Question

The International Space Station is orbiting at an altitude of about 370 km above the earth's surface. The mass of the earth is 5.976 × 1024 kg and the radius of the earth is 6.378 × 106 m. Assuming a circular orbit, what is the period of the International Space Station's orbit?

Answer 1

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, $M=5.976 \times 10^{24}\ kg$

Radius of Earth, $r=6.378\times 10^6\ m$

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 :

$T^2=\dfrac{4\pi^2}{GM}\times R^3$

R = r + h

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 5.976 \times 10^{24}}\times (370000+6.378\times 10^6)^3\\\\T^2=30433264.1641\ s\\\\T=5516.63\ s$

So, the period of the International Space Station's orbit is 5516.63 seconds.