) / (6.83 x 10⁶)² = 6,840 N (on Earth's surface, the force would be mg = 7,850 N).

b) Orbital speed v = √[G M_E / r] = √[(6.67 x 10^-11) (5.98 x 10²⁴) / (6.83 x 10⁶)] = 7,640 m/s (17,100 mph).

c) The orbital speed v is circumference of the orbit, 2π r, divided by the time T to complete one orbit, so T = 2π r / v = 2π (6.83 x 10⁶ m) / (7.64 x 10³ m/s) = 5620 s • (1 hour /3600 s) = 1.56 hour.