13.4 Satellite orbits and energy  (Page 5/8)

Solution

The total energy at Earth’s surface is

The change in energy is $\text{Δ}E=E_{\text{orbit}}-E_{\text{surface}}=2.98\times {10}^{11}\text{J}$ . To get the kinetic energy, we subtract the change in potential energy from [link] , $\text{Δ}U=3.32\times {10}^{10}\text{J}$ . That gives us $K_{\text{orbit}}=2.98\times {10}^{11}-3.32\times {10}^{10}=2.65\times {10}^{11}\text{J}$ . As stated earlier, the kinetic energy of a circular orbit is always one-half the magnitude of the potential energy, and the same as the magnitude of the total energy. Our result confirms this.

The second approach is to use [link] to find the orbital speed of the Soyuz , which we did for the ISS in [link] .

So the kinetic energy of the Soyuz in orbit is

the same as in the previous method. The total energy is just

Significance

Summary

• Orbital velocities are determined by the mass of the body being orbited and the distance from the center of that body, and not by the mass of a much smaller orbiting object.
• The period of the orbit is likewise independent of the orbiting object’s mass.
• Bodies of comparable masses orbit about their common center of mass and their velocities and periods should be determined from Newton’s second law and law of gravitation.

Conceptual questions