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

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

To find the mean distance of the asteroid Icarus from the Sun, you can use Kepler's Third Law of Planetary Motion. The equation is:

(T1^2 / T2^2) = (S1^3 / S2^3)

Where T1 is the period of Icarus (410 days), S1 is the distance of Icarus from the Sun, T2 is the period of Earth (365.25 days), and S2 is the distance of Earth from the Sun (approximately 1.5 x 10^11 m).

Let's substitute the values into the equation:

(410^2 / 365.25^2) = (S1^3 / (1.5 x 10^11)^3)

Simplifying the equation further:

(168100 / 133225.0625) = (S1^3 / 3.375 x 10^33)

Solving for S1:

S1^3 = (168100 / 133225.0625) * (3.375 x 10^33)

S1^3 = 4.177284 x 10^33

Taking the cube root of both sides:

S1 = ∛(4.177284 x 10^33)

S1 ≈ 1.62 x 10^11 m

So, the mean distance of the asteroid Icarus from the Sun is approximately 1.62 x 10^11 meters.

To find the mean distance of the asteroid Icarus from the Sun, we can use Kepler's Third Law, which relates the orbital periods and distances of celestial bodies.

Kepler's Third Law equation can be written as:
(T1^2 / T2^2) = (S1^3 / S2^3)

Where:
T1 = orbital period of Icarus
T2 = orbital period of Earth (reference point)
S1 = mean distance of Icarus from the Sun
S2 = mean distance of Earth from the Sun

We can rearrange the equation to solve for S1:
S1 = (T1^2 * S2^3) / (T2^2)

Given:
T1 (orbital period of Icarus) = 410 days = 3.542 x 10^7 seconds
T2 (orbital period of Earth) = 1 day = 8.64 x 10^4 seconds
S2 (mean distance of Earth from the Sun) = 1.5 x 10^11 meters

Substituting these values into the equation:
S1 = ((3.542 x 10^7 s)^2 * (1.5 x 10^11 m)^3) / ((8.64 x 10^4 s)^2)

Evaluating this expression, we get:
S1 ≈ 1.78 x 10^11 meters (rounded to two significant figures)

It seems there was an error in the calculations mentioned in the attempted solution. It's possible that the cube root (CUBEROOT) was not applied correctly. Recheck the calculation using the equation provided above with the correct numerical values to obtain the accurate mean distance of Icarus from the Sun.