Question 4.5: A satellite is circulating the Earth at an altitude h = 150 ......
A satellite is circulating the Earth at an altitude h = 150 km above its surface, where the free fall acceleration g is 9.4 m/s². The Earth’s radius is 6.4 × 10^{6} m. What is the orbital Speed and period of the satellite?
As shown in Fig. 4.15, the radius of the satellite’s circular motion equals the sum of the Earth’s radius R and the altitude h, i.e.
r = R + h
By using the centripetal acceleration given by Eq. 4.29,
a_{r} = \frac{v^{2}}{r}, (Radial acceleration) (4.29)
we find that the magnitude of the satellite’s acceleration can be written as:
a_{\mathrm{r}}={\frac{v^{2}}{r}}={\frac{v^{2}}{R+h}}For the uniform circular motion of the satellite around the Earth, the satellite’s centripetal acceleration is then equal to the free fall acceleration g at this altitude. That is:
a_{\mathrm{r}}=g=9.4\,{\mathrm{m/s}}^{2}From the preceding two equations we have:
g={\frac{v^{2}}{R+h}}Solving for v and taking the positive root gives:
\ v={\sqrt{g(R+h)}}
={\sqrt{(9.4\,{\mathrm{m/s}}^{2})(6.4\times10^{6}\ {\mathrm{m}}+150\times10^{3}\ {\mathrm{m}})}}=7,847\,{\mathrm{m/s}}\approx28,000\,{\mathrm{km/h}}
With this high speed, the satellite would take T = 2πr/v = 1.46 h to make one complete revolution around the Earth.
