Question

For a planet orbiting the sun, rprp is the distance from the sun to the perihelion and rara is the distance from the sun to the aphelion. What is a planet’s orbital eccentricity if rprp is equal to 0.45ra0.45ra ?(1 point) Responses

Answer 1

To find the orbital eccentricity \( e \) of a planet, we can use the relationship between the perihelion distance \( r_p \), aphelion distance \( r_a \), and eccentricity. The formulas are:

- Perihelion distance: \( r_p = a(1 - e) \)
- Aphelion distance: \( r_a = a(1 + e) \)

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

The problem states that \( r_p = 0.45 r_a \). We can express \( r_a \) in terms of \( r_p \):

1. Starting from the relationship \( r_p = 0.45 r_a \):

\[
r_p = 0.45 r_a
\]

2. Substitute the definitions of \( r_p \) and \( r_a \):

\[
a(1 - e) = 0.45 a(1 + e)
\]

3. We can divide both sides by \( a \) (assuming \( a \neq 0 \)):

\[
1 - e = 0.45(1 + e)
\]

4. Distributing the 0.45:

\[
1 - e = 0.45 + 0.45e
\]

5. Rearranging terms to isolate \( e \):

\[
1 - 0.45 = e + 0.45e
\]

\[
0.55 = e + 0.45e
\]

\[
0.55 = 1.45e
\]

6. Finally, solving for \( e \):

\[
e = \frac{0.55}{1.45} \approx 0.3793
\]

Therefore, the orbital eccentricity \( e \) is approximately \( 0.379 \).