Question

n. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? The planet's orbital period will be four times Earth's orbital period. The planet's orbital period will be one-half Earth's orbital period. The planet's orbital period will be one-fourth Earth's orbital period. The planet's orbital period will be equal to Earth's orbital period. The planet's orbital period will be twice Earth's orbital period.

Answer 1

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.