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 the rotation. Deriving an expression for the shortest period of rotation T (using G gravitational constant,M mass of the planet, p density of the planet and R radius of the planet) Calculate the rotation period in hours assuming a density of 7.3 x 10³ kg/m³. Your answer should be with 2 decimal points such as 1.23

Physics for Scientists and Engineers, Technology Update (No access codes included)
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Chapter1: Physics And Measurement
Section: Chapter Questions
Problem 1.35P: A rectangular plate has a length of (21.310.2) cm and a width of (9.810.1) cm. (Calculate the area...
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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 the rotation.
Deriving an expression for the shortest period of rotation T (using G gravitational
constant,M mass of the planet, p density of the planet and R radius of the planet)
Calculate the rotation period in hours assuming a density of 7.3 x 10³ kg/m³.
Your answer should be with 2 decimal points such as 1.23
Transcribed Image Text: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 the rotation. Deriving an expression for the shortest period of rotation T (using G gravitational constant,M mass of the planet, p density of the planet and R radius of the planet) Calculate the rotation period in hours assuming a density of 7.3 x 10³ kg/m³. Your answer should be with 2 decimal points such as 1.23
Where required use Mass of Earth 6 x1024 Kg and radius of earth = 6.4 x106 m. density of Earth p
=5200Kg/m³
Transcribed Image Text:Where required use Mass of Earth 6 x1024 Kg and radius of earth = 6.4 x106 m. density of Earth p =5200Kg/m³
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