Our Sun rotates about the center of our Galaxy (mG≈4×1041kg) at a distance of about 3×104 light-years

Question

[1ly=(3.00×108m/s)⋅(3.16×107s/yr)⋅(1.00yr)].

What is the period of the Sun’s orbital motion about the center of the Galaxy?

Express your answer using one significant figure.

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General Physics PRemastered 12 months 1 Answer 1268 views 0

Answer ( 1 )

  1. For the orbital period (in seconds), the expression is –

    T = 2*pi*sqrt(r^3/GM)

    Where G = gravitational constant and M = Central mass

    Now 1 ly = (speed of light in m/s) times (1 year in seconds)
    = (3 x 10^8) (3.15 x 10^7) = (3 x 3.15) x 10^(8+7) = 9.45 x 10^15 metres

    therefore, our orbit of 3 x 10^4 ly is a distance of (3 x 10^4)(9.45 x 10^15) = 28.35 x 10^19

    = 2.835 x 10^20 m

    Again value of Gravitational constant, G = 6.67 × 10^(-11)
    And the mass of the galaxy, M = 4 x 10^41 kg

    Now put the values –

    T = 2*pi*sqrt(r^3/GM) = (2)(3.14)SqRoot((2.835 x 10^20)^3 / (6.67 × 10^(-11) x 4 x 10^41))

    solving this we get –

    T = 2*3.14*SQRT(0.854 x 10^30)

    = 6.28 x (0.854)^(1/2) x 10^(15)
    = 5.8 x 10^15 seconds

    Answer in one significant figure –

    T = 6.0 x 10^15 sec.

    Best answer

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