Suppose the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the samerotation speed. How much longer would a solar day

Question

Suppose the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the samerotation speed. How much longer would a solar day be than a sidereal day? Mercury’s siderealday is 58.6 days and its orbital period is 88 days. What is the length of Mercury’s solar day?

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Thành Công 2 months 2021-07-31T23:52:52+00:00 1 Answers 1 views 0

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    2021-07-31T23:54:02+00:00

    Answer:

    A) 0.1955 days

    B) 35.18 days

    Explanation: Given that the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the same rotation speed. That is the orbital period will be

    365.24/12 = P/4

    P = 121.75 days

    We have three periods, the orbital period Po = 121.75 days,

    the sidereal period Ps = 23.93419 hours and

    The mean solar day Pd =24 hours.

    Since they had exactly the same rotation speed, the number of solar days in a year is Po/Pd.

    The number of sidereal days in a year would be Po/Pd + 1 = 121.75 + 1 = 122.75 days. Since there are more sidereal days in a year this means that the sidereal day is a little bit shorter, 121.75/122.75 = 0.99185 times shorter.

    0.99185 × 24 = 23.80448 hours.

    24 – 23.80448 = 0.1955 days

    Therefore, solar day is 0.99185 days longer than a sidereal day

    Given that the Mercury’s side real day is 58.6 days and its orbital period is 88 days. What is the length of Mercury’s solar day?

    Using the formula

    Ld = PoPs/(Po + Ps)

    Ld = Length of Mercury’s solar day

    Po = orbital period

    Ps = Mercury’s side real day

    Ld = (88 × 58.6)/ (88 + 58.6)

    Ld = 5156.8/146.6

    Ld = 35.18 days

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