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.

r₂    =   38.42 r₁

Step-by-step 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

idk

step-by-step explanation:

one of each metals

step-by-step explanation: