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

Question 24

Multiple Choice

Find the polar equation of the planet's orbit and the perihelion and aphelion distances. ​
Earth a=95.956×106a = 95.956 \times 10 ^ { 6 } miles e=0.0167e = 0.0167


A) 0.9593×10810.0167cosθ\frac { 0.9593 \times 10 ^ { 8 } } { 1 - 0.0167 \cos \theta } Perihelion distance: r=9.4354×108r = 9.4354 \times 10 ^ { 8 }
Aphelion distance: r=9.7558×108r = 9.7558 \times 10 ^ { 8 }
B) 0.9593×10810.0167sinθ\frac { 0.9593 \times 10 ^ { 8 } } { 1 - 0.0167 \sin \theta } Perihelion distance: r=9.7558×108r = 9.7558 \times 10 ^ { 8 }
Aphelion distance: r=9.4354×108r = 9.4354 \times 10 ^ { 8 }
C) 0.0167×1081+0.9593sinθ\frac { 0.0167 \times 10 ^ { 8 } } { 1 + 0.9593 \sin \theta } Perihelion distance: r=9.4354×108r = 9.4354 \times 10 ^ { 8 }
Aphelion distance: r=9.7558×108r = 9.7558 \times 10 ^ { 8 }
D) 0.9593×10710.0167cosθ\frac { 0.9593 \times 10 ^ { 7 } } { 1 - 0.0167 \cos \theta } Perihelion distance: r=9.7558×108r = 9.7558 \times 10 ^ { 8 }
Aphelion distance: r=9.4354×108r = 9.4354 \times 10 ^ { 8 }
E) 0.9593×1071+0.0167cosθ\frac { 0.9593 \times 10 ^ { 7 } } { 1 + 0.0167 \cos \theta } Perihelion distance: r=9.4354×107r = 9.4354 \times 10 ^ { 7 }
Aphelion distance: r=9.7558×107r = 9.7558 \times 10 ^ { 7 }

Correct Answer:

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