Answers - Question 3
Take the radius of the Earth as 6.4x106 meters and the radius of Jupiter as 7.15x107 meters. Jupiter turns on its axis once every 9.8 hours (0.41 days. The acceleration due to gravity at the surface of Jupiter is 24 m/s/s.
a Find the centripetal acceleration of an object at the equator on Earth, and at the equator on Jupiter.
b Find the percentage by which the acceleration due to gravity is reduced at the equator by the rotation of the Earth, and of Jupiter.
1a The acceleration is given by ... v2/R
... where R is the radius of the Planet. v = 2pR/T ... where T is the rotation period in seconds. At once ... a = 4p2R/T2 For earth the centripetal acceleration is given by ... = 4 x 9.86 x 6.4x106/ (24 x 60 x 60)2 = 252x106/ 74.68 = 0.034 m/s/s b........ which gives a 0.34% reduction in g at the surface.
2a For Jupiter the centripetal acceleration is given by ...
a = 4p2R/T2 = 4 x 9.86 x 7.15x107/ (9.8 x 60 x 60)2 = 196x107/ 12.48 = 2.27 m/s/s b........ which gives a 9.4% reduction in g at the surface.
c Explain briefly in physical terms why Jupiter has a more pronounced equatorial bulge than the Earth.
The planets are not rigid bodies - the shape deforms from spherical because of the rotation. The effect is more pronounced on Jupiter because the increased radius and rotation rate give a much higher centripetal acceleration as a percentage of g(and hence much greater distorting forces at the equator.