To find the orbital eccentricity \(e\) of a moon given that \(rp\) (the periapsis distance) is equal to \(0.27ra\) (the apoapsis distance), we can use the formula for eccentricity in terms of periapsis and apoapsis distances:
\[ e = \frac{ra - rp}{ra + rp} \]
From the problem, we know that:
\[ rp = 0.27ra \]
Now, we can express \(ra\) in terms of \(rp\):
Let \(rp = x\), then \(ra = \frac{rp}{0.27} = \frac{x}{0.27}\).
Substituting these into the equation for eccentricity:
\[ e = \frac{ \frac{x}{0.27} - x }{ \frac{x}{0.27} + x } \]
Next, we can factor \(x\) out:
\[ e = \frac{ \frac{x - 0.27x}{0.27} }{ \frac{x + 0.27x}{0.27} } \]
Simplifying further:
\[ e = \frac{ 1 - 0.27 }{ 1 + 0.27 } \]
Calculating the numerator and denominator gives:
\[ e = \frac{0.73}{1.27} \]
Now, calculating the numerical value:
\[ e \approx \frac{0.73}{1.27} \approx 0.574 \]
Since we can round this value, the closest option given is:
0.57.