A satellite is in elliptical orbit with a period of 7.20 104 s about a planet of mass 7.00 1024 kg. At aphelion, at radius 5.1 107 m, the sa

Question

A satellite is in elliptical orbit with a period of 7.20 104 s about a planet of mass 7.00 1024 kg. At aphelion, at radius 5.1 107 m, the satellite’s angular speed is 4.987 10-5 rad/s. What is its angular speed at perihelion?

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Tryphena 4 years 2021-07-28T11:15:50+00:00 1 Answers 17 views 0

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  1. Maris
    0
    2021-07-28T11:17:00+00:00
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    Answer:

    1.64x10^{-4}rad/s

    Explanation:

    Given:

    satellite is in elliptical orbit with a period T= 7.20 x 10^{4} s

    Mass of planet m=  7.00x 10^{24} kg

    Satellite’s angular speed ωa- = 4.987 10-5 rad/s

    At aphelion, at radius Ra= 5.1 x 10^7 m

    Assuming torque is zero for this system, therefore, the conservation of angular momentum can be defines as

    Lp = La

    Ip ωp = Ia ωa—>eq(1)

    Where,

    ‘a’ represents aphelion and ‘p’ represents perihelion

    ω represents  angular velocity

    and I represents the rotational inertia

    Since, I= 1/2 mR²

    Here R is the radius at  aphelion /  perihelion

    m = mass of planet

    eq(1)=> ωp= Ia ωa/ Ip

    ωp= (1/2 m Ra² ωa) / 1/2 m Rp²

    ωp= (Ra/Rp)² ωa —>eq(2)

    In order to find Rp, we use : 2a= Rp + Ra

    where a represents semimajor axis

    with the help of Kepler’s third law for elliptical orbits

    a= ∛(GmT²/4π²)

    a= ∛ 6.67x 10^{-11} x 7.00x 10^{24} x (7.20 x 10^{4})² / 4π²

    a=  3.97 x 10^{7}m

    2a= Rp + Ra

    Rp= 2a-Ra

    Rp= 2 x 3.97 x 10^{7}–  5.1x 10^{7}

    Rp= 2.84 x 10^{7}m

    Substituting all the required values in eq 2, we have

    ωp= (Ra/Rp)² ωa

    ωp= (5.1x 10^{7}/2.84 x 10^{7})² x  4.987 x 10^{-5

    ωp= 3.224 x 4.987 x 10^{-5

    ωp= 1.64x10^{-4}rad/s

    Therefore, angular speed at perihelion is 1.64x10^{-4}rad/s

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