91Ó°ÊÓ

Objects that orbit around other objects are classified as satellites. Moon orbits around the earth, and the earth orbits around the sun. Moon is the satellite of the earth, and the earth is a satellite of the sun.

There are satellites that are placed in orbits around planets. These are artificial satellites.

The connection that can be used to calculate a parent body's mass M from its satellites' orbits:

$\frac{r^3}{T^2}=\frac{G}{4{\mathrm{Ï€}}^2}M\mathrm{...............}\left(1\right)$

If r and T for a satellite are known, the parent's mass M can be determined. This approach has been widely applied to determining the masses of celestial entities with satellites.

When the earth is a parent body and the moon is its satellite then equation ( 1 ) can be written as,

$\frac{r^3}{T^2}=\frac{G}{4p^2}{M}_{earth}$

From the Table 6.2, the value of $\frac{{\text{r}}^{\text{3}}}{{\text{T}}^{\text{2}}}$is equal to ${\text{1.01×10}}^{\text{18}}\text{ }\frac{{\text{km}}^{\text{3}}}{{\text{y}}^{\text{2}}}$. As a result, using this number in the calculation above,

$1.01×{10}^{19}\text{ }\frac{{\text{km}}^{\text{3}}}{{\text{y}}^{\text{2}}}=\frac{G}{4{\mathrm{π}}^2}{M}_{earth}\mathrm{...................}\left(2\right)$

The universal gravitational constant, G, is used here.

When the Jupiter is a parent body and the Europa is its satellite then equation ( 1 ) can be written as

$\frac{r^3}{T^2}=\frac{G}{4{\mathrm{Ï€}}^2}{M}_{Jupiter}$

For the Jupiter and the Europa, From the Table 6.2, the value of $\frac{{\text{r}}^{\text{3}}}{{\text{T}}^{\text{2}}}$is equal to $1.01×{10}^{21}\text{ }\frac{{\text{km}}^{\text{3}}}{{\text{y}}^{\text{2}}}$.

As a result, using this number in the calculation above

$3.19×{10}^{21}\text{ }\frac{{\text{km}}^{\text{3}}}{{\text{y}}^{\text{2}}}=\frac{G}{4{\mathrm{π}}^2}{M}_{Jupiter}\mathrm{..................}\left(3\right)$

The universal gravitational constant, G, is used here.

To calculate the ratio of the mass of Jupiter and the mass of the earth, we are dividing equation (3) with equation (2).

$\begin{array}{c}\frac{{\displaystyle \frac{G}{4{\mathrm{π}}^2}}{M}_{Jupiter}}{{\displaystyle \frac{G}{4{\mathrm{π}}^2}}{M}_{Earth}}=\frac{3.19×{10}^{21}\text{ }{\displaystyle \frac{{\mathrm{km}}^3}{{\mathrm{y}}^2}}}{1.01×{10}^{19}\text{ }{\displaystyle \frac{{\mathrm{km}}^3}{{\mathrm{y}}^2}}}\\ \frac{M_{Jupiter}}{M_{Earth}}=315.8\end{array}$

As a result, the mass of Jupiter is roughly 316 times the mass of Earth.