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

0.38
0.38

2.64
2.64

1.45
1.45

0.55

1 answer

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 relationships are given by:

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

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

From these two equations, we can derive relationships between \( r_p \), \( r_a \), and \( e \):

  1. Rearranging the equations, we can express \( a \) in terms of \( r_p \) and \( r_a \):

    • From \( r_p \): \[ a = \frac{r_p}{1 - e} \]
    • From \( r_a \): \[ a = \frac{r_a}{1 + e} \]
  2. Setting these two expressions for \( a \) equal to one another, we have: \[ \frac{r_p}{1 - e} = \frac{r_a}{1 + e} \]

  3. We also know from the problem statement that \( r_p = 0.45 r_a \). Substituting this into the equation gives: \[ \frac{0.45 r_a}{1 - e} = \frac{r_a}{1 + e} \]

  4. Dividing both sides by \( r_a \) (assuming \( r_a \neq 0 \)): \[ \frac{0.45}{1 - e} = \frac{1}{1 + e} \]

  5. Cross-multiplying yields: \[ 0.45(1 + e) = 1 - e \]

  6. Expanding the left side: \[ 0.45 + 0.45e = 1 - e \]

  7. Rearranging terms gives: \[ 0.45 + 0.45e + e = 1 \] \[ 0.45 + 1.45e = 1 \]

  8. Subtracting 0.45 from both sides results in: \[ 1.45e = 0.55 \]

  9. Finally, solving for \( e \): \[ e = \frac{0.55}{1.45} \approx 0.3793 \]

Thus, the eccentricity \( e \) can be approximated as \( 0.38 \).

The correct answer is: 0.38.