Chapter 7.1, Problem 1PP

To determine

To Calculate: The orbital radius of Ganymede.

Answer to Problem 1PP

  $\text{2}\text{.07}\times {\text{10}}^7\text{m}$

Explanation of Solution

Given:

Orbital period of Ganymede, $T=32\text{days}$

Formula used:

From Kepler’s third law of planetary motion:

  $r^3=\frac{G{m}_J{T}^2}{4{\pi }^2}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_J$ is the mass of the Jupiter given as $1.89\times {10}^{27}\text{kg}$ and T is the orbital period.

Calculations:

Convert period from days to seconds:

  $\begin{array}{c}T=32\text{days}\times 24\frac{\text{hours}}{\text{day}}\times 60\frac{\text{min}}{\text{hour}}\times 60\frac{\text{s}}{\text{min}}\\ =2.76\times {10}^6\text{s}\end{array}$

Substitute the values and solve for orbital radius of Ganymede:

  $\begin{array}{c}r={\left(\frac{G{m}_J{T}^2}{4{\pi }^2}\right)}^{1/3}\\ ={\left(\frac{6.67\times {10}^{-11}\times 1.89\times {10}^{27}\times 2.76\times {10}^6}{4\times {\left(3.14\right)}^2}\right)}^{1/3}\\ =20662957.3\text{m}\\ \text{=2}\text{.07}\times {\text{10}}^7\text{m}\end{array}$

Conclusion:

Thus, the orbital radius of Ganymede is $\text{2}\text{.07}\times {\text{10}}^7\text{m}$ .