Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy.
The orbital speeds of the planets are determined to be 43.3 km/s and 58.6 km/s. The slower planet's orbital period is 7.60 years. a) What is the mass of the star? b) What is the orbital period of the faster planet, in years?
`V_1 =43.3 (km)/s` and `V_2 =58.6 (km)/s`
For the slower planet 1 the angular speed is
`omega_1 =(2*pi)/T =V_1/R_1`
`R1 = T_1*V_1/(2*pi) =7.6*365*24*3600*43300/(2*pi)=1.652*10^12 m`
The gravitational force between the star and the slower planet 1 is
This force needs to be equal to the centripetal force
`GM/R_1^2 = V_1^2/R_1`
`M =(R_1*V_1^2)/G =(1.652*10^12*43300^2)/(6.67*10^-11) =4.64*10^31 kg`
To compare, the mass of Sun is `M_(sun) = 1.99*10^30 kg`
For the faster planet 2 we also write the gravitational force of interaction with star and the centripetal force
and observe that these two forces need to be equal.
`R_2 =G*M/V_2^2 =6.67*10^-11*(4.64*10^31)/58600^2 =9*10^11 m`
Then the orbital period is simply
`T_2 =2*pi*R_2/V_2 =2*pi*(9*10^11)/58600 =9.67*10^7 s =3.066 years`