Assuming Earth behaves as a perfect sphere with a radius of 6380 km, the standard gravitational acceleration,g, at sea level has a value of 9.80 m/s^2.
a.) As Earth rotates on its axis, objects at the equator experience a slightly different value of g due to additional centripetal acceleration. Calculate this adjusted value for g.
b.) Is this difference due to centripetal forces present at the poles? If g is dependent upon the position on the surface of Earth, why do we use the standard value of 9.80 m/s^2 in our calculations? Explain.
c.) Whats the value of g on board a space shuttle in a stable orbit at a height of 350 km above the equator? Hint: You don't need to know the mass of the space shuttle.
physics - bobpursley, Tuesday, December 16, 2008 at 7:46pm
c)...in the orbit, folks are weightless.
a,b I will be happy to critique your thinking. YOu know velocity at the equator...r*2PI/24hrs