Let us conclude this chapter by thinking about the energetics of orbital motion. We found, with Equation 13.24, that a satellite in a circular orbit must have $v^2=GM\text{/}r$. A satellite’s speed is determined entirely by the size of its orbit. The satellite’s kinetic energy is thus

But $-GMm\text{/}r$ is the potential energy, $U_{\text{G}}$, so

This is an interesting result. In all our earlier examples, the kinetic and potential energy were two independent parameters. In contrast, a satellite can move in a circular orbit only if there is a very specific relationship between K and U. It is not that K and U have to have this relationship, but if they do not, the trajectory will be elliptical rather than circular.

Equation 13.30 gives us the mechanical energy of a satellite in a circular orbit:

The gravitational potential energy is negative, hence the total mechanical energy is also negative. Negative total energy is characteristic of a bound system, a system in which the satellite is bound to the central mass by the gravitational force and cannot get away. The total energy of an unbound system must be $\ge 0$ because the satellite can reach infinity, where $U=0$, while still having kinetic energy. A negative value of $E_{\text{mech}}$ tells us that the satellite is unable to escape the central mass.

FIGURE 13.19 shows the energies of a satellite in a circular orbit as a function of the orbit’s radius. Notice how $E_{\text{mech}}={\displaystyle \frac{1}{2}}{U}_{\text{G}}$. This figure can help us understand the energetics of transferring a satellite from one orbit to another. Suppose a satellite is in an orbit of radius $r_1$ and we’d like it to be in a larger orbit of radius $r_2$. The kinetic energy at $r_2$ is less than at $r_1$ (the satellite moves more slowly in the larger orbit), but you can see that the total energy increases as r increases. Consequently, transferring a satellite to a larger orbit requires a net energy increase $\text{Δ}E>0$. Where does this increase of energy come from?

FIGURE 13.19 The kinetic, potential, and total energy of a satellite in a circular orbit.

Artificial satellites are raised to higher orbits by firing their rocket motors to create a forward thrust. This force does work on the satellite, and the energy principle of Chapter 10 tells us that this work increases the satellite’s energy by $\text{Δ}{E}_{\text{mech}}=W_{\text{ext}}$. Thus the energy to “lift” a satellite into a higher orbit comes from the chemical energy stored in the rocket fuel.

You might think that the way to get a satellite into a larger orbit would be to point the thrusters toward the earth and blast outward. That would work fine if the satellite were initially at rest and moved straight out along a linear trajectory. But an orbiting satellite is already moving and has significant inertia. A force directed straight outward would change the satellite’s velocity vector in that direction but would not cause it to move along that line. (Remember all those earlier motion diagrams for motion along curved trajectories.) In addition, a force directed outward would be almost at right angles to the motion and would do essentially zero work. Navigating in space is not as easy as it appears in Star Wars!

To move the satellite in FIGURE 13.20 from the orbit with radius $r_1$ to the larger circular orbit of radius $r_2$, the thrusters are turned on at point 1 to apply a brief forward thrust force in the direction of motion, tangent to the circle. This force does a significant amount of work because the force is parallel to the displacement, so the satellite quickly gains kinetic energy $(\text{Δ}K>0)$. But $\text{Δ}{U}_{\text{G}}=0$ because the satellite does not have time to change its distance from the earth during a thrust of short duration. With the kinetic energy increased, but not the potential energy, the satellite no longer meets the requirement $K=-\frac{1}{2}{U}_{\text{G}}$ for a circular orbit. Instead, it goes into an elliptical orbit.

FIGURE 13.20 Transferring a satellite to a larger circular orbit.

In the elliptical orbit, the satellite moves “uphill” toward point 2 by transforming kinetic energy into potential energy. At point 2, the satellite has arrived at the desired distance from earth and has the “right” value of the potential energy, but its kinetic energy is now less than needed for a circular orbit. (The analysis is more complex than we want to pursue here. It will be left for a homework Challenge Problem.) If no action is taken, the satellite will continue on its elliptical orbit and “fall” back to point 1. But another forward thrust at point 2 increases its kinetic energy, without changing $U_{\text{G}}$, until the kinetic energy reaches the value $K=-\frac{1}{2}{U}_{\text{G}}$ required for a circular orbit. Presto! The second burn kicks the satellite into the desired circular orbit of radius $r_2$. The work $W_{\text{ext}}=\text{Δ}{E}_{\text{mech}}$ is the total work done in both burns. It takes a more extended analysis to see how the work has to be divided between the two burns, but even without those details you now have enough knowledge about orbits and energy to understand the ideas that are involved.

CHAPTER 13 CHALLENGE EXAMPLE A binary star system

Astronomers discover a binary star system with a period of $90\text{days}$. Both stars have a mass twice that of the sun. How far apart are the two stars?

MODEL Model the stars as spherical masses exerting gravitational forces on each other.

VISUALIZE An isolated system rotates around its center of mass. FIGURE 13.21 shows the orbits and the forces. If $r$ is the distance of each star to the center of mass—the radius of that star’s orbit—then the distance between the stars is $d=2r$.

FIGURE 13.21 The binary star system.

FIGURE 13.21 Full Alternative Text

SOLVE Star 2 has only one force acting on it, $\vec{F}{}_{1\text{on}2}$, and that force has to provide the centripetal acceleration $v^2\text{/}r$ of circular motion. Newton’s second law for star 2 is

$$F_{1\text{on}2}={\displaystyle \frac{G{M}_1{M}_2}{d^2}}={\displaystyle \frac{G{M}^2}{4r^2}}=M{a}_r={\displaystyle \frac{M{v}^2}{r}}$$

where we used $M_1=M_2=M$. The equation for star 1 is identical. The star’s speed is related to the period and the circumference of its orbit by $v=2\pi r\text{/}T$. With this, the force equation becomes

$${\displaystyle \frac{G{M}^2}{4r^2}}={\displaystyle \frac{4{\pi }^{2}Mr}{T^2}}$$

Solving for r gives

$$\begin{array}{lll}r\hfill & =\hfill & {\left[{\displaystyle \frac{GM{T}^2}{16{\pi }^2}}\right]}^{1\text{/}3}\hfill \\ \hfill & =\hfill & {\left[{\displaystyle \frac{(6.67\times {10}^{-11}{\text{Nm}}^2\text{/}{\text{kg}}^2)(2\times 1.99\times {10}^{30}\text{ kg}){(7.78\times {10}^6\text{s})}^2}{16{\pi }^2}}\right]}^{1\text{/}3}\hfill \\ \hfill & =\hfill & 4.67\times {10}^{10}\text{m}\hfill \end{array}$$

The distance between the stars is $d=2r=9.3\times {10}^{10}\text{m}$.

REVIEW The result is in the range of solar-system distances and thus is reasonable.