In the last section we defined the potential energy associated with the earth/ball system. The ball really represented any mass near earth's surface. We saw that the potential energy difference was given by [math]\Delta U_g=mg\Delta y[/math] That equation is only valid near earth's surface. By "near" I mean where the value of the gravitational constant g is not noticeably changing. Out at distances where satellites orbit that gravitational constant is getting smaller. Therefore, at distance where earth's gravitational field strength is decreasing, we need to be more careful in calculating work done by gravity, which will be used to define the associated potential energy of gravity. This is the goal of this section.

Gravity is a conservative vector field, and not just near earth's surface. As you probably recall, the force of universal gravitation may be written [math]\vec{F}_G=-\frac{GMm}{r^2}\hat{r}[/math] where the [math]-\hat{r}[/math] represents the attractive (inward radial) nature of the force. [br][br]Let's suppose that the big 'M' is planet earth's mass, and that the little 'm' represents some object's mass. Suppose the object starts at one separation distance from earth [math]r_i[/math] and ends at a different distance [math]r_f.[/math] Let's calculate the work done by the gravitational force in this process.[br][br][center][math][br]W=\int\vec{F_G}\cdot\vec{ds} \\[br]W=\int_{r_i}^{r_f}-\frac{GMm}{r^2}\hat{r}\cdot (dr\hat{r}+rd \theta \hat{\theta}) \\[br]W=\int_{r_i}^{r_f}-\frac{GMm}{r^2}dr \\[br]W=\frac{GMm}{r}\Big |_{r_i}^{r_f} \\[br]W=\frac{GMm}{r_f} - \frac{GMm}{r_i} \\[br]\text{Recall that \Delta U =}-W_{C} \\[br]\Delta U_G = - \frac{GMm}{r_f} + \frac{GMm}{r_i} \\[br]\text{A convention used everywhere is that $U(r=\infty)\equiv 0$. Therefore: } \\[br]U_{G,f}-U_{G,i} = - \frac{GMm}{r_f} + \frac{GMm}{r_i} \\[br]0-U_{G,i} = - \frac{GMm}{\infty} + \frac{GMm}{r_i} \\[br]-U_{G,i} = \frac{GMm}{r_i} \\[br]U_{G} = -\frac{GMm}{r} \\[br][/math][/center][br][br]In the last line we have an expression for the gravitational potential energy of a two-mass system separated by a distance r. It is really the value of the potential energy relative to infinite separation. In this sense while we write it without the delta, it really is a difference. It's the difference in the potential energy as compared with the value of that energy while at infinite separation distance. Mathematically, [math]\Delta U_G = U_G(r)-U_G(\infty)=U_G(r).[/math] Since the value at infinity is zero, the quantity is the same as the difference.[br][br]It's worth noting that the potential energy is negative relative to infinity. The meaning of this is that the little mass m is in a bound state - or is bound by gravity to exist close to the larger mass M. Unless it somehow acquires enough kinetic energy to overcome this negative potential energy, the little mass will always be bound to M. Freedom in that sense would only occur when [math]K+U\ge 0[/math], or when [math]K\ge -U.[/math] Under such conditions the mass would be free to roam anywhere it wishes... all the way out to infinity where the potential energy is zero.

Freedom (to a particle) is freedom to roam through space. Absolute freedom is freedom to - at least in principle - be able to roam to infinite distance. We know from above that infinite separation distance implies a potential energy of zero. Therefore massive objects like a satellite in orbit or our moon orbiting earth have negative potential energies. [br][br]Since the notion of freedom corresponds to having the ability to reach infinity, the object would have to have enough kinetic energy to get it there. This means an object nearby would need to possess enough kinetic energy to exchange for (or pay back) the potential energy deficit to get the object to freedom.[br] [br]If we use energy conservation (and ignore any non-conservative work terms), we get: [math]\Delta K+\Delta U_G = 0[/math] where the final kinetic and potential energies are both zero - since the mass would, in principle, be able to stop once it reached infinite distance. Expanding the deltas and plugging in the zeros just mentioned leads to [math]K_f-K_i + U_f-U_i = -K_i-U_i=0[/math] The amount of kinetic energy required to achieve freedom is therefore [math]K_i=-U_i = \frac{GMm}{r_i}.[/math][br][br]Given this amount of kinetic energy, the object has enough energy to reach infinite distance and to therefore be free.  Such an object will violate the common expression "What goes up must come down." Instead, it will go up and never come down again.  The necessary speed to accomplish this is called the [b]escape speed[/b]. Sometimes the term escape velocity is used, but I imagine you see that we are dealing with a scalar and not a vector, so using 'velocity' is technically incorrect.[br][br]Calculating this escape speed is simple. We write [math]\frac{1}{2}mv_{escape}^2-\frac{GMm}{r}=0[/math] and solve for the escape speed. The algebra for the escape speed from earth becomes:[br][br][center][math]v_{escape}=\sqrt{\frac{2GM_{earth}}{r_{earth}}}=11200m/s.[/math][/center][br][br]While earth's escape speed of 11,200m/s is a commonly cited result, the problem is that it won't work from earth. If we were to try to shoot a cannonball from earth's surface at this speed with the intent of it leaving the solar system, it would face two problems. The first is the rapid loss of speed due to air drag and the second, and related issue, is that the air drag would heat up the object and likely turn it into a fireball as is the fate of rocks (meteors) or space shuttles as they travel through the atmosphere going too fast. This escape speed, therefore, only makes sense for worlds without atmospheres.[br][br]A second problem is that we don't have only two bodies in our solar system. We don't usually think of the moon being bound to the sun or to Jupiter, but in an energy sense it is. There would be a negative gravitational potential energy between the cannonball and each mass in our solar system. To truly escape to interstellar space (outside the solar system) would require enough kinetic energy to overcome the sum of all of those negative potential energy terms. The reality is that most of those terms would be very tiny, but it's worth mentioning them.[br][br]We should wonder why nobody calculates the speed required to REALLY allow an object to escape while including air drag. The answer is they can't. First of all, the energy expression would depend on the shape and size (not just mass), on the melting (or burning) point, and perhaps the changing shape of the half-burnt portion of what survives the atmosphere and hopes to reach interstellar space. In other words, it is too hard, and depends too much on case-by-case specifics.[br][br]As a last comment, I want to say that reaching interstellar space from doesn't require an object to reach 11,200m/s upon launch. Things we launch to space have rockets onboard. The continual upward thrust provided by the rockets negates the need to give all the kinetic energy up front. Instead the fuel's combustion provides a continuous infusion of energy so that the ride can be much slower and the laugh much gentler. [br][br][br][br]

A simple, but not so rigorous way to define the required size and mass of a black hole is obtained by setting the escape speed from the object equal to the speed of light in vacuum, which is [math]c=3.0\times10^8m/s[/math]. Since nothing can exceed that speed, nothing could escape to freedom from that object. [br][br]Consider, for example, how small the earth would need to be compressed such that - if it retained all its [math]5.98\times10^{24}kg[/math] of mass - it would become a black hole. Plugging in numbers in the escape speed equation above, you should find that the result is [math]8.86\times10^{-3}m.[/math] That means all the earth's mass would need to be contained within a radius of less than a centimeter! If nothing else, this makes the point that black holes are not made of ordinary matter. Rather, through tremendous pressures that result in star-sized nuclear explosions, matter gets compressed into an exotic form in black holes.