The period of oscillation of a near-Earth satellite (neglecting atmospheric effects) is 84.3 min. What is the period of a near-Moon satellite? (R_E = 6.37 × 10^6 m; R_M = 1.74 × 10... The period of oscillation of a near-Earth satellite (neglecting atmospheric effects) is 84.3 min. What is the period of a near-Moon satellite? (R_E = 6.37 × 10^6 m; R_M = 1.74 × 10^6 m; M_E = 5.98 × 10^24 kg; M_M = 7.36 × 10^22 kg.) Read more

Steps to Solve

1. Understand the relationship of periods of satellites The period of a satellite in orbit depends on the radius of the orbit and the mass of the celestial body it orbits. The formula for the period $T$ of a satellite is given by:

$$ T = 2\pi \sqrt{\frac{R^3}{GM}} $$

where $R$ is the distance from the center of the celestial body to the satellite, $G$ is the gravitational constant ($6.674 \times 10^{-11} , \text{m}^3 \text{kg}^{-1} \text{s}^{-2}$), and $M$ is the mass of the celestial body.

2. Calculate the period for the near-Earth satellite Using the given period of the near-Earth satellite, we can relate the periods for satellites around Earth and the Moon. Since we have:

$$ T_E = 2\pi \sqrt{\frac{R_E^3}{GM_E}} = 84.3 \text{ min} $$

3. Calculate the period for the near-Moon satellite For the near-Moon satellite, we will use the same period formula but with parameters for the Moon:

$$ T_M = 2\pi \sqrt{\frac{R_M^3}{GM_M}} $$

To find the relationship, we can divide the two equations:

$$ \frac{T_M^2}{T_E^2} = \frac{R_M^3/M_M}{R_E^3/M_E} $$

Now, substitute the known values:

• $R_M = 6.37 \times 10^6 , \text{m} + 1.74 \times 10^6 , \text{m} = 8.11 \times 10^6 , \text{m}$
• $R_E = 6.37 \times 10^6 , \text{m}$
• $M_M = 7.36 \times 10^{22} , \text{kg}$
• $M_E = 5.98 \times 10^{24} , \text{kg}$

4. Substituting the values into the equation Now we have:

$$ \frac{T_M^2}{(84.3 \times 60)^2} = \frac{(8.11 \times 10^6)^3 / (7.36 \times 10^{22})}{(6.37 \times 10^6)^3 / (5.98 \times 10^{24})} $$

This can be solved to find $T_M$.

5. Solve for (T_M) and convert back to minutes First, compute each part of the right side, then multiply by $(84.3 \times 60)^2$ and take the square root to find $T_M$.