halley's comet has an elliptical orbit with the sun at 1 focus.its closest distance to the sun is about 9 x 10 to the seventh power, while farthest is about 5.34 x 10 to the ninth power..
A. FIND THE LENGTH of the major axis of halley's comet's orbit.
B. WHAT IS THE LENGTH of its minor axis?
2007-12-31 11:41:31 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Astronomy & Space
I think we are missing some information here. It seems like we need one more piece of information, like the semi-minor axis or eccentricity or something. Are you sure you have the problem stated correctly?
We can solve part A with what we have here.The major axis is given by
A:major axis = (5.34*10^9 km + 9*10^7 km) =5.43*10^9 km
Normally the semi-major axis is called for:
a:semi-major axis = (5.34*10^9 km + 9*10^7 km)/2
= 2.715 * 10^9 km = 18.14 AU
The problem is I can't determine the minor axis without the eccentricity. Now the eccentricity of Halley's comet is known to be e = 0.967 (see Ref 1). Knowing this we can compute the semi-minor axis as
b = a*(1-e^2)^1/2 = 1.383*10^9 km
This means the minor axis would be 2*b.
2007-12-31 13:09:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
2a = 9*10^7 + 5.34*10^9 = 5.43*10^9 km
a = 2.715*10^9 km
c = 2.715*10^9 - 9*10^7 = 2.625*10^9 km
b² = (2.715*10^9)² - (2.625*10^9)²
b â 6.932532*10^8 km
2b â 1.386506*10^9 km
2008-01-02 19:45:49 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋