Question

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If the period of the first planet P1 is 750 years what is the mass, in kg, of the star it orbits around?

Answer 1

Answer:
`3.66(10)^(33)kg`

Step-by-step explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity
`\omega` of the planet P1 with a period
`T=750years=2.36(10)^(10)s`:

`\omega=(2\pi)/(T)=(V_(1))/(R)` (1)

Where:

`V_(1)=40.2km/s=40200m/s` is the velocity of planet P1

`R` is the radius of the orbit of planet P1

Finding
`R`:

`R=(V_(1))/(2\pi)T` (2)

`R=(40200m/s)/(2\pi)2.36(10)^(10)s` (3)

`R=1.5132(10)^(14)m` (4)

On the other hand, we know the gravitational force
`F` between the star S with mass
`M` and the planet P1 with mass
`m` is:

`F=G(Mm)/(R^(2))` (5)

Where
`G` is the Gravitational Constant and its value is
`6.674(10)^(-11)(m^(3))/(kgs^(2))`

In addition, the centripetal force
`F_(c)` exerted on the planet is:

`F_(c)=\frac{m{V_(1)}^(2)}{R^(2)}` (6)

Assuming this system is in equilibrium:

`F=F_(c)` (7)

Substituting (5) and (6) in (7):

`G(Mm)/(R^(2))=\frac{m{V_(1)}^(2)}{R^(2)}` (8)

Finding
`M`:

`M=(V^(2)R)/(G)` (9)

`M=((40200m/s)^(2)(1.5132(10)^(14)m))/(6.674(10)^(-11)(m^(3))/(kgs^(2)))` (10)

Finally:

`M=3.66(10)^(33)kg` (11) This is the mass of the star S