For a moon orbiting its planet, rp

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**.