91影视

The diameter of the moon is,${\text{d}}_{\text{M}}=1.5\text{鈥塳尘}$

The diameter of the asteroid is,${\text{d}}_{\text{A}}=55\text{鈥塳尘}$

The distance of the moon from the asteroid is,$\mathrm{r}=100\text{鈥塳尘}={10}^5\text{鈥尘}$

The period of the moon鈥檚 orbit is,$\mathrm{T}=27\text{h}=9720\text{0s}$

The volume of the asteroid is,$\mathrm{V}=14100{\text{鈥塳尘}}^{\text{3}}=1.41脳{10}^{13}{\text{鈥尘}}^{\text{3}}$

Findthemass of the asteroid by using Kepler鈥檚 third law. Then using it, find the density of the asteroid. According to Kepler鈥檚 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:

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

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

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

Kepler鈥檚 third law gives,

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

$\begin{array}{c}{\left(97200\text{鈥�}\mathrm{s}\right)}^2=\frac{4{\left(3.142\right)}^2}{(6.67脳{10}^{鈭�11}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2)脳\mathrm{M}}\left({10}^{15}\text{鈥�}{\mathrm{m}}^3\right)\\ \mathrm{M}=\frac{4{\left(3.142\right)}^2}{6.67脳{10}^{鈭�11}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2脳{\left(97200\text{鈥�}\mathrm{s}\right)}^2}\left({10}^{15}\text{鈥�}{\mathrm{m}}^3\right)\\ \mathrm{M}=6.3脳{10}^{16}\text{鈥�}\mathrm{kg}\end{array}$

Therefore, the mass of the asteroid is $6.3脳{10}^{16}\text{鈥�}\mathrm{kg}$.

The density of the asteroid is,

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

$\mathrm{蚁}=\frac{6.3脳{10}^{16}\text{鈥�}\mathrm{kg}}{1.41脳{10}^{13}\text{鈥�}{\mathrm{m}}^3}$

$\mathrm{蚁}=4.4脳{10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

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

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