Problem 11.8

11.8.

We know that Kepler's third law states that
T² = k a^2/3
where T is the period and a is the semi-major axis length. In astronaumical units, that is, when the period is given in years and the distance a is given in A.U., then the value for k is
k = 1
So, plugging in the numbers,
$(75.6)^{2} = (\frac{.057 + x}{2})^{3}$
where we have used the geometry of the problem to get the distance a in terms of the distance x.
solving for x, we find
x = 2(75.6)^2/3 - 0.57  =  35.2 A.U.
which is near the orbit of Pluto.