To find the eccentricity \( e \) of the orbit of a planet, we can use the relationship between the perihelion and aphelion distances. The perihelion distance \( r_p \) and aphelion distance \( r_a \) are defined as follows:
\[ r_p = a(1 - e) \] \[ r_a = a(1 + e) \]
where \( a \) is the semi-major axis of the orbit.
Given that \( r_p = 0.45 r_a \), we can substitute the expressions for \( r_p \) and \( r_a \):
\[ a(1 - e) = 0.45 \cdot a(1 + e) \]
Since \( a \) is common on both sides, we can cancel it out (assuming it is non-zero):
\[ 1 - e = 0.45(1 + e) \]
Now, we'll distribute \( 0.45 \):
\[ 1 - e = 0.45 + 0.45e \]
Next, we'll isolate \( e \):
\[ 1 - 0.45 = e + 0.45e \]
\[ 0.55 = e + 0.45e \]
Combine like terms:
\[ 0.55 = 1.45e \]
Now, solve for \( e \):
\[ e = \frac{0.55}{1.45} \approx 0.3793 \]
Rounding gives us \( e \approx 0.38 \).
The orbital eccentricity of the planet is approximately 0.38.
Thus, the correct response is:
0.38