Gravitational problems are ideal for the conservation-law tools we developed in Chapters 9 through 11. Because gravity is the only force, and it is a conservative force, both the momentum and the mechanical energy of the system $m_1+m_2$ are conserved. To employ conservation of energy, however, we need to determine an appropriate form for the gravitational potential energy for two interacting masses.

The definition of potential energy that we developed in Chapter 11 is

where $W_{\text{c}}(\text{i}\to \text{f})$ is the work done by a conservative force as a particle moves from position i to position f. For a flat earth, we used $F=-mg$ and the choice that $U=0$ at the surface $(y=0)$ to arrive at the now-familiar $U_{\text{G}}=mgy$. This result for $U_{\text{G}}$ is valid only for $y\ll R_{\text{e}}$, when the earth’s curvature and size are not apparent. We now need to find an expression for the gravitational potential energy of masses that interact over large distances.

FIGURE 13.9 shows two particles of mass $m_1$ and $m_2$. Let’s calculate the work done on mass $m_2$ by the conservative force $\vec{F}{}_{1\text{on}2}$ as $m_2$ moves from an initial position at distance r to a final position very far away. The force, which points to the left, is opposite the displacement; hence this force does negative work. Consequently, due to the minus sign in Equation 13.12, $\text{Δ}U$ is positive. A pair of masses gains potential energy as the masses move farther apart, just as a particle near the earth’s surface gains potential energy as it moves to a higher altitude.

FIGURE 13.9 Calculating the work done by the gravitational force as mass $m_2$ moves from r to $\text{∞}$.

We can establish a coordinate system with $m_1$ at the origin and $m_2$ moving along the x-axis. The gravitational force is a variable force, so we need the full definition of work:

$\vec{F}{}_{1\text{on}2}$ points toward the left, so its x-component is $(F_{1\text{on}2}){}_x=-G{m}_1{m}_2\text{/}{x}^2$. As mass $m_2$ moves from $x_{\text{i}}=r$ to $x_{\text{f}}=\text{∞}$, the potential energy changes by

To proceed further, we need to choose the point where $U=0$. We would like our choice to be valid for any star or planet, regardless of its mass and radius. This will be the case if we set $U=0$ at the point where the interaction between the masses vanishes. According to Newton’s law of gravity, the strength of the interaction is zero only when $r=\text{∞}$. Two masses infinitely far apart will have no tendency, or potential, to move together, so we will choose to place the zero point of potential energy at $r=\text{∞}$. That is, $U_{\text{at}\text{∞}}=0$.

This choice gives us the gravitational potential energy of masses $m_1$ and $m_2:$

This is the potential energy of masses $m_1$ and $m_2$ when their centers are separated by distance r. FIGURE 13.10 is a graph of $U_{\text{G}}$ as a function of the distance r between the masses. Notice that it asymptotically approaches 0 as $r\to \text{∞}$.

FIGURE 13.10 The gravitational potential-energy curve.

It may seem disturbing that the potential energy is negative, but we encountered similar situations in Chapter 10. All a negative potential energy means is that the potential energy of the two masses at separation r is less than their potential energy at infinite separation. Only the change in U has physical significance, and the change will be the same no matter where we place the zero of potential energy.

To illustrate, suppose two masses a distance $r_1$ apart are released from rest. How will they move? From a force perspective, you would note that each mass experiences an attractive force and accelerates toward the other. The energy perspective of FIGURE 13.11 tells the same thing. By moving toward smaller r (that is, $r_1\to r_2$), the system loses potential energy and gains kinetic energy while conserving $E_{\text{mech}}$. The system is “falling downhill,” although in a more general sense than we think about on a flat earth.

FIGURE 13.11 Two masses gain kinetic energy as their separation decreases.

EXAMPLE 13.1 Crashing into the sun

Suppose the earth suddenly came to a halt and ceased revolving around the sun. The gravitational force would then pull it directly into the sun. What would be the earth’s speed as it crashed?

MODEL Model the earth and the sun as spherical masses. This is an isolated system, so its mechanical energy is conserved.

VISUALIZE FIGURE 13.12 is a before-and-after pictorial representation for this gruesome cosmic event. The “crash” occurs as the earth touches the sun, at which point the distance between their centers is $r_2=R_{\text{s}}+R_{\text{e}}$. The initial separation $r_1$ is the radius of the earth’s orbit about the sun, not the radius of the earth.

FIGURE 13.12 Before-and-after pictorial representation of the earth crashing into the sun (not to scale).

FIGURE 13.12 Full Alternative Text

SOLVE Strictly speaking, the kinetic energy is the sum $K=K_{\text{earth}}+K_{\text{sun}}$. However, the sun is so much more massive than the earth that the lightweight earth does almost all of the moving. It is a reasonable approximation to consider the sun as remaining at rest. In that case, the energy conservation equation $K_2+U_2=K_1+U_1$ is

$$\frac{1}{2}{M}_{\text{e}}{v}_2{}^2-\frac{G{M}_{\text{s}}{M}_{\text{e}}}{R_{\text{s}}+R_{\text{e}}}=0-\frac{G{M}_{\text{s}}{M}_{\text{e}}}{r_1}$$

This is easily solved for the earth’s speed at impact. Using data from Table 13.2, we find

$$v_2=\sqrt{2G{M}_{\text{s}}(\frac{1}{R_{\text{s}}+R_{\text{e}}}-\frac{1}{r_1})}=6.13\times {10}^5\text{ m/s}$$

REVIEW The earth would be really flying along at over 1 million miles per hour as it crashed into the sun! It is worth noting that we do not have the mathematical tools to solve this problem using Newton’s second law because the acceleration is not constant. But the solution is straightforward when we use energy conservation.

EXAMPLE 13.2 Escape speed

A 1000 kg rocket is fired straight away from the surface of the earth. What speed does the rocket need to “escape” from the gravitational pull of the earth and never return? Assume a nonrotating earth.

MODEL In a simple universe, consisting of only the earth and the rocket, an insufficient launch speed will cause the rocket eventually to fall back to earth. Once the rocket finally slows to a halt, gravity will ever so slowly pull it back. The only way the rocket can escape is to never stop $(v=0)$ and thus never have a turning point! That is, the rocket must continue moving away from the earth forever. The minimum launch speed for escape, which is called the escape speed, will cause the rocket to stop $(v=0)$ only as it reaches $r=\text{∞}$. Now $\text{∞}$, of course, is not a “place,” so a statement like this means that we want the rocket’s speed to approach $v=0$ asymptotically as $r\to \text{∞}$.

VISUALIZE FIGURE 13.13 is a before-and-after pictorial representation.

FIGURE 13.13 Pictorial representation of a rocket launched with sufficient speed to escape the earth’s gravity.

FIGURE 13.13 Full Alternative Text

SOLVE Energy conservation $K_2+U_2=K_1+U_1$ is

$$0+0=\frac{1}{2}m{v}_1{}^2-\frac{G{M}_{\text{e}}m}{R_{\text{e}}}$$

where we used the fact that both the kinetic and potential energy are zero at $r=\text{∞}$. Thus the escape speed is

$$v_{\text{escape}}=v_1=\sqrt{\frac{2G{M}_{\text{e}}}{R_{\text{e}}}}=11\text{,}200\text{ m/s}\approx 25\text{,}000\text{ mph}$$

REVIEW It is difficult to assess the answer to a problem with which we have no direct experience, although we certainly expected the escape speed to be very large. Notice that the problem was mathematically easy; the difficulty was deciding how to interpret it. That is why—as you have now seen many times—the “physics” of a problem consists of thinking, interpreting, and modeling. We will see variations on this problem in the future, with both gravity and electricity, so you might want to review the reasoning involved.