Planets travel in elliptical orbits with the sun at one focus. Assume that the focus is at the pole, the major axis lies on the polar axis, and the length of the major axis is $2 a$ (see figure). The polar equation of the orbit of a planet is given below, where $e$ is the eccentricity. If $a = 88.908 \times 10 ^ { 6 }$ miles and $e = 0.0260$, find the perihelion distance (the minimum distance from the sun to the planet). Round your answer to the nearest mile.

A) $86,596,392$ miles
B) $91,219,608$ miles
C) $8,659,639$ miles
D) $9,121,961$ miles
E) $865,963,920$ miles

Question

Planets travel in elliptical orbits with the sun at one focus. Assume that the focus is at the pole, the major axis lies on the polar axis, and the length of the major axis is $2 a$ (see figure). The polar equation of the orbit of a planet is given below, where $e$ is the eccentricity. If $a = 88.908 \times 10 ^ { 6 }$ miles and $e = 0.0260$, find the perihelion distance (the minimum distance from the sun to the planet). Round your answer to the nearest mile.

A) $86,596,392$ miles 
B) $91,219,608$ miles 
C) $8,659,639$ miles 
D) $9,121,961$ miles 
E) $865,963,920$ miles

Quizplus · Accepted Answer

The polar equation given is $$r = \dfrac{2a(1-e^2)}{1+e \cos 	heta}.$$

At perihelion, $	heta = 0$, so we have $$r = \dfrac{2a(1-e^2)}{1+e \cos 0}.$$

Since $\cos 0 = 1$, we simplify to get $$r = \dfrac{2a(1-e^2)}{1+e}.$$

Plugging in $a = 88.908 	imes 10 ^ { 6 }$ miles and $e = 0.0260$, we get $$r = \dfrac{2 \cdot 88.908 	imes 10 ^ { 6 } (1 - (0.0260)^2)}{1+0.0260} \approx 86,596,392.$$

Therefore, the perihelion distance is approximately $86,596,392$ miles, which is the closest to answer choice $	extbf{(A)}$.

Planets Travel in Elliptical Orbits with the Sun at One 2a2 a2a

Planets Travel in Elliptical Orbits with the Sun at One $2 a$