How fast would an object have to travel on the surface of Jupiter at the equator to keep up with the Sun (that is, so the Sun would appear to remain in the same position in the sky)? Use the facts that the radius of Jupiter is approximately 44,360 miles and its revolution is approximately 10 hours.

27872.2 miles per hour

Explanation:

Given that :

Radius of Jupiter is approximately = 44,360 miles

Revolution is 10 hours ;

Jupiter makes one revolution in 10 hours :

Using the relation to obtain the velocity :

V = re

r = radius

w = 2π/T

Hence,

V = r * 2π/ T

V =44360 * 2 * π/10

V = 88720 * π/10

V = 278722.10 / 10

V = 27872.210

V = 27872.2 miles per hour

force tends to be low and long diustance masses are high