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Observations of this planet over time show that it is in a nearly circular orbit around its star and completes one orbit in only 9.5 days. How many times the orbital radius r of the earth around our sun is this exoplanet's orbital radius around its sun? Assume that the earth is also in a nearly circular orbit.

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Answer:

r₂ = 38.42 r₁

Explanation:

Eccentricity of Ellipse is defined as

e = c /a

Where e is eccentricity, c is distance between the center of the ellipse and the focus, and a is a semi major axis.

Eccentricity is a measure of how divorce of a circular shape is ellipse and its value goes from 0 (circular shape) up to 1 very far away of circular shape

The first Kepler´s law, establishes that planets orbit sun in an elliptical path, having a sun as one focus. But in fact, even that planets move in an ellipse path most of planets in a solar system have a very close circular orbit. Our planet Earth for instance has an eccentricity close to 0,0167, and Pluto which has a biggest one is just 0,25.

If a planet complete one orbit in only 9.5 days, that planet has traveled

2π*r₁ then 9.5 ⇒ 2π*r₁ (length of the orbit)

Earth complete one orbit in 365 days so

365 ⇒ 2*π*r₂

We have a proportion

9.5/2πr₁ = 365 /2πr₂

r₂/r₁ = 365/9.5

r₂/r₁ = 38.42

r₂ = 38.42 r₁

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