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 41.8 km/s and 64.6 km/s. The slower planet's orbital period is 8.84 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?
Any help will be appreciated!
To answer this question, we can use Kepler's third law of planetary motion, which states that the square of the orbital period of a planet is proportional to the cube of its average distance from the star. The formula for Kepler's third law is:
T^2 = (4π^2 / G * M) * r^3
where T is the orbital period of the planet, M is the mass of the star, r is the average distance from the planet to the star, G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2), and π is pi (approximately 3.14159).
Let's calculate the mass of the star (M) first:
Given:
Orbital speed of the slower planet (v1) = 41.8 km/s
Orbital speed of the faster planet (v2) = 64.6 km/s
Orbital period of the slower planet (T1) = 8.84 years
We know that the orbital speed of a planet can be calculated using the formula:
v = 2πr / T
where v is the orbital speed, r is the average distance from the planet to the star, and T is the orbital period of the planet.
We can rearrange this formula to solve for r:
r = (v * T) / (2π)
Let's calculate the average distance (r1) of the slower planet from the star:
r1 = (41.8 km/s * 8.84 years) / (2π)
= (41.8 * 10^3 m/s * 8.84 * 365.25 * 24 * 60 * 60 s) / (2 * 3.14159)
≈ 2.209 * 10^12 m
Now, let's calculate the mass (M) of the star:
T1^2 = (4π^2 / G * M) * r1^3
M = (4π^2 / G) * (r1^3 / T1^2)
= (4 * 3.14159^2 / (6.67430 × 10^-11 m^3 kg^-1 s^-2)) * ((2.209 * 10^12 m)^3 / (8.84 years)^2)
Using this formula, we get:
M ≈ 2.431 * 10^30 kg
Therefore, the mass of the star is approximately 2.431 * 10^30 kg.
Now, let's calculate the orbital period of the faster planet (T2) using the same formula:
T2^2 = (4π^2 / G * M) * r2^3
We need to find the average distance (r2) of the faster planet from the star. We can rearrange the formula for r:
r2 = (v2 * T2) / (2π)
T2 = (2π * r2) / v2
Since we already know the value of v2 and T2 can be the time, we can now calculate,
T2 = (2π * r2) / v2
= (2π * (v2 * T2) / (2π)) / v2
= T2
It follows that the orbital period of the faster planet is approximately 8.84 years.