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B.1 Derivation of Elliptical Orbits

Starting from the energy equation for a central force (Chapter 10), we have

E=12mr˙2+UeffE = \frac{1}{2} m\dot{r}^{2}+ U_{eff}

where UeffU_{eff} is the effective potential. Re-arranging the equation gives:

r˙2=2m(EUeff)(B.1)\dot{r}^{2}= \frac{2}{m} (E - U_{eff}) (\mathrm{B}.1)

where r˙=drdt\dot{r} = \frac{\mathrm{d}r}{\mathrm{d}t}. We want to get r(θ)r(\theta) to describe the orbit. To do this, we will use the ratio of r˙\dot{r} and θ˙\dot{\theta} ,

r˙θ˙=dr/dtdθ/dt=drdθ(B.2)\frac{\dot{r}}{\dot{\theta}} = \frac{\mathrm{d}r/\mathrm{d}t}{\mathrm{d}\theta /\mathrm{d}t} = \frac{\mathrm{d}r}{\mathrm{d}\theta} (\mathrm{B}.2)

So we need to get an equation for θ˙\dot{\theta} .

For θ˙\dot{\theta} , we can use the conservation of angular momentum. Recall that central forces conserve angular momentum by definition (see Chapter 10). The magnitude of angular momentum is:

L=mr2θ˙=θ˙=Lmr2(B.3)L = mr^{2}\dot{\theta} =\Rightarrow \dot{\theta} = \frac{L}{mr^{2}} (\mathrm{B}.3)

with LL = constant.

We can then combine the r˙\dot{r} and θ˙\dot{\theta} equations as follows:

(drdθ)2=r˙2θ˙2=fromEquationB.2\Bigg(\frac{\mathrm{d}r}{\mathrm{d}\theta} \Bigg)^{2}= \frac{\dot{r}^{2}}{\dot{\theta}^{2}} =\Rightarrow \mathrm{from} \mathrm{Equation} \mathrm{B}.2
(drdθ)2=2m(EUeff)(Lmr2)2=subinEquationsB.1andB.3\Bigg(\frac{\mathrm{d}r}{\mathrm{d}\theta} \Bigg)^{2}= \frac{\frac{2}{m} (E - U_{eff})}{\Big(\frac{L}{mr^{2}} \Big)^{2}} =\Rightarrow \mathrm{sub} \mathrm{in} \mathrm{Equations} \mathrm{B}.1 \mathrm{and} \mathrm{B}.3
(drdθ)2=2m(EUeff)L2(m2r4)\Bigg(\frac{\mathrm{d}r}{\mathrm{d}\theta} \Bigg)^{2}= \frac{2}{m} \frac{(E - U_{eff})}{L^{2}} (m^{2}r^{4})
1r4(drdθ)2=2m(EUeff)L2=bringallrtermstotheotherside\frac{1}{r^{4}} \Bigg(\frac{\mathrm{d}r}{\mathrm{d}\theta} \Bigg)^{2}= \frac{2m(E - U_{eff})}{L^{2}} =\Rightarrow \mathrm{bring} \mathrm{all} r \mathrm{terms} \mathrm{to} \mathrm{the} \mathrm{other} \mathrm{side}
(1r2drdθ)2=2m(EUeff)L2=simplifytheLHSoftheequation(B.4)\Bigg(\frac{1}{r^{2}} \frac{\mathrm{d}r}{\mathrm{d}\theta} \Bigg)^{2}= \frac{2m(E - U_{eff})}{L^{2}} =\Rightarrow \mathrm{simplify} \mathrm{the} \mathrm{LHS} \mathrm{of} \mathrm{the} \mathrm{equation} (\mathrm{B}.4)

Equation B.4 applies to any effective potential. Since we’re talking about orbits, the central force is gravity. That makes the effective potential:

Ueff=L22mr2γr(B.5)U_{eff}= \frac{L^{2}}{2mr^{2}} - \frac{\gamma}{r} (\mathrm{B}.5)

where γ=GMm\gamma = GMm for simplicity. Substituting Equation B.5 into Equation B.4 gives,

(1r2drdθ)2=2mEL2(2mL2)(L22mr2)+(2mL2)(γr)\Bigg(\frac{1}{r^{2}} \frac{\mathrm{d}r}{\mathrm{d}\theta} \Bigg)^{2}= \frac{2mE}{L^{2}} - \bigg(\frac{2m}{L^{2}} \bigg)\Bigg(\frac{L^{2}}{2mr^{2}} \Bigg) + \bigg(\frac{2m}{L^{2}} \bigg)\bigg(\frac{\gamma}{r} \bigg)
(1r2drdθ)2=2mEL21r2+2mγrL2=simplify(B.6)\Bigg(\frac{1}{r^{2}} \frac{\mathrm{d}r}{\mathrm{d}\theta} \Bigg)^{2}= \frac{2mE}{L^{2}} - \frac{1}{r^{2}} + \frac{2m\gamma}{rL^{2}} =\Rightarrow \mathrm{simplify} (\mathrm{B}.6)

To solve Equation B.6, there is a substitution trick we can use to help simplify the problem. Instead of solving for rr, we’re going to solve for y=1ry = \frac{1}{r}. The reason for this substitution is that

dydθ=1r2drdθ=rememberthaty=1r\frac{\mathrm{d}y}{\mathrm{d}\theta} = - \frac{1}{r^{2}} \frac{\mathrm{d}r}{\mathrm{d}\theta} =\Rightarrow \mathrm{remember} \mathrm{that} y = \frac{1}{r}

Thus, we can write Equation B.6 in terms of y=1ry = \frac{1}{r} instead of rr.

(dydθ)2=2mEL2y2+2mγL2y=suby=1rintoEquationsB.6(B.7)\Bigg(\frac{\mathrm{d}y}{\mathrm{d}\theta} \Bigg)^{2}= \frac{2mE}{L^{2}} - y^{2}+ \frac{2m\gamma}{L^{2}} y =\Rightarrow \mathrm{sub} y = \frac{1}{r} \mathrm{into} \mathrm{Equations} \mathrm{B}.6 (\mathrm{B}.7)

Already this looks simpler. We can simplify more, however, by “completing the square”,

(ymγL2)2=y2+2mγL2y(mγL2)2(B.8)-\bigg(y - \frac{m\gamma}{L^{2}} \bigg)^{2}= -y^{2}+ \frac{2m\gamma}{L^{2}} y - \bigg(\frac{m\gamma}{L^{2}} \bigg)^{2}(\mathrm{B}.8)

Note that the first two terms on the righthand side of Equation 8 are present in the righthand side of Equation B.7. From Equation B.8, we can say that

y2+2mγL2=(ymγL2)2+(mγL2)2(B.9)-y^{2}+ \frac{2m\gamma}{L^{2}} = -\bigg(y - \frac{m\gamma}{L^{2}} \bigg)^{2}+ \bigg(\frac{m\gamma}{L^{2}} \bigg)^{2}(\mathrm{B}.9)

And we can substitute Equation B.9 into Equation B.7 to give:

(dydθ)2=2mEL2(ymγL2)2+(mγL2)2(B.10)\Bigg(\frac{\mathrm{d}y}{\mathrm{d}\theta} \Bigg)^{2}= \frac{2mE}{L^{2}} - \bigg(y - \frac{m\gamma}{L^{2}} \bigg)^{2}+ \bigg(\frac{m\gamma}{L^{2}} \bigg)^{2}(\mathrm{B}.10)

With Equation B.10, we have an equation with one yy term and two constant terms. To make the equation a bit simpler, we will introduce a couple more substitutions for the constants.

First, we will use z=ymγL2z = y - \dfrac{m\gamma}{L^{2}}. Note that dzdθ=dydθ\dfrac{\mathrm{d}z}{\mathrm{d}\theta} = \dfrac{\mathrm{d}y}{\mathrm{d}\theta} because mγL2\dfrac{m\gamma}{L^{2}} is a constant.

(dzdθ)2=z2+2mEL2+(mγL2)2(B.11)\Bigg(\frac{\mathrm{d}z}{\mathrm{d}\theta} \Bigg)^{2}= -z^{2}+ \frac{2mE}{L^{2}} + \bigg(\frac{m\gamma}{L^{2}} \bigg)^{2}(\mathrm{B}.11)

Second, we will substitute for the sum of the remaining constants as B2B^{2},

B2=2mEL2+(mγL2)2(B.12)B2=(mγL2)2(2EL2mγ2+1)\begin{aligned} B^{2}&= \frac{2mE}{L^{2}} + \bigg(\frac{m\gamma}{L^{2}} \bigg)^{2}(\mathrm{B}.12) \\ B^{2}&= \bigg(\frac{m\gamma}{L^{2}} \bigg)^{2}\Bigg(\frac{2EL^{2}}{m\gamma ^{2}} + 1\Bigg) \end{aligned}
B=mγL21+2EL2mγ2(B.13)B = \frac{m\gamma}{L^{2}}\sqrt{1 + \frac{2EL^{2}}{m\gamma^{2}}} (\mathrm{B}.13)

Substituting Equation B.12 into Equation B.11, we get:

(dzdθ)2=z2+B2\Bigg(\frac{\mathrm{d}z}{\mathrm{d}\theta} \Bigg)^{2}= -z^{2}+ B^{2}

Now we have a fairly straightforward integral to solve. The solution to this problem is available in Appendix A.

Mathematical expression from source PDF page 285

We can select the initial conditions such that the initial angle is anything we want because the motion is periodic. For ellipses, it helps to have the angle in terms of cos instead of sin, so we can set C=π2C = \frac{\pi}{2}. Thus, we get the solution:

z=Bsin(θ+π2)=Bcosθ(B.14)z = B\sin \bigg(\theta + \frac{\pi}{2} \bigg) = B\cos \theta (\mathrm{B}.14)

Now we need to put back all of those substitutions that we made!

1.z=ymγL2=ymγL2=Bcosθ1. z = y - \frac{m\gamma}{L^{2}} =\Rightarrow y - \frac{m\gamma}{L^{2}} = B\cos \theta
2.y=1r=1rmγL2=Bcosθ2. y = \frac{1}{r} =\Rightarrow \frac{1}{r} - \frac{m\gamma}{L^{2}} = B\cos \theta
  1. From Equation B.13: B=mγL21+2EL2mγ2B = \dfrac{m\gamma}{L^{2}}\sqrt{1 + \dfrac{2EL^{2}}{m\gamma^{2}}}

1rmγL2=mγL21+2EL2mγ2cosθ1r=mγL2+mγL21+2EL2mγ2cosθ1r=mγL2(1+1+2EL2mγ2cosθ)(B.15)\begin{aligned} \frac{1}{r} - \frac{m\gamma}{L^{2}} &= \frac{m\gamma}{L^{2}}\sqrt{1 + \frac{2EL^{2}}{m\gamma^{2}}}\,\cos\theta \\ \frac{1}{r} &= \frac{m\gamma}{L^{2}} + \frac{m\gamma}{L^{2}}\sqrt{1 + \frac{2EL^{2}}{m\gamma^{2}}}\,\cos\theta \\ \frac{1}{r} &= \frac{m\gamma}{L^{2}}\left(1 + \sqrt{1 + \frac{2EL^{2}}{m\gamma^{2}}}\,\cos\theta\right) (\mathrm{B}.15) \end{aligned}

We now define the eccentricity of the orbit, ε\varepsilon as:

ε=1+2EL2mγ2(B.16)\varepsilon = \sqrt{1 + \frac{2EL^{2}}{m\gamma^{2}}} (\mathrm{B}.16)

Substitute Equation B.16 into Equation B.15:

1r=mγL2(1+εcosθ)(B.17)\frac{1}{r} = \frac{m\gamma}{L^{2}} (1 + \varepsilon \cos \theta) (\mathrm{B}.17)

Finally, let’s solve for rr.

r=(L2mγ)11+εcosθ(B.18)r = \Bigg(\frac{L^{2}}{m\gamma} \Bigg) \frac{1}{1 + \varepsilon \cos \theta} (\mathrm{B}.18)

We now have an equation of r(θ)r(\theta) that defines our orbit. You will note that Equation B.18 has a similar structure to the equation for a generic ellipse (Equation 11.2), but with different constants out front.

If ε\varepsilon = 0, then we have a circular orbit with a radius of

rc=L2mγ(B.19)r_{c}= \frac{L^{2}}{m\gamma} (\mathrm{B}.19)

We define rcr_{c} as the radius of a circular orbit for a given angular momentum, LL. See also Chapter 11.2 for the derivation of rcr_{c} using Newton’s laws.

So substituting in rcr_{c}, we recover Equation 11.7.

r=rc1+εcosθ(B.20)r = \frac{r_{c}}{1 + \varepsilon \cos \theta} (\mathrm{B}.20)

B.2 Approximations

Many times, mathematical functions can be simplified by using limits and making reasonable approximations. There are many ways to simplify a function. Here, we will look at a few of them.

B.2.1 Binomial Approximation

For simple power functions, such as f(x)f(x) = (1 + x)ax)^{a}, you can apply the binomial or linear expansion to approximate their value. The approximation works as follows:

f(x)f(x0)+df(x0)dx(xx0)f(x) \approx f(x_{0}) + \frac{\mathrm{d}f(x_{0})}{\mathrm{d}x} (x - x_{0})

where x0x_{0} is the point about which you are measuring xx. That is, you are expanding the series about the point where x=x0x = x_{0}.

At first glance, this approximation seems reasonable. The approximation states that the value of a function at position xx near position x0x_{0} can be approximated by the value at position x0x_{0} with a modification given by the slope at x0x_{0} and the distance between x0x_{0} and xx. For values of xx0x \approx x_{0}, this approximation should be reasonable.

For the function, f(x)f(x) = (1 + x)ax)^{a} with x0x_{0} = 0, the binomial approximation would be,

f(x)f(0)+df(0)dx(x)f(x)1+ax\begin{aligned} f(x) &\approx f(0) + \frac{\mathrm{d}f(0)}{\mathrm{d}x} (x) \\ f(x) &\approx 1 + ax \end{aligned}

In this case, the value of xx must be close to zero for the approximation to be valid. If xx is not close to zero, then we need to shift x0x_{0}.

B.2.2 Taylor Series

For more general functions, you can approximate the function using its Taylor series,

f(x)=f(x0)+df(x0)dx(xx0)+12d2f(x0)dx2(xx0)2+13!d3f(x0)dx3(xx0)3++1n!dnf(x0)dxn(xx0)n\begin{aligned} f(x) &= f(x_{0}) + \frac{\mathrm{d}f(x_{0})}{\mathrm{d}x} (x - x_{0}) + \frac{1}{2} \frac{\mathrm{d}^{2}f(x_{0})}{\mathrm{d}x^{2}} (x - x_{0})^{2}+ \\ \frac{1}{3!} \frac{\mathrm{d}^{3}f(x_{0})}{\mathrm{d}x^{3}} (x - x_{0})^{3}+ \cdot \cdot \cdot + \frac{1}{n!} \frac{\mathrm{d}^{n}f(x_{0})}{\mathrm{d}x^{n}} (x - x_{0})^{n} \end{aligned}

where x0x_{0} = 0 has the same meaning as before.

For example, the Taylor series expansion of 1+x\sqrt{1 + x} (where x0x_{0} = 0) is equal to:

1+x=1+12x+12!(14)x2+13!(38)x3+1+x=1+12x18x2+116x3\begin{aligned} \sqrt{1 + x} &= 1 + \frac{1}{2}x + \frac{1}{2!}\left(-\frac{1}{4}\right)x^{2} + \frac{1}{3!}\left(\frac{3}{8}\right)x^{3} + \cdots \\ \sqrt{1 + x} &= 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{1}{16}x^{3} - \cdots \end{aligned}

If xx is small (x(x is close to zero), then x2x^{2} and x3x^{3} are very small and don’t change the result much. For example, if x=0.1x = 0.1, then x3=0.001x^{3}= 0.001 and 116x3=6.25×105\frac{1}{16} x^{3}= 6.25\times 10^{-5}. That term is much smaller than 1 and all subsequent terms will be even smaller, so they are negligible. That means, for small values of xx, we can approximate 1+x\sqrt{1 + x} as:

1+x1+12x\sqrt{1 + x} \approx 1 + \frac{1}{2}x

It is key that xx is small (for x0x_{0} = 0). If xx is larger, then the higher order terms are more significant and cannot be considered negligible.

The Taylor series for the sinx\sin x and cosx\cos x (with x0x_{0} = 0) are:

sinx=xx33!+x55!x77!+x99!cosx=1x22!+x44!x66!+x88!\begin{aligned} \sin x &= x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \frac{x^{9}}{9!} - \cdot \cdot \cdot \\ \cos x &= 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - \cdot \cdot \cdot \end{aligned}

Note that xx must be in radians for the Taylor series approximation to hold. You cannot use xx in degree. If xx (in radians) is small, then the higher order terms again become negligible and sinxx\sin x \approx x and cosx1x22\cos x \approx 1 - \frac{x^{2}}{2} . There are several functions with well established Taylor series approximations based on the above definition. Appendix A.3 lists the expansions for many common equations.