3 years ago

What is the sidereal period used in Kepler’s third law?

1 answer:

3 0

His law exaplins/shows that the average distance of a planet from the Sun cubed is directly proportional to the orbital period squared.

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Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1

zheka24 [161]

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$v_{f}=1721.1m/s$

Explanation:

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$dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\ (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s$

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