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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 barely provides the centripetal force needed for the rotation. (Why?) (a) Show then that the corresponding shortest period of rotation is given by where ρ is the density of the planet, assumed to be homogeneous. (b) Evaluate the rotation period assuming a density of 3.0 g/cm², typical of many planets, satellites, and asteroids. No such object is found to be spinning with a period shorter than found by this analysis.

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator barely provides the centripetal force needed for the rotation. (Why?) (a) Show then that the corresponding shortest period of rotation is given by
where ρ is the density of the planet, assumed to be homogeneous. (b) Evaluate the rotation period assuming a density of 3.0 g/cm², typical of many planets, satellites, and asteroids. No such object is found to be spinning with a period shorter than found by this analysis.
 

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Give data:
Mass of the Moon, M0 = 7.36 × 1022 kg
Mean radius of the Moon, R0 = 1.74 × 106 m
Gravitational constant, G = 6.67× 10-11 N\cdot m / kg2
Weight of the object on the surface of the Earth, WEarth = 100 N
Free-fall acceleration on the surface of the Earth, g = 9.83 m / s2
Mass of the Earth, M = 5.98 × 1024 kg

234-1836_f23.JPG
(b) Weight of a body is equal to the mass of the object times the free-fall acceleration.Consider the mass of the object to be .
Thus, the weight of the body on the Moon is
234-1527_f24.JPG
234-1882_f25.JPG
(c) Let R1 be the corresponding radius of the Earth.If the weight on the surface of the Earth is equal to the weight on the surface of the Moon, then the weight of the object is
234-729_f26.JPG

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