The space shuttle orbits the earth 16 times per day. mearth = 5.975 x 1024 kg ...

The space shuttle orbits the earth 16 times per day.

m_earth = 5.975 x 10²⁴ kg r_earth = 6.371 x 10⁶ m G = 6.67 x 10^-11 N^.m²/kg²

How high (in miles) above the earth does it orbit? Height “h” above the Earth’s surface?
What is the acceleration due to gravity at this height? a_Centrip = a_Grav
What is the period (in hours) of revolution? T = ? 1/f f Þ rev/hr

w = 16 rev/dy = 16rev/24h = 0.667 rev/h

= 0.667rev/h(2prad/rev)(1h/3600s) = 1.16x10^-3 rad/s

F_Centrip = F_Grav m_Sa_Centrip = (Gm_Sm_E)/r²

Expert Solution

genius_generous answered 3 years ago

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