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 about the sun, estimate the closest possible approach of a planet for which $\mathrm { a } = 18$ and $\mathrm { e } = 0.6$ to a planet for which $a = 45$ and $\mathrm { e } = 0.030$.
A) 18 astronomical units
B) 0 astronomical units
C) 75 astronomical units
D) 36 astronomical units

Question

Quizplus · Accepted Answer

The closest possible approach occurs when $\theta = 0$. Substituting this value into the polar equation gives:$\mathrm { r } = \frac { \mathrm { a } \left( 1 - \mathrm { e } ^ { 2 } \right) } { 1 + \mathrm { e } \cos 0 } = \frac { \mathrm { a } \left( 1 - \mathrm { e } ^ { 2 } \right) } { 1 + \mathrm { e } }$For the first planet, $\mathrm { a } = 18$ and $\mathrm { e } = 0.6$, so the closest possible approach is:$\mathrm { r } = \frac { 18 \left( 1 - 0.6 ^ { 2 } \right) } { 1 + 0.6 } = 10.8$ astronomical unitsFor the second planet, $\mathrm { a } = 45$ and $\mathrm { e } = 0.030$, so the closest possible approach is:$\mathrm { r } = \frac { 45 \left( 1 - 0.030 ^ { 2 } \right) } { 1 + 0.030 } = 43.7$ astronomical unitsThe closest possible approach of the first planet is 10.8 astronomical units, and the closest possible approach of the second planet is 43.7 astronomical units. The closest possible approach of the two planets is therefore 10.8 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 $\mathrm { r } = \frac { \mathrm { a } \left( 1 - \mathrm { e } ^ { 2 } \right) } { 1 + \mathrm { e } \cos \theta }$