Recall, that we can, in principle, solve any problem where the force depends only on time, F(t) = ma = m[vdot] = m[xddot]. Basically, since the left side is a pure function of time, and the right side is a pure time derivative, we can integrate both sides with respect to time, once to get the velocity, and twice to get the position.
It turns out that we can do the same thing for one dimensional problems where the force depends only on x. But this requires a trick that transforms the time derivative into an x derivative. Here's the trick:
[xddot] = d/dt[xdot] = dv/dt = dv/dx * dx/dt = dv/dx * v = v dv/dx = 1/2 d/dx(v2)
The step in the middle where dv/dt becomes dv/dx * dx/dt is an application of the chain rule. The last step is just a clever rewriting of the expression and is easily verified by expanding the derivative of v2. These are just mathematical "tricks", no basic physics involved here.
Physics enters when we look at the "new" equation of motion, i.e. using the transformed expression for [xddot]. We have:
F(x) = m/2 d/dx(v2) = d/dx (1/2 mv2) = dT/dx
where we use the letter T for the kinetic energy, 1/2 mv2 . This expression says that the force is equal to the change of kinetic energy with position. This part is true for any force. But, if the force is a function only of x, then we can integrate both sides of the equation with respect to x. Recall that work is the integral of force over the path travelled. Therefore:
W = [Int x0 to x] F(x) dx = T(x) - T(x0) = T - T0
This relation says that the work done on the object equals the change in its kinetic energy. Not so surprising maybe, but this isn't true for all forces -- frictional forces for example. Forces for which the work done equals the change in kinetic energy (for any possible path) are called conservative forces. (They are conservative because they conserve mechanical energy.) Since F(x) is integrable, we can define a potential function, V(x), such that F(x) = -dV/dx . Now, the above integral becomes:
W = [Int x0 to x] F(x) dx = -[Int from x0 to x] dV = V(x0) - V(x) = T - T0
Rewriting, we get T0 + V(x0) = constant = T + V(x) = E . The left side is constant because both pieces are constant -- x0 is a defined point, so the kinetic and potential at that point have a fixed value. The constant is E , the total energy, and the relation is an expression of the consesrvation of energy.
But, remember that we started out trying to solve for the motion of an object; what we have now is conservation of energy, an important concept, but not the motion. However, we can use that equation to find the motion. Note that E is constant, and normally can be determined from the initial conditions, V(x) is determined from the force function, F(x), so all that's left is T = 1/2 mv2. So, this can be regarded as an equation for v.
1/2 mv2 = E - V(x) or v = ±[Sqrt]{2/m( E - V(x) )}
Since v= dx/dt it seems possible that it can be integrated to solve for x.
[Int from x0 to x] dx / ±[Sqrt]{2/m( E - V(x) )} = t - t0
In principle, this provides a definite solution for the position as
a function of time in any one dimensional problem where the force is a
function of position only. In practice, the square root in the integral
makes for an ugly calculation. The usual approach in these types
of problems is to use the energy equation as much as possible to get a
solution and avoid explicitly doing the integral given above.
Let's look at some examples where these ideas can be applied.
1/2 mv2 + mgx = E
Where E is the total energy and must be determined from the initial conditions -- what is the inital velocity and position for instance. In this case, if the energy is known, then the energy equation gives an algebraic relationship between position and velocity.
For instance, if an object is shot upward from x=0 with initial velocity v0, we can easily determine the maximum height. First, in this case, the initial conditions tell us that the energy is E = 1/2 mv02. At the maximum height, the velocity will be zero, and the energy equation tells us that E = mgxmax. Relating these two equations, we can solve for xmax in terms of v0 -- this is what the statement of the problem implicitly asks for.
xmax = v02/2g
F(r) = -GMm/r2
where G is the gravitational constant, M is the mass of the Earth, and m is the mass of the space probe. I want to determine the potential function directly (this part is done somewhat differently in the text, if you can understand both methods, then all the better). Since the force is the derivative of the potential function, the potential must be the integral of the force:
V(r) = -[Int from r0 to r] F(r) dr = GMm [Int from r0 to r] dr/r2 = -GMm( 1/r - 1/r0 ) = GMm( 1/r0 - 1/r )
where r0 is the radius of the Earth, the starting distance between the space probe and the center of the Earth. (Note: If I had used r0=0 as the lower limit of the integral, the the result would be infinite: 1/r0 = 1/0 = infinity ! This is a known problem with the 1/r2 forces. I anticipated this problem, and chose a starting radius different from zero, and then there are no infinities, and no problems. I believe this is also the reason for the approach followed in the text.) Applying this result to the energy equation gives:
GMm( 1/r0 - 1/r ) + 1/2 mv2 = E
[Can you see that this equation is the same as the result reached in the text -- the equation in example 2.3.2 appearing just before the heading "Maximum Height: Escape Speed"?
To solve this problem we need to dissect and understand the statement
of the problem. First, there are two specific events, the launch
of the rocket, and its arrival at Jupiter. At launch, we are told
that the rocket has a velocity v0, and it is implied that the rocket
is at the surface of the Earth. The arrival at Jupiter is slightly
more subtle. What we want is to get the probe out of Earth's gravity;
but the gravity force has infinite extent, it slowly goes to zero like
one over the distance squared, but the distance can be infinitely large.
However, for practical purposes, if the probe gets far enough away, the
Earth's gravity becomes negligible, and in fact the sun's gravity becomes
more important. For the purposes of this problem, escaping Earth's
gravity can be interpreted as letting r tend to infinity.
If the probe has just enough initial velocity (energy) to escape Earth's
gravity, then it will have zero velocity (kinetic energy) at r = [infinity].
These two events are summarized below:
| event | position | velocity | energy |
| launch | r = re | v0 = ve | 1/2 mve2 |
| arrival | r = [infinity] | v = 0 | GMm/re |
The energy at arrival is a known value (all the parameters in the expression are known constants), and the only unknown quantity in the expression for the energy at launch is the escape velocity. Applying conservation of energy, we can equate these two expressions, and solve for the escape velocity:
1/2 mve2 = GMm/re or solving for the escape velocity ve = [sqrt]{2GM/re} .
This answer can be simplified (or written in terms of other parameters) by noting that at the Earth's surface, the force due to gravity is mg, and it is also, using the law of gravitation, GMm/re2. Therefore, GM/re = gre, and
ve = [sqrt]{2gre} = [sqrt]{2*(9.8m/s2)*(6.4*106m)} = 11km/s