In 1993 the spacecraft Galileosent an image (Fig. 13-48) of asteroid 243 Ida and a tiny orbiting moon (now known as Dactyl), the first confirmed example of an asteroid–moon system. In the image, the moon, which is $\mathbf{\text{1.5km}}$wide, is$\mathbf{\text{100km}}$ from the center of the asteroid, which is role="math" localid="1661157158474" $\mathbf{\text{55km}}$long. Assume the moon’s orbit is circular with a period of $\mathbf{27}\mathbf{}\mathbf{h}$.

(a) What is the mass of the asteroid?

(b) The volume of the asteroid, measured from the Galileoimages, is${\mathbf{\text{14100\hspace{0.17em}\hspace{0.17em}km}}}^{\mathbf{\text{3}}}$ . What is the density (mass per unit volume) of the asteroid? was sent spinning out of control. Just before the collision and in

The diameter of the moon is,${\text{d}}_{\text{M}}=1.5\text{\hspace{0.17em}km}$

The diameter of the asteroid is,${\text{d}}_{\text{A}}=55\text{\hspace{0.17em}km}$

The distance of the moon from the asteroid is,$\mathrm{r}=100\text{\hspace{0.17em}km}={10}^5\text{\hspace{0.17em}m}$

The period of the moon’s orbit is,$\mathrm{T}=27\text{h}=9720\text{0s}$

The volume of the asteroid is,$\mathrm{V}=14100{\text{\hspace{0.17em}km}}^{\text{3}}=1.41\times {10}^{13}{\text{\hspace{0.17em}m}}^{\text{3}}$

Findthemass of the asteroid by using Kepler’s third law. Then using it, find the density of the asteroid. According to Kepler’s third law, the squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of their orbits.

Formulae are as follows:

${\mathrm{T}}^2=\frac{4{\mathrm{\pi }}^2}{\mathrm{GM}}{\mathrm{r}}^3$

$\mathrm{\rho }=\frac{\mathrm{M}}{\mathrm{V}}$

where T is time, G is gravitational constant, M is mass, V is volume, $\mathrm{\rho }$ is density and r is the radius.

Kepler’s third law gives,

${\mathrm{T}}^2=\frac{4{\mathrm{\pi }}^2}{\mathrm{GM}}{\mathrm{r}}^3$

$\begin{array}{c}{\left(97200\mathrm{s}\right)}^2=\frac{4{\left(3.142\right)}^2}{(6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2)\times \mathrm{M}}\left({10}^{15}{\mathrm{m}}^3\right)\\ \mathrm{M}=\frac{4{\left(3.142\right)}^2}{6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2\times {\left(97200\mathrm{s}\right)}^2}\left({10}^{15}{\mathrm{m}}^3\right)\\ \mathrm{M}=6.3\times {10}^{16}\mathrm{kg}\end{array}$

Therefore, the mass of the asteroid is $6.3\times {10}^{16}\mathrm{kg}$.

The density of the asteroid is,

$\mathrm{\rho }=\frac{\mathrm{M}}{\mathrm{V}}$

$\mathrm{\rho }=\frac{6.3\times {10}^{16}\mathrm{kg}}{1.41\times {10}^{13}{\mathrm{m}}^3}$

$\mathrm{\rho }=4.4\times {10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

Therefore, the density of the asteroid is ${\text{4.4×10}}^{\text{3}}\frac{\text{kg}}{{\text{m}}^{\text{3}}}$.

Therefore, using Kepler’s third law, the mass of the body around which the other body is orbiting can be found.