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regarding elips

the orbit of the earth around the sun is an ellipse  sun  is on one foci. if length of major axis is 300 million km
and eccentricity is 0.0167 find minimum and maximum distance of the earth from the sun
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2 Answers

For an ellipse  the distance, d, from the center to a focal point is d = a e,  where a is the semi-major axis length and e is the eccentricity.   Given the data, a = 150 x 10^6  and e = 0.0167  .  So  d = 2.505 x 10^6.
The distance of closest approach is   a - d =  147.495 km and the maximum distance to the sun is
a + d = 152.505 .

Comments

Good alternate technique, but off by 1000.  ;-S
Xp: min distance from sun (Perihelion)
Xa: max distance from sun (Aphelion)
a: Semi-major axis =150 million km
 
Xp = a(1-e) = 150,000,000(1-.0167)~147.5 Million Km
Xa = a(1+e) = 150,000,000(1+.0167)~152.5 Million Km
 
Since e~0, it's nearly a circular orbit.
 
Fun Fact: Winter in the Northern Hemisphere occurs at Perihelion (Closest to sun)
 
This is because distance is not as important as Solar Constant (Earth axis tilt)