what is the magnitude of centripetal accelerations of an object on Earth's equator due to the rotation of earth? what would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration of magnitude 9.8m/s^2?
2006-10-14 15:28:19 · 3 answers · asked by Galaxy D 2 in Science & Mathematics ➔ Physics
The magnitude of centripetal acceleration (acceleration due to gravity) is 9.8 m/s^2
(for the second part of the question you are going to have to find out what the rotation period of the earth actually is) sorry i dont know i heard it is 4s or something but i could be wrong...
2006-10-14 15:33:03 · answer #1 · answered by shelly 4 · 0⤊ 0⤋
a careful reading of http://curious.astro.cornell.edu/question.php?number=310 shows that the centripetal acceleration is 0.3% that of g at the equator
2006-10-14 16:42:39 · answer #2 · answered by arbiter007 6 · 0⤊ 0⤋
shelly_lynn90, the rotation period of the earth is maybe 1 per day?
Look at this site: http://curious.astro.cornell.edu/question.php?number=310
The conclusion is "Taking into account both of the above effects, the gravitational acceleration is 9.78 m/s2 at the equator and 9.83 m/s2 at the poles, so you weigh about 0.5% more at the poles than at the equator."
2006-10-14 16:11:21 · answer #3 · answered by sofarsogood 5 · 0⤊ 0⤋