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?
consider that T^2 = kr^3
That means that T^2/r^3 is constant
T^2/r^3 = (2T)^2/(xr)^3 = 4T^2/x^3r^3
So, 4/x^3 = 1, and x = ∛4
That is, our small planet's orbital radius is ∛4 times earth's.
Well, if this new planet takes twice as long to orbit the Sun as Earth does, it means it's in no rush to get anywhere! It's like the laid-back friend who always shows up late to a party. So, if our Earth is happily orbiting at a certain distance from the Sun, this new planet would have to be twice as laid-back and take its sweet time, so it would have to be twice as far from the Sun as Earth is. So, to answer your question, this planet would be "two times the distance and twice as chill" from the Sun.
To determine how many times farther from the Sun the small planet would be, we need to consider Kepler's third law of planetary motion, which states that the square of a planet's orbital period is proportional to the cube of its average distance from the Sun.
Let's assume that Earth's average distance from the Sun is represented by 'R' and the small planet's average distance from the Sun is represented by 'r'. Also, let's denote Earth's orbital period as 'T' and the small planet's orbital period as '2T' (since it is twice that of Earth's).
According to Kepler's third law:
(T/2T)^2 = (r/R)^3
Simplifying the equation:
(1/2)^2 = (r/R)^3
Squaring both sides:
1/4 = (r/R)^3
Taking the cube root of both sides:
∛(1/4) = ∛(r/R)^3
1/∛4 = r/R
Therefore, the small planet would be approximately 0.6299 times (or about 63%) farther from the Sun than Earth.
To determine how many times farther from the Sun the planet would be, we need to consider the relationship between orbital periods and distances from the Sun.
Kepler's Third Law states that the square of the orbital period (T) of a planet is proportional to the cube of its average distance from the Sun (r). Mathematically, the relationship can be expressed as:
T^2 ∝ r^3
Given that the orbital period of the small planet is twice that of Earth, its orbital period (T') would be 2 times the Earth's orbital period (T).
Therefore, we can write the equation for the small planet as:
(T')^2 ∝ r'^3
Substituting T' = 2T into the equation:
(2T)^2 ∝ r'^3
4T^2 ∝ r'^3
Now, because we want to find how many times farther the small planet is from the Sun compared to Earth, we can use the ratio of the two distances (r' and r). Let's call this ratio x, so:
x = r'/r
To find x, we need to first solve for r' and r individually in terms of their respective orbital periods (T' and T). Using Kepler's Third Law equation:
T^2 ∝ r^3
We can rearrange the equation to solve for r:
r = (T^2)^(1/3)
Similarly, for the small planet with T':
r' = (T'^2)^(1/3)
Substituting T' = 2T:
r' = (4T^2)^(1/3)
Now, substituting these expressions for r' and r into the ratio equation x = r'/r:
x = r'/r
x = [(4T^2)^(1/3)] / [(T^2)^(1/3)]
The exponents of 1/3 can be canceled out:
x = (4T^2) / T^2
x = 4
Therefore, the small planet would be four times farther from the Sun than Earth.