# 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?

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### 1 Answer

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`

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