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January 30, 2015

January 30, 2015

Posted by **Michelle** on Tuesday, December 16, 2008 at 7:12pm.

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 7:46pmc)...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

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