# Answer to Question #6048 in Mechanics | Relativity for Erica Evoli

Question #6048

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

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

Expert's answer

(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.

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.

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