To find the orbital radius of the satellite, we can use the formula for circular orbital motion:
\[ v = \sqrt{\frac{GM}{r}} \]
where:
- \( v \) is the orbital speed,
- \( G \) is the gravitational constant \((6.674 \times 10^{-11} , \text{m}^3/\text{kg/s}^2)\),
- \( M \) is the mass of the Earth \((5.972 \times 10^{24} , \text{kg})\),
- \( r \) is the orbital radius (the distance from the center of the Earth to the satellite).
Rearranging the formula to solve for \( r \):
\[ r = \frac{GM}{v^2} \]
Substituting the values:
\[ r = \frac{(6.674 \times 10^{-11} , \text{m}^3/\text{kg/s}^2)(5.972 \times 10^{24} , \text{kg})}{(4000 , \text{m/s})^2} \]
Calculating the numerator:
\[ GM = 6.674 \times 10^{-11} \times 5.972 \times 10^{24} \approx 3.986 \times 10^{14} , \text{m}^3/\text{s}^2 \]
Calculating \( v^2 \):
\[ v^2 = (4000 , \text{m/s})^2 = 16,000,000 , \text{m}^2/\text{s}^2 \]
Now plug these values into the formula for \( r \):
\[ r = \frac{3.986 \times 10^{14}}{16,000,000} \approx 2.49125 \times 10^7 , \text{m} \]
This is approximately:
\[ r \approx 2.49 \times 10^7 , \text{m} \quad \text{or} \quad 24,912 , \text{km} \]
Now, to find the altitude above the Earth's surface, we need to subtract the Earth's average radius (~6371 km):
\[ \text{Altitude} = r - R_{\text{Earth}} = 24,912 , \text{km} - 6371 , \text{km} \approx 18,541 , \text{km} \]
Thus, the orbital radius of the satellite's motion is approximately 24,912 km from the center of the Earth, and around 18,541 km above the Earth's surface.
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