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...

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?

Expert Answers
valentin68 eNotes educator| Certified Educator

a)

We have

`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

`F_1 =G*(m1*M)/R_1^2`

This force needs to be equal to the centripetal force

`F_(1cp)= m1*V_1^2/(R_1)`

Therefore

`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`

b)

For the faster planet 2 we also write the gravitational force of interaction with star and the centripetal force

`F_2 =G*(m2*M)/R_2^2`

`F_(2cp) =m2*V_2^2/R_2`

and observe that these two forces need to be equal.

`G*M/R_2^2 =V_2^2/R_2`

`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`