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

0.55
0.55

1.45
1.45

2.64
2.64

0.38

Answer 1

The eccentricity \( e \) of an orbiting body can be calculated with the formula:

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

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

Given that \( r_p \) is equal to \( 0.45 r_a \), we can denote \( r_a \) as \( r_a \) and \( r_p \) as \( 0.45 r_a \).

Now, let's substitute these values into the eccentricity formula:

1. Calculate \( r_a - r_p \): \[ r_a - r_p = r_a - 0.45 r_a = 0.55 r_a \]

2. Calculate \( r_a + r_p \): \[ r_a + r_p = r_a + 0.45 r_a = 1.45 r_a \]

3. Plug these into the eccentricity formula: \[ e = \frac{0.55 r_a}{1.45 r_a} \]

The \( r_a \) terms cancel out: \[ e = \frac{0.55}{1.45} \]

Now, calculating the value: \[ e \approx \frac{0.55}{1.45} \approx 0.3793 \]

Therefore, the orbital eccentricity of the planet is approximately 0.38.

The answer is 0.38.