To determine the velocity required for the International Space Station (ISS) to stay in orbit, we can use the formula for the orbital velocity (\(v\)) of an object in circular orbit around a planet:
\[ v = \sqrt{\frac{GM}{r}} \]
Where:
- \(G\) is the universal gravitational constant \((6.674 \times 10^{-11} , \text{m}^3 \text{kg}^{-1} \text{s}^{-2})\),
- \(M\) is the mass of the Earth \((5.972 \times 10^{24} , \text{kg})\),
- \(r\) is the distance from the center of the Earth to the object in orbit (which is the radius of the Earth plus the altitude of the orbit).
We can find \(r\) as follows:
\[ r = \text{radius of the Earth} + \text{altitude of the ISS} \] \[ r = (6.37 \times 10^6 , \text{m}) + (4.08 \times 10^5 , \text{m}) = 6.77 \times 10^6 , \text{m} \]
Now we can calculate the orbital velocity:
\[ v = \sqrt{\frac{(6.674 \times 10^{-11} , \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) \times (5.972 \times 10^{24} , \text{kg})}{6.77 \times 10^6 , \text{m}}} \]
Calculating the gravitational force:
\[ GM = (6.674 \times 10^{-11}) \times (5.972 \times 10^{24}) \approx 3.986 \times 10^{14} , \text{m}^3 \text{s}^{-2} \]
Now plug in the values into the velocity formula:
\[ v = \sqrt{\frac{3.986 \times 10^{14}}{6.77 \times 10^6}} \approx \sqrt{5.88 \times 10^7} \approx 7.66 \times 10^3 , \text{m/s} \]
Thus, the velocity at which the ISS must travel to stay in its orbit is approximately:
\[ \boxed{7.66 \times 10^3 , \text{m/s}} \]
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