For a planet orbiting the sun, rp

is the distance from the sun to the perihelion and ra
is the distance from the sun to the aphelion. What is a planet’s orbital eccentricity if rp
is equal to 0.45ra
?(1 point)
Responses

2.64
2.64

0.38
0.38

1.45
1.45

0.55

1 answer

To find the eccentricity \( e \) of the orbit of a planet, we can use the relationship between the perihelion and aphelion distances. The perihelion distance \( r_p \) and aphelion distance \( r_a \) are defined as follows:

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

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

Given that \( r_p = 0.45 r_a \), we can substitute the expressions for \( r_p \) and \( r_a \):

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

Since \( a \) is common on both sides, we can cancel it out (assuming it is non-zero):

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

Now, we'll distribute \( 0.45 \):

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

Next, we'll isolate \( e \):

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

\[ 0.55 = e + 0.45e \]

Combine like terms:

\[ 0.55 = 1.45e \]

Now, solve for \( e \):

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

Rounding gives us \( e \approx 0.38 \).

The orbital eccentricity of the planet is approximately 0.38.

Thus, the correct response is:

0.38