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24 September, 01:44

Use Kepler's third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and the Sun is 1.496 * 1011 m. Earth's orbital period around the Sun is 365.26 days. 6.34 * 1029 kg 1.99 * 1030 kg 6.28 * 1037 kg 1.49 * 1040 kg

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Answers (2)
  1. 24 September, 02:51
    0
    its 1.99 x10^30 just answered it
  2. 24 September, 03:27
    0
    Kepler’s third law formula: T^2=4pi^2*r^3 / (GM)

    We’re trying to find M, so:

    M=4pi^2*r^3 / (G*T^2)

    M=4pi^2 * (1.496 * 10^11 m) ^3 / ((6.674 * 10^-11N*m^2/kg^2) * (365.26days) ^2)

    M=1.48 * 10^40 (m^3) / ((N*m^2/kg^2) * days^2))

    Let’s work with the units:

    (m^3) / ((N*m^2/kg^2) * days^2)) =

    = (m^3*kg^2) / (N*m^2*days^2)

    = (m*kg^2) / (N*days^2)

    = (m*kg^2) / ((kg*m/s^2) * days^2)

    = (kg) / (days^2/s^2)

    = (kg*s^2) / (days^2)

    So:

    M=1.48 * 10^40 (kg*s^2) / (days^2)

    Now we need to convert days to seconds in order to cancel them:

    1 day=24 hours=24*60minutes=24*60*60s=86400s

    M=1.48 * 10^40 (kg*s^2) / ((86400s) ^2)

    M=1.48 * 10^40 (kg*s^2) / (86400^2*s^2)

    M=1.48 * 10^40kg/86400^2

    M=1.98x10^30kg

    The closest answer is 1.99 * 10^30

    (it may vary a little with rounding - the difference is less than 1%)
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