You observe an asteroid with a near-circular orbit which takes 3.2 (Earth) years to go around the Sun. What is its average distance from the Sun?
Added by Jasmine A.
Step 1
Given: T_A = 3.2 Earth years, T_E = 1 Earth year, D_E = 149.6 x 10^9 m We know that (T_A/T_E)^2 = (D_A/D_E)^3 Substitute the values: (3.2/1)^2 = (D_A/149.6 x 10^9)^3 10.24 = (D_A/149.6 x 10^9)^3 D_A = 149.6 x 10^9 * cuberoot(10.24) D_A = 149.6 x 10^9 * 2 D_A = Show more…
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