0

# 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

### 2 Answers by Expert Tutors

Tutors, sign in to answer this question.
Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (627 lesson ratings) (627)
0
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
Tom D. | Very patient Math Expert who likes to teachVery patient Math Expert who likes to te...
0
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)