Question

n. 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?

Answer 1

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

                  $T_2 = \frac{T_1}{2}$