To calculate the mass of the planet, we can use the formula derived from Newton's law of gravitation and centripetal force:

$v = \sqrt{\frac{GM}{r}}$

Where:

• $v$ is the orbital speed of the moon (8.0 km/s or 8000 m/s),
• $G$ is the gravitational constant ($6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}$),
• $M$ is the mass of the planet (which we need to find),
• $r$ is the orbital radius of the moon.

We also know that the moon takes 28 hours to complete one full orbit, which we will convert into seconds: $28 \, \text{hours} = 28 \times 3600 \, \text{seconds} = 100800 \, \text{seconds}.$

The distance covered in one full orbit (the circumference of the orbit) is: $\text{circumference} = 2\pi r.$

Using the relation between speed and distance, we have: $v = \frac{\text{distance}}{\text{time}} = \frac{2\pi r}{T},$ where $T$ is the orbital period (100800 seconds). Solving for $r$:

$r = \frac{vT}{2\pi}.$

Now, substitute the values of $v$ and $T$:

$r = \frac{(8000 \, \text{m/s})(100800 \, \text{s})}{2\pi} \approx 1.28 \times 10^9 \, \text{m}.$

Next, we use the first equation $v = \sqrt{\frac{GM}{r}}$ to solve for $M$:

$M = \frac{v^2 r}{G}.$

Substitute the values:

$M = \frac{(8000 \, \text{m/s})^2 (1.28 \times 10^9 \, \text{m})}{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}}.$

Now, calculate the mass:

$M \approx 9.79 \times 10^{25} \, \text{kg}.$

Thus, the mass of the planet is approximately $\boxed{9.79 \times 10^{25} \, \text{kg}}$.

Would you like further details or have any questions?

Here are some related questions to expand on this topic:

1. What factors affect the orbital speed of a moon around its planet?
2. How does the mass of a planet influence the orbits of its moons?
3. What would happen if the moon's orbital speed were to increase?
4. How can you derive Kepler's Third Law from this setup?
5. How do astronomers measure the mass of distant planets using their moons' orbits?

Tip: When calculating orbital dynamics, always make sure units are consistent, particularly when converting time and distances.