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
[tex]M_{star } = 4M_{sun}[/tex]
Mathematically gravitational potential is given as
[tex]F = \frac{K M m}{r^2}[/tex]
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
[tex]F_{planet } = 4 F_{earth }[/tex]
The centrifugal force that balances this attraction Force is
[tex]F = a_c* m[/tex]
Where [tex]a_c[/tex] is the centrifugal acceleration which can be mathematically represented as [tex]a_c = \frac{r}{T^2}[/tex]
and m is the mass
Substituting this into the equation for centrifugal force
[tex]F = \frac{r}{T^2}*m[/tex]
substituting into the equation above
[tex]\frac{r}{T_2^2} *m = 4(\frac{r}{T_1^2} *m)[/tex]
Given that the diameter is the same and assuming that the mass is constant
Then
[tex]\frac{1}{T_2^2} = 4(\frac{1}{T_1^2} )[/tex]
[tex]T_2 ^2 = \frac{T_1^2}{4}[/tex]
Take square root of both sides
[tex]T_2 = \frac{T_1}{2}[/tex]
