Posted by **Adam** on Friday, May 14, 2010 at 2:58pm.

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?

- college physics -
**tchrwill**, Sunday, May 16, 2010 at 10:02am
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.

- college physics -
**find any other answers let me know!!!**, Saturday, November 27, 2010 at 12:33am
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

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