The space

The space rocket orbits the Earth at an altitude of 400 km above the Earth's surface. Determine the magnitude of the centrifugal acceleration of the missile in its orbit.

Correct answer:

a =  8.6863 m/s²

Step-by-step explanation:

$$\begin{gathered} g=9.81{\text{m/s}}^2 \\ h=400\text{km} \\ R=6378\text{km} \\ {F}_1=F_2=a\cdot m \\ {F}_1=k\cdot \frac{m\cdot M}{r^2} \\ g=k\cdot M/R^2 \\ a=k\cdot M/(R+h{)}^2 \\ a=k\cdot \frac{g\cdot R^2}{k}/(R+h{)}^2 \\ a=g\cdot (\frac{R}{R+h}{)}^2=9.81\cdot (\frac{6378}{6378+400}{)}^2=8.6863{\text{m/s}}^2 \end{gathered}$$

Try another example