The earth-sun mean distance is S1 = 1.496*10^11 m
The earth's period of revolution is 365.25 cdays or 3.156*10^7 s
The Kepler's formula is correct but you used a wrong number.
(410)^2/(365.25)^2 = 1.260 =
= S1^3/(1.496*10^11 m)^3
(1.26)^1/3 = 1.0801 = S1/1.496*10^11
S1 = 1.616 * 10^7 m
1. The problem statement, all variables and given/known data
The asteroid Icarus, though only a few hundred meters across, orbits the Sun like the other planents. Its period is about 410 d. What is its mean distance from the Sun?
2. Relevant equations
Keplers Law
T_1 ^2 / T_2 ^2 = S_1 ^3 / S_2 ^3
3. The attempt at a solution
I chose my second reference point to be the earth
the distance from the earth to the sun is about 1.5 E 11 m
the period of the Earht is one day or 8.64E4 s
410 d is equal to 3.542E7 s
from keplers law
S_1 = CUBEROOT( (T_1^2 S_2^3)/T_2^2
CUBEROOT( ((3.542 E 7 s)^2 (1.5 E 11 m)^3)/(8.64 E 4 s)^2
I'm getting about 1.87 E 11 m
the book says 1.62 E 11 m
what am I doing wrong
1 answer
1 answer