Jupiter's rotation period is about 10 hours and its radius is 70, 400km. Work out the speed of a cloud near the equator in km/s, recognizing that it must move around the circumference of Jupiter in one rotation period. Compare that speed with the speed of a point on the equator at Earth. What can you say about hte relative importance of ratational effects in the two systems?Whats the actual calculation that relates Jupiter's rotation%26amp;speed of cloud near equartor?
Hi. Radius * 2 = diameter. Diameter times pi = circumference. Circumference / 10 equals km/h.
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment