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?

The answer for this one is 1.6 earth radii, but I can't seems to get solution to find the answer. I would appreciate for any hint or help.

To find the answer, we can start by using Kepler's third law, which relates the orbital period of a planet to its distance from the Sun. Kepler's third law mathematically states that the square of the orbital period (T) is proportional to the cube of the average distance between the planet and the Sun (r).

So, let's assume the Earth's distance from the Sun is r₁, and its orbital period is T₁. The small planet's distance from the Sun is r₂, and its orbital period is T₂.

From the information given, we know that T₂ = 2 * T₁ (the small planet's orbital period is twice that of Earth).

Using Kepler's third law, we can write an equation to compare the two planets' distances from the Sun:

(T₂/T₁)² = (r₂/r₁)³

Substituting 2 * T₁ for T₂ in the equation, we get:

(2 * T₁ / T₁)² = (r₂ / r₁)³

Simplifying the equation gives:

4 = (r₂ / r₁)³

Taking the cube root of both sides gives:

∛4 = ∛(r₂ / r₁)³

Simplifying further:

∛4 = r₂ / r₁

To find how many times farther the small planet is from the Sun compared to the Earth, we need to compare the ratios of the distances r₂ and r₁.

r₂ / r₁ = ∛4

Calculating the cube root of 4:

r₂ / r₁ ≈ 1.587

So, the small planet would be approximately 1.587 times farther from the Sun compared to Earth, or in other words, about 1.6 times the Earth-Sun distance.