Two \(50-\mathrm{kg}\) satellites move in circular orbits about the Earth at altitudes of \(1000 \mathrm{~km}\) (about \(620 \mathrm{mi}\) ) and \(36000 \mathrm{~km}\) (about \(22000 \mathrm{mi}\) ), respectively. The lower one monitors particles about to enter the atmosphere, and the higher, geosynchronous one takes weather pictures from its stationary position with respect to the Earth's surface over the equator (see Example 7.13). What is the difference in the gravitational potential energies of the two satellites in their respective orbits?
THINKING IT THROUGH. The potential energies of the satellites are given by Equation 7.19. Since an increase in altitude \((h)\) results in a less negative value of \(U\), the satellite with the greater \(h\) is higher in the gravitational-potential energy well and has more gravitational potential energy.
 Example 7.13: Some communication and weather satellites are launched into circular orbits above the Earth's equator so they are synchronous (from the Greek syn-, same, and chronos, time) with the Earth's rotation. That is, they remain "fixed" or "hover" over one point on the equator. At what altitude are these geosynchronous satellites?
THINKING IT THROUGH. To remain above one location at the equator, the period of the satellite's revolution must be the same as the Earth's period of rotation, that is, \(24 \mathrm{~h}\). Also, the centripetal force keeping the satellite in orbit is supplied by the gravitational force of the Earth, \(F_{\mathrm{g}}=F_{c}\). The distance between the center of the Earth and the satellite is \(r=R_{\mathrm{E}}+h .\left(R_{\mathrm{E}}\right.\) is the radius of the Earth and \(h\) is the height or altitude of the satellite above the Earth's surface.)
Equation(7.19)

Transcribed Image Text:

U= Gmm r GmME RE +h (7.19)