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3 January, 09:25

Two very small planets orbit a much larger star. Planet A orbits the star with a period TA. Planet B orbits the star at four times the distance of planet A, but planet B is four times as massive as planet A. Assume that the orbits of both planets are approximately circular.

Planet B orbits the star with a period TB equal to

a. TA/4.

b. 16TA.

c. 8TA.

d. 4TA

d. TA

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Answers (1)
  1. 3 January, 09:32
    0
    Given that,

    Period of Planet A around the star

    Pa = Ta

    Let the semi major axis of planet A from the star be

    a (a) = x

    Given that, the distance of Planet B from the star (i. e. it semi major axis) is 4 time that of Planet A

    a (b) = 4 * a (a) = 4 * x

    a (b) = 4x

    Also planet B is 4 times massive as planet A

    Using Kepler's law

    P² ∝a³

    P²/a³ = k

    Where

    P is the period

    a is the semi major axis

    Then,

    Pa² / a (a) ³ = Pb² / a (b) ³

    Pa denotes Period of Planet A and it is Pa = Ta

    a (a) denoted semi major axis of planet A and it is a (a) = x

    Pb is period of planet B, which is the required question

    a (b) is the semimajor axis of planet B and it is a (b) = 4x

    So,

    Ta² / x³ = Pb² / (4x) ³

    Ta² / x³ = Pb² / 64x³

    Cross multiply

    Ta² * 64x³ = Pb² * x³

    Divide both sides by x³

    Ta² * 64 = Pb²

    Then, Pb = √ (Ta² * 64)

    Pb = 8Ta

    Then, the period of Planet B is eight times the period of Planet A.

    The correct answer is C
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