# A planet orbits a nearby star. The mass of the planet is 1.12 Jupiter masses and the mass of the star is 1 solar mass. The distance between planet and star is 0.023 AU. The orbits are circular. (1...

A planet orbits a nearby star. The mass of the planet is 1.12 Jupiter masses and the mass of the star is 1 solar mass. The distance between planet and star is 0.023 AU. The orbits are circular. (1 Mjupiter = 1.89 x 10^27 kg, 1 Msun = 1.99 x 10^30 kg, 1 AU = 1.496 x 10^8 km)

a. Determine the distance from the center of the star to the center of the mass of the system and the distance from the center of the planet to the center of mass of the system.

b. The period of the star's orbit is 1.34 Earth days. What is the speed of the star in its orbit in km/s? (Hint, remember that speed=distance/time and the circumference of a circle=2piR. What is the radius of the star's orbit?)

*print*Print*list*Cite

a) Let's denote the distance from the center of the star to the center of mass of the system by `r_s ` and the distance from the center of the planet to the center of mass of the system by `r_p ` .

Then, the sum of these distances will equal the distance between the planet and a star, d:

` r_s + r_p = d =1 AU `

By property of the center of mass, the following has to be true:

`m_s*r_s = m_p*r_p `

So, we have two equations with two variables that can be solved by substitution:

`r_s = m_p/m_s r_p `

Plugging this into the first equation, we get

`m_p/m_s r_p + r_p = 1 `

`r_p (m_p/m_s + 1) = 1 ` (AU)

`r_p = m_s/(m_p + m_s) ` (AU)

Then, `r_s = m_s/m_p * m_s/(m_p + m_s) = m_p/(m_p + m_s) ` (AU)

The sum of masses is `1.12*1.89*10^27 + 1.99*10^30 = 1.992*10^30 ` kg

Then

`r_p = (1.99*10^30)/(1.992*10^30) = 0.999 ` AU

` r_s = (1.12*1.89*10^27)/(1.992*10^30) =0.002 ` AU

As we can see, the distance between the planet and the center of mass of the system is almost equal to the distance between the planet and the star. This means the center of mass of the system is very close to the center of the star. This is because the star is a lot (about 1000 times) heavier than the planet. For the next question, we can consider the center of mass of the system to be located at the center of the star.

b) Then, the radius of the orbit of the planet is the distance between the planet and the star,

`1 AU = 1.496*10^8 ` km

The speed of the planet can be calculated by dividing the distance (length, or circumference of the orbit) by the time this distance is traversed (period).

The circumference of the orbit is

` 2pi*1.496*10^8 = 9.39*10^8` km

The period is 1.34 Earth days, which can be converted to seconds using

1 day =24 hours = 24*60 minutes = 24*60*60 seconds = 86,400 seconds

1.34 days = 1.34*86,400 = 115,776 seconds

The speed of the planet is then

`(9.39*10^8)/(115,776) = 8,110 ` km/s.

_