The physics of orbital radius and orbital period of a planet.
If a small planet were discovered whose orbital period was twice that of Earth, how many times farther from the sun would this planet be?
Kepler: the ratio of the squares of the period is equal to the ratio of the cubes of the mean distances.
Period^2/1year= (radius/re)^3
2^2=4=(Radius/re)^3
radius= re*cubroot4
To determine the orbital radius and orbital period relationship for planets, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the orbital period (T) of a planet is directly proportional to the cube of its average distance from the sun (R).
Mathematically, this relationship can be represented as:
T^2 ∝ R^3
Now, let's solve the problem.
Given that the orbital period of the small planet is twice that of Earth, we can denote the orbital period of Earth as Te and the orbital period of the new planet as Tp.
Thus, we have Tp = 2Te.
Using Kepler's Third Law, we can rewrite the equation as:
(Tp/Te)^2 = (Rp/Re)^3
Substituting Tp = 2Te and solving for Rp/Re:
(2Te/Te)^2 = (Rp/Re)^3
2^2 = (Rp/Re)^3
4 = (Rp/Re)^3
Now, let's take the cube root of both sides to solve for Rp/Re:
(Rp/Re)^3 = 4
Rp/Re = ∛4
Rp/Re = 1.587
Therefore, the new planet would be approximately 1.587 times farther from the sun than Earth.
To determine how many times farther from the sun the planet would be, we need to understand the relationship between orbital period and orbital radius for planets.
Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of its average distance from the sun. Mathematically, this relationship is expressed as:
T^2 ∝ R^3
Where T is the orbital period and R is the orbital radius.
Now, let's assume the orbital period of the small planet is twice that of Earth. If we denote the orbital period of Earth as TE and the orbital period of the small planet as Tsmall, we can express this relationship as:
Tsmall = 2 * TE
According to Kepler's third law, the relationship between the orbital period and the orbital radius is the same for both Earth and the small planet. Therefore, we can write:
(Tsmall)^2 ∝ (Rsmall)^3
(2 * TE)^2 ∝ (Rsmall)^3
4 * (TE^2) ∝ (Rsmall)^3
Since we know that (TE^2) ∝ RE^3 (Kepler's third law for Earth), we can substitute this into the equation:
4 * (TE^2) ∝ (Rsmall)^3
4 * RE^3 ∝ (Rsmall)^3
Here, the cube of the orbital radius of Earth is proportional to the cube of the orbital radius of the small planet.
To find the ratio of their orbital radii, we can take the cube root of both sides of the equation:
∛(4 * RE^3) = ∛(Rsmall)^3
(∛4) * RE = Rsmall
Therefore, the radius of the small planet's orbit (Rsmall) would be (∛4) times the radius of Earth's orbit (RE), since the cube root of 4 is approximately 1.5874.
Hence, the small planet would be approximately 1.5874 times farther from the sun than Earth.