You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m an

Question

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 3 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 44.6 m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2)

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Thạch Thảo 5 years 2021-08-25T14:13:25+00:00 1 Answers 459 views 1

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  1. thuthuy
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    2021-08-25T14:15:06+00:00
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    Answer:

    a) M_{planet} = 5.416\times 10^{25}\,kg, b) M_{star} = 1.325\times 10^{31}\,kg

    Explanation:

    a) Gravity constant on the surface is equal to:

    g = \frac{G\cdot M_{planet}}{r_{planet}^{2}}

    The mass of the planet can be determined by clearing its variable in the expression:

    M_{planet} = \frac{g\cdot r_{planet}^{2}}{G}

    M_{planet} = \frac{(44.6\,\frac{m}{s^{2}} )\cdot(0.9\times 10^{7}\,m)^{2}}{6.67\cdot 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} }

    M_{planet} = 5.416\times 10^{25}\,kg

    b) By using Newton’s Laws, the equation of equilibrium of the planet with respect to the star is:

    \Sigma F_{r} = G\cdot\frac{M_{planet}\cdot M_{star}}{r^{2}} = M_{planet}\cdot \frac{v^{2}}{r}

    G\cdot \frac{M_{star}}{r} = v^{2}

    M_{star} = \frac{v^{2}\cdot r}{G}

    The translation speed of the planet is:

    v = \frac{2\pi\cdot (3\times 10^{11}\,m)}{(402\,days)\,(\frac{24\,h}{1\,day} )\,(\frac{3600\,s}{1\,h} )}

    v = 54270.188\,\frac{m}{s}

    The mass of the star is:

    M_{star} = \frac{(54270.188\,\frac{m}{s} )^{2}\cdot (3\times 10^{11}\,m)}{6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} }

    M_{star} = 1.325\times 10^{31}\,kg

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