To find the orbital eccentricity \( e \) of a planet, we can use the following relationships for perihelion distance (\( r_p \)) and aphelion distance (\( r_a \)):
\[ r_p = a(1 - e) \] \[ r_a = a(1 + e) \]
where \( a \) is the semi-major axis of the orbit.
From the problem, we know that \( r_p = 0.45 r_a \). Substituting the formulas for \( r_p \) and \( r_a \):
\[ a(1 - e) = 0.45 \cdot a(1 + e) \]
Since \( a \) is common on both sides, we can divide both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 1 - e = 0.45(1 + e) \]
Now, distribute the 0.45 on the right side:
\[ 1 - e = 0.45 + 0.45e \]
Next, combine the terms involving \( e \):
\[ 1 - 0.45 = e + 0.45e \] \[ 0.55 = e + 0.45e \] \[ 0.55 = 1.45e \]
Now, solve for \( e \):
\[ e = \frac{0.55}{1.45} \] \[ e \approx 0.3793 \]
Which can be rounded to \( e \approx 0.38 \).
So, the planet's orbital eccentricity is approximately 0.38.