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swin2013

  • 3 years ago

If the value for k for planets moving around our sun is 3.35 *10^18 m^3s^2, what would the period of a planet be if its average radius to the sun is 3.2*10^8?

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  1. Shadowys
    • 3 years ago
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    I believe you're familiar with Kepler's third law, \(\frac{T^2 }{r^3}=k\)?

  2. swin2013
    • 3 years ago
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    T^2 = 4pi^2/GM a^3?

  3. Shadowys
    • 3 years ago
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    Yeah, but you don't need that complicated equation here since k is given as 3.35 *10^18 m^3s^2.

  4. swin2013
    • 3 years ago
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    will it be about 1.05 ?

  5. Shadowys
    • 3 years ago
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    You used this? \(r^3 = \frac{10.24 \times 10^{16}}{3.35 \times 10^{18}}\)

  6. swin2013
    • 3 years ago
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    T = K a^3 sqrt/3.35E18 * (3.2E8)^3

  7. swin2013
    • 3 years ago
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    because they want to find T right?

  8. swin2013
    • 3 years ago
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    yea my answer is wrong -_- lol

  9. Shadowys
    • 3 years ago
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    lol yh sorry about that. Eyes get blurry in a hurry. \(1.05 \times 10^{21} s\) yup you're right.

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