A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that s

Question

A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth’s orbital period? A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth’s orbital period?

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Bơ 5 years 2021-08-14T04:45:17+00:00 1 Answers 181 views 0

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  1. ngockhue
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    2021-08-14T04:47:08+00:00
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    Answer:

    The orbital period of the  planet would be half that of the earth

    Explanation:

    From the question we are told that

             The mass of star is four times the mass of sun which can be mathematically represented as

                            M_{star } = 4M_{sun}

    Mathematically gravitational potential is given as

                              F = \frac{K M m}{r^2}

      Where M is mass one

                   m is mass two

      From this equation we see that the attraction force is directly proportional to the mass of the star

        Thus we can say that

                  F_{planet } = 4 F_{earth }

    The centrifugal force that balances this attraction Force  is  

                    F = a_c* m

    Where  a_c is the centrifugal acceleration which can be mathematically represented as a_c = \frac{r}{T^2}

              and  m is the mass

        Substituting this into the equation for centrifugal force

                             F = \frac{r}{T^2}*m

           substituting  into the equation above

                        \frac{r}{T_2^2} *m = 4(\frac{r}{T_1^2} *m)

    Given that the diameter is the same and assuming that the mass is constant

             Then

                        \frac{1}{T_2^2} = 4(\frac{1}{T_1^2} )

                       T_2 ^2 = \frac{T_1^2}{4}

    Take square root of both sides

                      T_2 = \frac{T_1}{2}

                     

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