Find the polar equation of the planet's orbit and the perihelion and aphelion distances.
Earth $a = 95.956 \times 10 ^ { 6 }$ miles $e = 0.0167$
A) $\frac { 0.9593 \times 10 ^ { 8 } } { 1 - 0.0167 \cos \theta }$ Perihelion distance: $r = 9.4354 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.7558 \times 10 ^ { 8 }$
B) $\frac { 0.9593 \times 10 ^ { 8 } } { 1 - 0.0167 \sin \theta }$ Perihelion distance: $r = 9.7558 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.4354 \times 10 ^ { 8 }$
C) $\frac { 0.0167 \times 10 ^ { 8 } } { 1 + 0.9593 \sin \theta }$ Perihelion distance: $r = 9.4354 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.7558 \times 10 ^ { 8 }$
D) $\frac { 0.9593 \times 10 ^ { 7 } } { 1 - 0.0167 \cos \theta }$ Perihelion distance: $r = 9.7558 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.4354 \times 10 ^ { 8 }$
E) $\frac { 0.9593 \times 10 ^ { 7 } } { 1 + 0.0167 \cos \theta }$ Perihelion distance: $r = 9.4354 \times 10 ^ { 7 }$ Aphelion distance: $r = 9.7558 \times 10 ^ { 7 }$

Question

Find the polar equation of the planet's orbit and the perihelion and aphelion distances. 
Earth $a = 95.956 \times 10 ^ { 6 }$ miles $e = 0.0167$ 
A) $\frac { 0.9593 \times 10 ^ { 8 } } { 1 - 0.0167 \cos \theta }$ Perihelion distance: $r = 9.4354 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.7558 \times 10 ^ { 8 }$
B) $\frac { 0.9593 \times 10 ^ { 8 } } { 1 - 0.0167 \sin \theta }$ Perihelion distance: $r = 9.7558 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.4354 \times 10 ^ { 8 }$
C) $\frac { 0.0167 \times 10 ^ { 8 } } { 1 + 0.9593 \sin \theta }$ Perihelion distance: $r = 9.4354 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.7558 \times 10 ^ { 8 }$
D) $\frac { 0.9593 \times 10 ^ { 7 } } { 1 - 0.0167 \cos \theta }$ Perihelion distance: $r = 9.7558 \times 10 ^ { 8 }$ Aphelion distance: $r = 9.4354 \times 10 ^ { 8 }$
E) $\frac { 0.9593 \times 10 ^ { 7 } } { 1 + 0.0167 \cos \theta }$ Perihelion distance: $r = 9.4354 \times 10 ^ { 7 }$ Aphelion distance: $r = 9.7558 \times 10 ^ { 7 }$

Quizplus · Accepted Answer

The polar equation of a planet's orbit in terms of its semi-major axis $a$ and eccentricity $e$ is given by $\frac{a(1-e^2)}{1+e\cos\theta}$. For Earth, $a = 95.956 \times 10^6$ miles and $e = 0.0167$, so the equation simplifies to $\frac{0.9593 \times 10^8}{1 - 0.0167\cos\theta}$ after adjusting for the correct form and scale. The perihelion distance (closest approach) occurs when $\cos\theta = 1$, leading to the smallest denominator and thus the smallest $r$, and the aphelion distance (farthest point) occurs when $\cos\theta = -1$, leading to the largest $r$. The distances provided in option A correctly reflect the calculation based on these principles.

Find the Polar Equation of the Planet's Orbit and the Perihelion