Question

The International Space Station is orbiting at an altitude of about 370 km above the
earth's surface. The mass of the earth is 5.976 × 10
24
kg and the radius of the earth is
6.378 × 10
6 m.
(a) Assuming a circular orbit, what is the period of the International Space Station's
orbit?
(b) Assuming a circular orbit, what is the speed of the International Space Station in its
orbit

Answer 1

(a)& Assuming a circular orbit, what is the period of the International Space Station's
orbit?

The gravitation force is equal to the centripetal force:

GMm/r² = mV²/r, or GM/r = V².

T = 2pi*r / V
V² = [ 2pi r / T ]²
GM/r = [ 2pi r / T ]²
6.67*10^(-11) * 5.98*10^24 / [ (6.378*10^6 + 370*10³ ) ]³ = [ 2pi / T]²
T = 5514s

(b)& Assuming a circular orbit, what is the speed of the International Space Station in its
orbit

V = sqrt [ 6.67*10^(-11) * 5.98*10^24 / [ [ (6.378*10^6 + 370*10³ )]
V = 7690 m/s.