Sure! Let's solve this question step-by-step.
### Variables and Constants:
1. Mass of the Moon, [tex]\( M \)[/tex]: [tex]\( 7.36 \times 10^{22} \)[/tex] kilograms
2. Radius of the Moon, [tex]\( R \)[/tex]: [tex]\( 1.738 \times 10^6 \)[/tex] meters
3. Gravitational Constant, [tex]\( G \)[/tex]: [tex]\( 6.67430 \times 10^{-11} \)[/tex] m[tex]\(^3\)[/tex] kg[tex]\(^{-1}\)[/tex] s[tex]\(^{-1}\)[/tex]
### Formula:
The formula to calculate the period [tex]\( T \)[/tex] of a satellite in a circular orbit just above the surface of a celestial body is given by:
[tex]\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \][/tex]
where:
- [tex]\( r \)[/tex] is the orbital radius (which would be just above the Moon's surface, so [tex]\( r = R \)[/tex])
- [tex]\( G \)[/tex] is the gravitational constant
- [tex]\( M \)[/tex] is the mass of the celestial body
### Step-by-Step Solution:
1. Identify the orbital radius: The satellite is in a circular orbit just above the Moon's surface. Therefore, the orbital radius [tex]\( r \)[/tex] is the same as the radius of the Moon, [tex]\( R \)[/tex]. So,
[tex]\[ r = 1.738 \times 10^6 \text{ meters} \][/tex]
2. Substitute the known values into the formula:
[tex]\[ T = 2\pi \sqrt{\frac{(1.738 \times 10^6)^3}{(6.67430 \times 10^{-11}) \times (7.36 \times 10^{22})}} \][/tex]
3. Calculate the numerator (Orbital radius cubed):
[tex]\[ (1.738 \times 10^6)^3 = 5.244896 \times 10^{18} \][/tex]
4. Calculate the denominator (Gravitational constant times the mass of the Moon):
[tex]\[ (6.67430 \times 10^{-11}) \times (7.36 \times 10^{22}) = 4.9138768 \times 10^{12} \][/tex]
5. Divide the numerator by the denominator and take the square root:
[tex]\[ \frac{5.244896 \times 10^{18}}{4.9138768 \times 10^{12}} = 1.067525106 \times 10^6 \][/tex]
[tex]\[ \sqrt{1.067525106 \times 10^6} = 1032.3208 \text{ seconds} \][/tex]
6. Multiply by [tex]\( 2\pi \)[/tex]:
[tex]\[ T = 2\pi \times 1032.3208 \text{ seconds} \][/tex]
[tex]\[ T \approx 6495.5025 \text{ seconds} \][/tex]
### Conclusion:
The period of a satellite in a circular orbit just above the surface of the Moon is approximately [tex]\( 6495.5025 \)[/tex] seconds or roughly [tex]\( 108.26 \)[/tex] minutes.
