Solve the problem.
-Given that the polar equation $\mathrm { r } = \frac { \mathrm { a } \left( 1 - \mathrm { e } ^ { 2 } \right) } { 1 + \mathrm { e } \cos \theta }$ models the orbits of the planets, estimate the smallest possible distance between Earth, for which $\mathrm { a } = 1.00$ and $\mathrm { e } = 0.017$, and Venus, for which $\mathrm { a } = 0.78$ and $\mathrm { e } = 0.007$.
A) Zero astronomical units
B) About $0.2$ astronomical units
C) About $0.4$ astronomical units
D) About 2 astronomical units

Question

Quizplus · Accepted Answer

The smallest distance between two planets in their orbits occurs when they are at their closest points, which is when they are at their perihelion (the point in their orbit closest to the Sun). The perihelion distance for each planet can be calculated from the given polar equation by setting $\theta = 0$, which simplifies the equation to $r = \frac{a(1 - e^2)}{1 + e}$. For Earth, $a = 1.00$ and $e = 0.017$, and for Venus, $a = 0.78$ and $e = 0.007$. Calculating the perihelion distances and subtracting them gives a value close to $0.2$ astronomical units, indicating that the smallest possible distance between Earth and Venus is about $0.2$ astronomical units.

Solve the Problem r=a(1−e2)1+ecos⁡θ\mathrm { r } = \frac { \mathrm { a } \left( 1 - \mathrm { e } ^ { 2 } \right) } { 1 + \mathrm { e } \cos \theta }r=1+ecosθa(1−e2)​

Solve the Problem $\mathrm { r } = \frac { \mathrm { a } \left( 1 - \mathrm { e } ^ { 2 } \right) } { 1 + \mathrm { e } \cos \theta }$