The period of a body orbiting the Earth is defined by T = 2Pisqrt(r^3/µ) where T = the orbit period in minutes, r = the orbital radius in meters and µ = the Earth's gravitational constant of 3.986365x10^14.
I'll let you punch out the numbers.
A space shuttle is in a circular orbit at a height of 250km, where the acceleration of Earth’s gravity is 93% of its surface value. What is the period of its orbit?
2 answers
Given:a = 0.93g
you have to find the radius of the earth which is 6.37*10^6m
R = REarth+ 250km =6.62v 106 m
v=2*pi*r/T a=v^2/R=(2*pi*R/T)^2=4*pi^2R/T
T=sqrt(4*pi^2*R/a
T=sqrt((4*pi^2)(6.62*10^6m) / (.93)(9.8m/s^2)
=5355s
=89min
you have to find the radius of the earth which is 6.37*10^6m
R = REarth+ 250km =6.62v 106 m
v=2*pi*r/T a=v^2/R=(2*pi*R/T)^2=4*pi^2R/T
T=sqrt(4*pi^2*R/a
T=sqrt((4*pi^2)(6.62*10^6m) / (.93)(9.8m/s^2)
=5355s
=89min