Question 13.8: KEPLER'S THIRD LAW The asteroid Pallas has an orbital period...

IDENTIFY and SET UP:

We need Kepler’s third law, which relates the period T and the semi-major axis a for an orbiting object (such as an asteroid). We use Eq. (13.17) to determine a; from Appendix F we have m_S = 1.99 × 10^{30} kg, and a conversion factor from Appendix E gives T=(4.62 \mathrm{yr})\left(3.156 \times 10^{7} \mathrm{~s} / \mathrm{yr}\right)=1.46 \times 10^{8} \mathrm{~s}. Note that we don’t need the value of the eccentricity.

T=\frac{2 \pi a^{3 / 2}}{\sqrt{G m_{\mathrm{S}}}}                    (13.17)

EXECUTE:

From Eq. (13.17), a^{3 / 2}=\left[\left(G m_{\mathrm{S}}\right)^{1 / 2} T\right] / 2 \pi. To solve for a, we raise both sides of this expression to the \frac{2}{3} power and then substitute the values of G, m_S, and T:

(Plug in the numbers yourself to check.)

EVALUATE: Our result is intermediate between the semi-major axes of Mars and Jupiter (see Appendix F). Most known asteroids orbit in an “asteroid belt” between the orbits of these two planets.