For a planet orbiting the sun, rp

Question

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

1.45
1.45

0.55
0.55

2.64
2.64

0.38

Answer 1

The eccentricity \( e \) of an orbit can be calculated using the formula:

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

where \( r_a \) is the distance at aphelion and \( r_p \) is the distance at perihelion.

From the problem, we know:

\[ r_p = 0.45 r_a \]

Now, we can substitute \( r_p \) into the equation for the eccentricity:

1. First, substitute \( r_p \):

\[ e = \frac{r_a - 0.45 r_a}{r_a + 0.45 r_a} \]

2. Simplify the expression:

\[ e = \frac{r_a (1 - 0.45)}{r_a (1 + 0.45)} = \frac{0.55 r_a}{1.45 r_a} \]

3. Notice that \( r_a \) cancels out:

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

4. Now, perform the division:

\[ e \approx 0.3793 \]

Rounding to two decimal places gives us \( e \approx 0.38 \).

Thus, the planet's orbital eccentricity is 0.38.