In recent years, a number of nearby stars have been found to possess planets. Suppose, the orbital radius of such a planet is found to be 3.0 1011 m, with a period of 980 days. Find the mass of the star.
To find the mass of the star, we need to use Kepler's Third Law of Planetary Motion, which relates the orbital radius and the period of a planet around a star to the mass of the star.
Kepler's Third Law states:
(T₁/T₂)² = (R₁/R₂)³
In this case, T₁ is the period of the given planet (980 days), R₁ is the orbital radius of the planet (3.0 x 10^11 m), T₂ is the period of Earth (365 days), and R₂ is the orbital radius of Earth (1.5 x 10^11 m).
Let's plug in the values and solve for the mass of the star:
(980/365)² = (3.0 x 10^11 / 1.5 x 10^11)³
Simplifying the equation, we have:
(2.6849) = (2)³
Taking the cube root of both sides of the equation, we have:
1.6236 = 2
Now, we can set up the proportions to find the mass of the star:
(R₁/R₂)³ = (M₁/M₂)
Plugging in the values for R₁, R₂, and M₂ (mass of Earth), we get:
[(3.0 x 10^11 m) / (1.5 x 10^11 m)]³ = (M₁ / 5.972 x 10^24 kg)
Simplifying the equation, we get:
(2)³ = (M₁ / 5.972 x 10^24 kg)
Solving for M₁ (mass of the star), we have:
M₁ = (2)³ x (5.972 x 10^24 kg)
M₁ = 8 x 5.972 x 10^24 kg
M₁ = 47.776 x 10^24 kg
M₁ = 4.7776 x 10^25 kg
Therefore, the mass of the star is approximately 4.7776 x 10^25 kg.