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 \( e \):
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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 \( r_a \): \[ a = \frac{r_a}{1 + e} \]
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Setting these two expressions for \( a \) equal to one another, we have: \[ \frac{r_p}{1 - e} = \frac{r_a}{1 + e} \]
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We also know from the problem statement that \( r_p = 0.45 r_a \). Substituting this into the equation gives: \[ \frac{0.45 r_a}{1 - e} = \frac{r_a}{1 + e} \]
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Dividing both sides by \( r_a \) (assuming \( r_a \neq 0 \)): \[ \frac{0.45}{1 - e} = \frac{1}{1 + e} \]
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Cross-multiplying yields: \[ 0.45(1 + e) = 1 - e \]
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Expanding the left side: \[ 0.45 + 0.45e = 1 - e \]
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Rearranging terms gives: \[ 0.45 + 0.45e + e = 1 \] \[ 0.45 + 1.45e = 1 \]
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Subtracting 0.45 from both sides results in: \[ 1.45e = 0.55 \]
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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.