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Arturiano [62]
3 years ago
11

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 1 AU would have an orbital period of about

__________ year(s).
Physics
1 answer:
Lunna [17]3 years ago
6 0

Explanation:

Orbital radius, r=1\ AU=1.496\times 10^{11}\ m

According to Kepler's law :

T^2\propto r^3

T^2=\dfrac{4\pi^2r^3}{GM}

Where

M is the mass of sun, M=1.98\times 10^{30}\ kg

T^2=\dfrac{4\pi^2\times (1.496\times 10^{11})^3}{6.67\times 10^{-11}\times 1.98\times 10^{30}}

T=\sqrt{1.0008\times 10^{15}}

T = 31635423.18 s

or

T = 1.0003 years

So, a  solar-system planet that has an orbital radius of 1 AU would have an orbital period of about 1.0003 year(s).

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Answer:

x = A cos (w \sqrt{2y_{o}/g})

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For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

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In the initial state

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the phase constant is included to take into account possible changes due to the collision of the mass.

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        cos (w’t + fi) = 0

        w’t + Ф = π / 2

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        Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

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