Suppose an asteroid with a period of revolution of 0.567 yr is discovered.What is the average distance of this asteroid from the Sun?
2007-03-06 13:35:34 · 2 answers · asked by Lilith_Angel 2 in Science & Mathematics ➔ Astronomy & Space
It would be somewhere between Mercury (orbital period .241 years) and Venus (orbital period .615 years).
2007-03-06 13:52:24 · answer #1 · answered by Anonymous · 1⤊ 0⤋
The average distance is 0.6850... astronomical units or
1.025... x 10^11 metres.
Here's how to find this result:
Kepler's 3rd law is P^2 = a^3, where P is in years and ' a ' is in A.U. or astronomical units, the "average distance" of the Earth from the Sun. (It's actually the value of the "semi-major axis," but to all intents and purposes that's the "average distance.")
So, if P = 0.567, a = P^(2/3) = 0.6850... A.U. = 0.6850 x 1.496 x 10^11metres
= 1.025... x 10^11 metres.
Live long and prosper.
2007-03-06 15:32:22 · answer #2 · answered by Dr Spock 6 · 0⤊ 0⤋