To solve for the radius of the satellite's orbit around the Sun, we can use Kepler's Third Law of Planetary Motion, which relates the orbital period of a satellite to the radius of its orbit around the Sun. Kepler's Third Law is given by:

$T^2 = \frac{4\pi^2}{GM_{\odot}}r^3$

Where:

• $T$ is the orbital period of the satellite (in seconds),
• $G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}$),
• $M_{\odot}$ is the mass of the Sun ($1.989 \times 10^{30} \, \text{kg}$),
• $r$ is the radius of the orbit (which we want to find).

We can rearrange the equation to solve for $r$:

$r^3 = \frac{GMT^2}{4\pi^2}$

Step 1: Convert the orbital period $T$ to seconds.

Given that the period $T$ is 2.00 days, we need to convert this to seconds:

$T = 2.00 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour}$ $T = 2.00 \times 86400 \, \text{s} = 172800 \, \text{s}$

Step 2: Substitute known values into the equation.

Now, substitute $T = 172800 \, \text{s}$, $G = 6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}$, and $M_{\odot} = 1.989 \times 10^{30} \, \text{kg}$ into the rearranged Kepler's Third Law:

$r^3 = \frac{(6.674 \times 10^{-11})(1.989 \times 10^{30})(172800)^2}{4\pi^2}$

Step 3: Calculate the value of $r$.

Let's compute this value.The radius of the satellite's orbit around the Sun is approximately $4.65 \times 10^9$ meters, or 4.65 million kilometers.

Would you like more details or have any questions?

Here are 5 related questions to expand on this concept:

1. How would the radius change if the satellite's period was 5 days instead of 2 days?
2. How is Kepler's Third Law applied to objects orbiting planets like Earth?
3. How does the mass of the central object (the Sun in this case) affect the orbit of a satellite?
4. What assumptions are made in Kepler’s Laws about the shapes of orbits?
5. How does the gravitational force between two objects depend on their distance?

Tip: The orbital period and the radius of an orbit are strongly related. Increasing the orbital period significantly increases the orbital radius due to the cubic relationship in Kepler's Third Law.