Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars’s orbital p

Question

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars’s orbital period is 687 days, and Mars’s distance from the Sun is 2.279 × 1011 m. 1.49 × 1011 m 1.49 × 1033 m 3.34 × 1011 m 3.34 × 1033 m

The answer is A 1.49x1011m

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Orla Orla 5 years 2021-08-29T20:46:53+00:00 2 Answers 19 views 0

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  1. ngocdiep
    0
    2021-08-29T20:47:56+00:00
    Reply

    Answer:

    1.49 x 10^11

    Explanation:

    I just did it EDG 2020 Option A

  2. mit
    0
    2021-08-29T20:48:42+00:00
    Reply

    Answer:

    1.49 x 10^{11}

    Explanation:

    Kepler’s third law states that The square of the orbital period of a planet is directly proportional to the cube of its orbit.

    Mathematically, this can be stated as

    T^{2} ∝ R^{3}

    to remove the proportionality sign we introduce a constant

    T^{2} = kR^{3}

    k = \frac{T^{2} }{R^{3} }

    Where T is the orbital period,

    and R is the orbit around the sun.

    For mars,

    T = 687 days

    R = 2.279 x 10^{11}

    for mars, constant k will be

    k = \frac{687^{2} }{(2.279*10^{11}) ^{3} } = 3.987 x 10^{-29}

    For Earth, orbital period T is 365 days, therefore

    365^{2} = 3.987 x 10^{-29} x R^{3}

    R^{3} = 3.34 x 10^{33}

    R = 1.49 x 10^{11}

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