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