To find the period [tex]\( T \)[/tex] of a satellite in a circular orbit just above the surface of the Moon, we will use Kepler's Third Law adapted for circular orbits. The period [tex]\( T \)[/tex] is given by the formula:
[tex]\[
T^2 = \frac{4 \pi^2}{G M} r^3
\][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)[/tex])
- [tex]\( M \)[/tex] is the mass of the Moon ([tex]\( 7.36 \times 10^{22} \, \text{kg} \)[/tex])
- [tex]\( r \)[/tex] is the radius of the Moon ([tex]\( 1.738 \times 10^6 \, \text{m} \)[/tex])
Let's go through the solution step-by-step:
1. Substitute the known values into the formula:
[tex]\[
T^2 = \frac{4 \pi^2 (1.738 \times 10^6)^3}{6.67 \times 10^{-11} \times 7.36 \times 10^{22}}
\][/tex]
2. Calculate the numerator and the denominator separately:
- Numerator: [tex]\( 4 \pi^2 (1.738 \times 10^6)^3 \)[/tex]
- Denominator: [tex]\( G M = 6.67 \times 10^{-11} \times 7.36 \times 10^{22} \)[/tex]
3. Simplify the expression to find [tex]\( T^2 \)[/tex]:
Evaluate the numerator:
[tex]\[
(1.738 \times 10^6)^3 = 5.241 \times 10^{18} (\text{reasoning from given constants})
\][/tex]
Continuing the calculation:
[tex]\[
4 \pi^2 \cdot 5.241 \times 10^{18}
\][/tex]
Evaluate:
[tex]\[
4 \cdot 9.8696 \cdot 5.241 \times 10^{18} \approx 2.06 \times 10^{20}
\][/tex]
Evaluate the denominator:
[tex]\[
6.67 \times 10^{-11} \times 7.36 \times 10^{22} = 4.91 \times 10^{12}
\][/tex]
4. Divide the numerator by the denominator to find [tex]\( T^2 \)[/tex]:
[tex]\[
T^2 = \frac{2.06 \times 10^{20}}{4.91 \times 10^{12}} \approx 4.20 \times 10^{7}
\][/tex]
5. Take the square root of [tex]\( T^2 \)[/tex] to find [tex]\( T \)[/tex]:
[tex]\[
T = \sqrt{4.20 \times 10^{7}} \approx 6.50 \times 10^{3}
\][/tex]
The period [tex]\( T \)[/tex] of the satellite is approximately [tex]\( 6.50 \times 10^3 \)[/tex] seconds.
Therefore, the correct answer is:
[tex]\[
\boxed{6.50 \times 10^3 \, \text{seconds}}
\][/tex]
This corresponds to option C.
