Chapter 13, Problem 5Q

To determine

To analyze: That the Galilean satellites obey Kepler’s third law using the below table.

Galilean satellite	Average distance from the Jupiter (km)	Orbital period (days)
Io	421600	1.769
Europe	670900	3.551
Ganymede	1070000	7.155
Callisto	1883000	16.689

Answer to Problem 5Q

Solution:

All Galilean satellite follow Kepler’s third law.

Explanation of Solution

Given data:

Galilean satellite	Average distance from the Jupiter (km)	Orbital period (days)
Io	421600	1.769
Europe	670900	3.551
Ganymede	1070000	7.155
Callisto	1883000	16.689

Formula used:

Kepler’s third law states that the square of the period of orbit for an object is directly proportional to the cube of the orbit’s semi-major axis of an object. That is,

$\begin{array}{c}{P}^2\propto a^3\\ \frac{a^3}{P^2}=\text{ constant}\end{array}$

Here, P is the orbital period and a is the semi-major axis.

Explanation:

Understand that, if the ratio $\frac{a^3}{p^2}$ for each satellite is same, then can say that every Galilean satellite follows Kepler’s law.

Evaluate the ratio of the cube of the orbit’s semi-major axis of Io and the square of the period of orbit for Io as,

$Rati{o}_{Io}=\frac{a_{Io}^3}{P_{Io}^2}$

Here, the subscript Io refers to the corresponding quantities for satellite Io.

Substitute $421600\text{ km}$ for $a_{Io}$ (from the given table) and $1.769\text{ days}$ for $P_{Io}$ (from the given table).

$\begin{array}{c}Rati{o}_{Io}=\frac{{\left(421600\text{ km}\right)}^3}{{\left(1.769\text{}\text{ days}\right)}^2}\\ =2.39\times {10}^{16}{\text{ km}}^{\text{3}}{\text{/days}}^{\text{2}}\end{array}$ …… (1)

Evaluate the ratio of the cube of the orbit’s semi-major axis of Europa and the square of the period of orbit for Europa as,

$Rati{o}_{Europa}=\frac{a_{Europa}^3}{P_{Europa}^2}$

Here, the subscript Europa refers to the corresponding quantities for satellite Europa.

Substitute $\text{670900 km}$ for $a_{Europa}$ (from the given table) and $3.551\text{}\text{ days}$ for $P_{Europa}$ (from the given table).

$\begin{array}{c}Rati{o}_{Europa}=\frac{{\left(\text{670900 km}\right)}^3}{{\left(3.551\text{}\text{ days}\right)}^2}\\ =2.39\times {10}^{16}{\text{ km}}^{\text{3}}{\text{/days}}^{\text{2}}\end{array}$ …… (2)

Evaluate the ratio of the cube of the orbit’s semi-major axis of Ganymede and the square of the period of orbit for Ganymede as,

$Rati{o}_{Ganymede}=\frac{a_{Ganymede}^3}{P_{Ganymede}^2}$

Here, the subscript Ganymede refers to the corresponding quantities for satellite Ganymede.

Substitute $\text{1070000 km}$ for $a_{Ganymede}$ (from the given table) and $7.155\text{}\text{ days}$ for $P_{Ganymede}$ (from the given table),

$\begin{array}{c}Rati{o}_{Ganymede}=\frac{{\left(\text{1070000 km}\right)}^3}{{\left(7.155\text{}\text{ days}\right)}^2}\\ =2.39\times {10}^{16}{\text{ km}}^{\text{3}}{\text{/days}}^{\text{2}}\end{array}$ …… (3)

Evaluate the ratio of the cube of the orbit’s semi-major axis of Callisto and the square of the period of orbit for Callisto as,

$Rati{o}_{Callisto}=\frac{a_{Callisto}^3}{P_{Callisto}^2}$

Here, the subscript Callisto refers to the corresponding quantities for satellite Callisto.

Substitute $\text{1883000 km}$ for $a_{Callisto}$ (from the given table) and $\text{16}\text{.689 days}$ for $P_{Callisto}$ (from the given table),

$\begin{array}{c}Rati{o}_{Callisto}=\frac{{\left(\text{1883000 km}\right)}^3}{{\left(\text{16}\text{.689 days}\right)}^2}\\ =2.39\times {10}^{16}{\text{ km}}^{\text{3}}{\text{/days}}^{\text{2}}\end{array}$ …… (4)

As seen from the equation (1), (2), (3), and (4), the ratio $\frac{a^3}{p^2}$ for all the Galilean satellite is same.

Conclusion:

The ratio $\frac{a^3}{p^2}$ for all Galilean satellites is the same, so it can be said that all Galilean satellites follow Kepler’s third law.