Answer
(a) Solve Kepler's third law (Equation 12-7) for the mass of 243 Ida, using the orbit distance and period given in the problem.
(b) $8.9\times 10^{16}Kg$
Work Step by Step
(a) We will use Kepler's third law to find the mass of 243 Ida; that is: $M=(\frac{2\pi}{T})^2\frac{r^3}{G}$. Using this equation, we can find the required mass by substituting the values of the orbit distance and period, which are given in the problem.
(b) We know that
$M_{243 Ida}=(\frac{2\pi}{T})^2\frac{r^3}{G}$
We plug in the known values to obtain:
$M_{243}Ida=(\frac{2\pi}{19h\times 3600s/h})^2(\frac{(89\times 10^3)^3}{6.67\times 10^{-11}})$
$M_{243}Ida=8.9\times 10^{16}Kg$
