Question

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for rotation. A. Show that the corresponding shortest period of rotation is: 3п T= | Gр Here, p is the uniform density (mass per unit volume) of the spherical planet. I B. Compute the rotation period assuming a density of 3.0 g/cm”, typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis.

Answer 1

® The gravitational acceleration is: g= GM of centripetal acceleration R2 a=Rw? =R (21)? a= uOR T2 7 GM T2 =) T²- unar? R2 = UnR GM = 3.14 R3 ) GM 3 un Gm = 31.1 M TE 311 GP B T 30 667X10 "x3000 4.708 no-4+11 =6.8618103 T=6861 second