Question

astronomy: orbital motion online practice complete this assessment to review what youve learned. it will not count toward your grade. for a planet orbiting the sun, ( r_p ) is the distance from the sun to the perihelion and ( r_a ) is the distance from the sun to the aphelion. what is a planet’s orbital eccentricity if ( r_p ) is equal to ( 0.45r_a )? (1 point) ( \bigcirc ) 1.45 ( \bigcirc ) 2.64 ( \bigcirc ) 0.38 ( \bigcirc ) 0.55

Explanation:

Step1: Recall the formula for eccentricity (e)

The formula for the orbital eccentricity \( e \) is \( e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}} \), where \( r_{p} \) is the perihelion distance and \( r_{a} \) is the aphelion distance. We know that \( r_{p} = 0.45r_{a} \).

Step2: Substitute \( r_{p} \) into the formula

Substitute \( r_{p}=0.45r_{a} \) into the formula:
\[

$$\begin{align*} e&=\frac{r_{a}- 0.45r_{a}}{r_{a}+0.45r_{a}}\\ &=\frac{0.55r_{a}}{1.45r_{a}} \end{align*}$$

\]
The \( r_{a} \) terms cancel out (since \( r_{a}
eq0 \)).

Step3: Calculate the value of e

\[
e=\frac{0.55}{1.45}\approx0.38
\]

Answer:

0.38