Question

For a moon orbitant its planet xp is the shortest distance between the Moon and its planet and XA is the longest distance between the Moon and its planet what is the Moon's orbital eccentricle if XP is equal to 0.27 x8

Answer 1

To find the orbital eccentricity of a moon, we can use the formula relating the shortest distance (periapsis, \( r_p \)) and the longest distance (apoapsis, \( r_a \)) to the eccentricity \( e \):

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

In your case, you mentioned that the shortest distance \( r_p \) is equal to \( 0.27 r_a \) (where \( r_a \) is the longest distance).

Let's denote:
- \( r_p = 0.27 r_a \)
- \( r_a = r_a \)

Now we can plug these values into the formula for eccentricity:

1. Substitute \( r_p \) into the eccentricity formula:

\[
e = \frac{r_a - 0.27 r_a}{r_a + 0.27 r_a}
\]

2. Simplify the expression:

\[
e = \frac{1r_a - 0.27r_a}{1r_a + 0.27r_a} = \frac{0.73 r_a}{1.27 r_a}
\]

3. The \( r_a \) cancels out:

\[
e = \frac{0.73}{1.27}
\]

4. Calculate the value:

\[
e \approx 0.5732
\]

Thus, the orbital eccentricity \( e \) of the moon is approximately \( 0.5732 \).