Question

The International Space Station is orbiting at an altitude of about 3270 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? [T=(2*pi*r)/(GM/r)^0.5, r=R_e+h, G=6.67*10^-11 N*m²/kg²]

Group of answer choices

Answer 1

$$T=2\pi r\sqrt{\frac{r}{GM}}\\ T=2\pi (6.378 \cdot 10^{6} )\sqrt{\frac{(6.378 \cdot 10^{6} )}{(5.976 \cdot 10^{24} )(6.67 \cdot 10^{-11} )}}=5069\ s$$