Respuesta :
Answer:
T = 5516.63 seconds
Explanation:
Given that,
The International Space Station is orbiting at an altitude of about 370 km above the earth's surface.
Mass of the Earth, [tex]M=5.976 \times 10^{24}\ kg[/tex]
Radius of Earth, [tex]r=6.378\times 10^6\ m[/tex]
We need to find the period of the International Space Station's orbit. It is a case of Kepler's third law. Its mathematical form is given by :
[tex]T^2=\dfrac{4\pi^2}{GM}\times R^3[/tex]
R = r + h
[tex]T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 5.976 \times 10^{24}}\times (370000+6.378\times 10^6)^3\\\\T^2=30433264.1641\ s\\\\T=5516.63\ s[/tex]
So, the period of the International Space Station's orbit is 5516.63 seconds.
