Question

The International Space Station is orbiting at an altitude of about 370 km above Earth's surface. The mass of the earth is 5.976 x 10^24 kg and the radius of earth is6.378 x 10^6 m.
a) Assuming circular orbit, what is the period of the International Space Station's Orbit?
b) Assuming circular orbit , what is the speed of the International Space Station in it's orbit?

Answer 1

R=distance from center of the earth to the Space Station = 6.378e06+370e03 m
R= 6.748e06 m

The circular orbit of the Space Station resulted from the attraction force between the Earth and the Space Station

GMm / R² = mv² / R

where m = mass of the Space Station, M = mass of the Earth, G = 6.6726 x 10^-11 N m²/kg²

v = √(GM/R) = √(6.6726 x 10^-11*5.976 x 10²⁴/6.748 x 10⁶) = 7687 m/s -> part (b)

v= ωR , ω = v/R = 7687 / 6.748e06 = 0.0011392 rad/s

T = 2∏/ω = 5515.44 s -> part (a)