Question

If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m.

Answer 1

Assuming that earth is a perfect circle. First solve the circumference of the earth,

C = 2pi( r )

Where are is the radius

C = 2pi ( 6.38 x 10^6 m)

C = 4.01 x 10^7 m

O the orbital period

T = 4.01 x 10^7 m / 7,750 m/s

T = 5172.48 s