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What is the period of a satellite in a circular orbit just above the surface of the Moon? The Moon's mass is [tex]$7.36 \times 10^{22}$[/tex] kilograms and its radius is [tex]$1.738 \times 10^6$[/tex] meters.

[tex] G = 6.67 \times 10^{-11} \frac{Nm^2}{kg^2} \quad T^2 = \frac{4 \pi^2}{GM} r^3 [/tex]

A. [tex]1.94 \times 10^4[/tex] seconds
B. [tex]1.30 \times 10^4[/tex] seconds
C. [tex]6.50 \times 10^3[/tex] seconds
D. [tex]3.24 \times 10^3[/tex] seconds
E. [tex]1.62 \times 10^3[/tex] seconds


Sagot :

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.
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