For a moon orbiting its planet, rp is the shortest distance between the moon and its planet and ra is the longest distance between the moon and its planet. What is a moon's orbital eccentricity if rp is equal to 0.27ra ?(1 point)

The orbital eccentricity \( e \) of an orbiting body can be calculated using the relationships between the periapsis distance \( r_p \), apoapsis distance \( r_a \), and eccentricity given by the equations:

\[
r_p = a(1 - e)
\]
\[
r_a = a(1 + e)
\]

where \( a \) is the semi-major axis of the orbit.

From these equations, we can express the semi-major axis \( a \) in terms of \( r_p \) and \( r_a \):

\[
a = \frac{r_p + r_a}{2}
\]

Now, substituting \( r_a \) in terms of \( r_p \), we have:

\[
r_p = 0.27 r_a
\]

Let \( r_a = r_a \). Then we can write:

\[
r_p = 0.27 r_a
\]

Using the two previous equations for \( r_p \) and \( r_a \):

1. \( r_p = a(1 - e) \)
2. \( r_a = a(1 + e) \)

Now let's substitute \( r_p \) into the second equation in terms of \( r_a \):

From \( r_p = 0.27 r_a \):

Substituting \( r_a \):

\[
0.27 r_a = a(1 - e)
\]
\[
r_a = a(1 + e)
\]

From \( a \):

\[
a = \frac{r_p + r_a}{2} = \frac{0.27r_a + r_a}{2} = \frac{(0.27 + 1)r_a}{2} = \frac{1.27r_a}{2}
\]

Now we know both \( r_p \) and \( r_a \):

Substituting back to find \( e \):

1. From periapsis:
\[
0.27 r_a = \left(\frac{1.27r_a}{2}\right)(1 - e)
\]

2. From apoapsis:
\[
r_a = \left(\frac{1.27r_a}{2}\right)(1 + e)
\]

You can now divide both equations to isolate \( e \):

By solving either equation, we can get \( e \) directly:

Combining these equations leads to:

After simplifying, we find that:

Using:

\[
e = \frac{(r_a - r_p)}{(r_a + r_p)}
\]

Substituting:

\[
e = \frac{r_a - 0.27 r_a}{r_a + 0.27 r_a} = \frac{(1 - 0.27)r_a}{(1 + 0.27)r_a} = \frac{0.73}{1.27} \approx 0.57
\]

So the moon's orbital eccentricity is approximately **0.57**.

The correct choice is:

**0.57**