From 1a1b919c5aa107780837457e58c4324e9847cd4e Mon Sep 17 00:00:00 2001 From: Indigo5684 <159226326+Indigo5684@users.noreply.github.com> Date: Tue, 30 Sep 2025 13:19:43 -0500 Subject: [PATCH] Mechanics Chapter 8 --- .../8-two-body-central-force-problems.md | 279 ++++++++++++++++++ 1 file changed, 279 insertions(+) create mode 100644 docs/physics/mechanics/8-two-body-central-force-problems.md diff --git a/docs/physics/mechanics/8-two-body-central-force-problems.md b/docs/physics/mechanics/8-two-body-central-force-problems.md new file mode 100644 index 0000000..b073fc3 --- /dev/null +++ b/docs/physics/mechanics/8-two-body-central-force-problems.md @@ -0,0 +1,279 @@ +# Chapter 8 - Two-Body Central Force Problems + +## Section 8.1 - The Problem + +Consider two objects in an inertial frame, acted upon by a central force. We know that the Lagrangian is + +$$\mathcal{L} = \frac{1}{2}m_1 \dot{\mathbf{r}}_1^2 + \frac{1}{2} m_2 \dot{\mathbf{r}}_2^2$$ + +## Section 8.2 - CM and Relative Coordinates; Reduced Mass + +We know that with $M = m_1 + m_2$, the center of mass becomes + +$$\mathbf{R} = \frac{m_1\mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{M}$$ + +We know that the total momentum thus can be written as + +$$\mathbf{P} = M \dot{\mathbf{R}}$$ + +As total momentum is conserved, $\dot{\mathbf{R}}$ is constant. Thus, we can choose a reference frame where the center of mass is at rest. + +We can thus rewrite our coordinates as + +$$\mathbf{r}_1 = \mathbf{R} + \frac{m_2}{M} \mathbf{r} \; \text{and} \; \mathbf{r}_2 = \mathbf{R} - \frac{m_1}{M} \mathbf{r}$$ + +This uses the center of mass position $\mathbf{R}$ and relative position $\mathbf{r}$ + +Then, via substitution, we see that kinetic energy becomes + +$$T = \frac{1}{2}(M \dot{\mathbf{R}}^2 + \frac{m_1 m_2}{M} \dot{\mathbf{r}}^2)$$ + +We can introduce $\mu = \frac{m_1 m_2}{M} = \frac{m_1 m_2}{m_1 + m_2}$, or the *reduced mass*, to write $T$ as + +$$T = \frac{1}{2}(M \dot{\mathbf{R}}^2 + \mu \dot{\mathbf{r}}^2)$$ + +Then, we can see that + +$$\mathcal{L} = \frac{1}{2} M \dot{\mathbf{R}}^2 + (\frac{1}{2}\mu \dot{\mathbf{r}}^2 - U(R)) = \mathcal{L}_{cm} + \mathcal{L}_{rel}$$ + +This allows us to solve this as two separate problems. + +## Section 8.3 - The Equations of Motion + +$\mathcal{L}$ is independent of $\mathbf{R}$, so we know that $\dot{\mathbf{R}}$ is a constant. In other words, $\ddot{\mathbf{R}} = 0$, which means $M \ddot{\mathbf{R}} = 0$. + +We can rewrite this as $\mathcal{L}_{cm} = \frac{1}{2} M \dot{\mathbf{R}}$, or the Lagrangian of a free particle. + +Additionally, $\mathcal{L}_{rel}$ is indistinguishable from the Lagrangian of a single particle with mass $\mu$ and position $\mathbf{r}$. Then, the corresponding Lagrangian equation becomes + +$$\mu \ddot{\mathbf{r}} = -\nabla U(\mathbf{r})$$ + +**Definition**. In the *CM frame*, the center of mass is at rest. We know that this is an inertial frame as it is not accelerating. + +That is, $\ddot{\mathbf{R}} = 0$. Then, the Lagrangian is simply $\mathcal{L} = \mathcal{L}_{rel} = \frac{1}{2}\mu \dot{\mathbf{r}}^2 - U(r)$. It immediately follows that $\mathbf{R}$ is an ignorable coordinate. + +**Note**. Know that $\mathbf{r}$ is not the position of either particle, but the position of particle 1 with respect to particle 2. That is, $\mathbf{r}_1 = \frac{m_2}{M} \mathbf{r}$. and $\mathbf{r}_2 = -\frac{m_1}{M}\mathbf{R}$ + +Since total momentum is zero, we see that both particles must either be still or be moving with opposing momenta. + +We can now apply the conservation of angular momentum. + +$$\begin{align} +\mathbf{L} &= \mathbf{r}_1 \times \mathbf{p}_1 + \mathbf{r}_2 + \mathbf{p}_2 \\ +&= m_1 \mathbf{r}_1 \times \dot{\mathbf{r}}_1 + m_2 \mathbf{r}_2 \times \dot{\mathbf{r}}_2 \\ +&= m_1 \frac{m_2}{M} \mathbf{r} \times \frac{m_2}{M}\dot{\mathbf{r}} + m_2 -\frac{m_1}{M}\mathbf{r} \times (-\frac{m_1}{M}) \dot{\mathbf{r}} \\ +&= frac{m_1}{m_2}{M^2} (m_2 \mathbf{r} \times \dot{\mathbf{r}} + m_1 \mathbf{r} \times \dot{\mathbf{r}}) \\ +&= \mu \mathbf{r} \times \dot{\mathbf{r}} +\end{align}$$ + +Since angular momentum $\mathbf{L}$ is conserved, we can also state the quantity $\mathbf{r} \times \dot{\mathbf{r}}$ is constant. Particularly, its direction is constant, which means the components remain in a fixed plane. We choose this plane, so this now becomes a two-body problem. + +Let us switch to polar coordinates. We know that $\dot{\mathbf{r}}^2 = \dot{r}^2 + r^2 \dot{\phi}^2$. Thus, the Lagrangian becomes + +$$\mathcal{L} = \frac{1}{2} \mu \dot{\mathbf{r}}^2 - U(r) = \frac{1}{2}({\mathbf{r}}^2 = \dot{r}^2 + r^2 \dot{\phi}^2) - U(R)$$ + +We can see that $\phi$ is ignorable, meaning $\mu r^2 \dot{\phi}$ is constant. We know that this quantity is the angular momentum, so $\ell = \ell_z = \mu r^2 \dot{\phi}$. + +**Definition**. The Lagrangian equation for $r$, or the *radial equation*, becomes + +$$\mu r \dot{\phi}^2 - \frac{dU}{dr} = \mu \ddot{r}$$ + +We can rewrite this to see that + +$$-\frac{dU}{dr} = \mu (\ddot{r} - r \dot{\phi}^2)$$ + +This then becomes the radial component of $\mathbf{F} = \mu \mathbf{a}$. + +## Section 8.4 - The Equivalent One-Dimensional Problem + +The angular momentum equation can be solved to see that $\dot{\phi} = \frac{\ell}{\mu r^2}$. We can thus eliminate $\dot{\phi}$ in the radial equation to see that + +$$\mu \ddot{r} = - \frac{dU}{dr} + \mu r \dot{\phi}^2 = -\frac{dU}{dr} + F_{cr}$$ + +Here, we define $F_{cr}$ as some fictitious centrifugal force defined such that + +$$F_{cr} = \mu r \dot{\phi}^2 = \frac{\ell^2}{\mu r^3} = -\frac{d}{dr}(\frac{\ell^2}{2\mu r^2}) = - \frac{dU_{cf}}{dr}$$ + +Thus, with $U_{cr}(r) = \frac{\ell^2}{2\mu r^2}$ this force is conservative. + +**Definition**. *Effective potential energy* is defined as $U_{eff}(r) = U(r) + U_{cf}(r)$, so that $\mu \ddot{r} = -\frac{d}{dr} U_{eff}(r)$. Then, the radial motion of the particle is exactly the same as if the particle were moving in one dimension with the effective potential energy. + +We can thus multiply through by $\dot{r}$ to see that + +$$\begin{align} +\mu \ddot{r} \dot{r} &= -\dot{r}\frac{d}{dr} U_{eff}(r) \\ +\frac{d}{dt}(\frac{1}{2} \mu \dot{r}^2) &= -\frac{dr}{dt} \frac{d}{dr} U_{eff}(r) \\ +&= -\frac{d}{dt}U_{eff}(r) +\end{align}$$ + +For this to be true for any positions $\dot{r}$ and $r$, we see that $\frac{1}{2} \mu \dot{r} + U_{eff}(r)$ must be a constant. We can substitute to see that + +$$\begin{align} +\frac{1}{2} \mu \dot{r} + U_{eff}(r) &= \frac{1}{2} \mu \dot{r} + U(r) + \frac{\ell^2}{2\mu r^2} \\ +&= \frac{1}{2} \mu \dot{r} + U(r) + \frac{(\mu r^2 \dot{\phi})^2}{2\mu r^2} \\ +&= \frac{1}{2} \mu \dot{r} + U(r) + \frac{1}{2}\mu r^2 \dot{\phi}^2 \\ +&= \frac{1}{2} \mu \dot{r} + \frac{1}{2}\mu r^2 \dot{\phi}^2 \\ +&= E +\end{align}$$ + +This is simply conservation of energy. + +## Section 8.5 - The Equation of the Orbit + +This gives us an equation for $r(t)$. However, we are also interested in $r = r(\phi)$. + +We know that $\mu \ddot{r} = F(r) + F_{cr}(r) = F(r) + \frac{\ell}{\mu r^3}$ This is a bit annoying to work with, so we will replace $r$ with $u$ by + +$$u = \frac{1}{r}$$ + +Now, we can use the chain rule to see that, since $\ell = \mu r^2 \dot{\phi}$, we can write + +$$\frac{d}{dt} = \frac{d\phi}{dt} \frac{d}{d\phi} = \dot{\phi} \frac{d}{d\phi} = \frac{\ell}{\mu r^2} \frac{d}{d\phi} = \frac{\ell u^2}{\mu} \frac{d}{d\phi}$$ + +Then, we can see that + +$$\dot{r} = \frac{d}{dt}r = \frac{\ell u^2}{\mu} \frac{d}{d\phi} \frac{1}{u} = -\frac{\ell}{\mu} \frac{du}{d\phi}$$ + +Subsequently, + +$$\ddot{r} = \frac{d}{dt}\dot{r} = \frac{\ell u^2}{\mu} \frac{d}{d\phi} (-\frac{\ell}{\mu} \frac{du}{d\phi}) = -\frac{\ell^2 u^2}{\mu^2} \frac{d^2 u}{d\phi^2}$$ + +We can substitute this into the differential equation to see that + +$$-\frac{\ell^2 u^2}{\mu} \frac{d^2 u}{d\phi^2} = F + \frac{\ell^2 u^3}{\mu}$$ + +We can then rewrite this as + +$$u''(\phi) = -u(\phi) = \frac{\mu}{\ell^2 u(\phi)^2} F$$ + +## Section 8.6 - The Kepler Orbits + +Let us assume our force can be written in the form + +$$F(r) = -\frac{\gamma}{r^2} = -\gamma u^2$$ + +Here, $\gamma$ is the force constant. For gravity, $\gamma = G m_1 m_2$. For the Coulomb force, $\gamma = k q_1 q_2$ + +This then tells us that + +$$u''(\phi) = -u(\phi) = \frac{\gamma\mu}{\ell^2}F$$ + +We can then substitute $w(\phi) = u(\phi) - \gamma \mu/\ell^2$, and as $\ddot{w} = \ddot{u}$, see that $w'' = -w$ and so + +$$w(\phi) = A \cos(\phi - \delta)$$ + +For a choice of $\phi = 0$, we can cancel $\delta$, so that + +$$u(\phi) = \frac{\gamma \mu}{\ell^2} + A \cos \phi = \frac{\gamma \mu}{\ell^2}(1 + \epsilon \cos \phi)$$ + +Here, we define $\epsilon$ as $\epsilon = A \ell^2 / \gamma \mu$. + +We then define some constant $c$ such that $1/c = \gamma \mu / \ell^2$. Then, + +$$u = \frac{1}{r} = \frac{1}{c}(1 + \epsilon \phi)$$ + +This allows us to solve for $r(\phi)$ as + +$$r(\phi) = \frac{c}{1 + \epsilon \cos \phi}$$ + +Given $\epsilon < 1$, we see the orbit is bounded as the denominator never vanishes. Then, we can see minimum and maximum values for $\cos \phi = \pm 1$ Then, + +$$r_{min} = \frac{c}{1 + \epsilon} \; \text{ and } r_{max} = \frac{c}{1-\epsilon}$$ + +**Definition**. The *perihelion* is where $r = r_{min}$. The *aphelion* is where $r = r_{max}$. + +We can transform this into an ellipse with the form + +$$\frac{(x+d)^2}{a^2} + \frac{y^2}{b^2} = 1$$ + +This assumes that $a = \frac{c}{1-\epsilon^2}$, $b = \frac{c}{\sqrt{1-\epsilon^2}}$, $d = a\epsilon$. + +**Definition**. Now, we see that $b/a = \sqrt{1-\epsilon^2}$, This allows us to define $\epsilon$ as the *eccentricity*. Note if $\epsilon=0$, we recover a circle. + +**Theorem**. *Kepler's First Law*. The position of the sun is one of the ellipse's foci. + +**Theorem**. *Kepler's Second Law*. The rate at which the line from the sun to a comet sweeps out area is + +$$\frac{dA}{dt} = \frac{\ell}{2\mu}$$ + +We know that $A_{ellipse} = \pi a b$, so the period becomes + +$$\tau = \frac{A}{dA/dt} = \frac{(\pi a b) 2\mu}{\ell}$$ + +We can square both sides, then replace $b^2$ with $a^2(1-\epsilon^2)$, and replace $a(1-\epsilon^2)$ with $c$. + +$$\tau^2 = 4\pi^2 \frac{a^2(1-\epsilon^2)\mu^2}{\ell^2} = 4\pi^2 \frac{a^3 c\mu^2}{\ell^2}$$ + +We then know that $c = \ell^2 / \gamma \mu$, telling us that + +$$\tau^2 = 4\pi^2 \frac{a^e \mu}{\gamma}$$ + +Now, consider energy. We know that $E = U_{eff}(r_{min})$, as we don't want to consider velocity at $r_{max}$ - that's complicated and annoying. Then, we know that + +$$U_{eff}(r_{min}) = -\frac{\gamma}{r_{min}} + \frac{\ell^2}{2\mu r_{min}^2} = \frac{1}{2r_{min}}$$ + +We can recall that $r_{min} = c/(1 + \epsilon)$, and that $c = \ell^2 / \gamma \mu$. Then, we can see that + +$$r_{min} = \frac{\ell^2}{\gamma \mu (1+\epsilon)}$$ + +We can then apply some annoying algebra to see that + +$$E = \frac{\gamma \mu(1 + \epsilon)}{2\ell^2}(\gamma(1+\epsilon) - 2\gamma) = \frac{\gamma^2\mu}{2\ell^2}(\epsilon^2-1)$$ + +Note that negative energies correspond with eccentricities $\epsilon < 1$, which in turn correspond to bounded orbits. + +## Section 8.7 - The Unbounded Kepler Orbits + +When $\epsilon \geq 1$, then we see that + +$$r(\phi) = \frac{c}{1 + \epsilon \cos \phi}$$ + +has an asymptote. When $\epsilon = 0$ (or $E = 0$), we see that when $\epsilon = \pm \pi$, $r \rightarrow \infty$. Converting to cartesian equations allows us to recover + +$$y^2 = c^2 - 2cx$$ + +This is a parabola. + +When $\epsilon > 1$ or $E > 0$, we see that the denominator vanishes when $\epsilon \cos \phi = -1$. Thus, we can find the maximum allowed angle $\phi_{max}$ as + +$$\epsilon \cos(\phi_max) = -1$$ + +We are then confined to the angles $\phi \in (-\phi_{max}, \phi_{max})$. We can then rewrite the equation of motion as + +$$\frac{(x-\delta)^2}{\alpha^2}-\frrac{y^2}{\beta^2}=1$$ + +## Section 8.8 - Changes of Orbit + +Let us consider a satellite in orbit. + +**Definition**. For a satellite orbiting Earth, the closest and furthest points are referred to as the *perigee* and *apogee* (instead of perihelion and aphelion for the sun). + +We know our orbit will follow the path + +$$r(\theta) = \frac{c}{1 + \epsilon \cos(\phi - \delta)}$$ + +We cannot eliminate the $\delta$ parameter via a clever choice of $\phi$ as we are dealing with multiple orbits. + +Now, assume the spacecraft is initially in an orbit with energy $E_1$, angular momentum $\ell_1$, and orbital parameters $c_1$, $\epsilon_1,$ and $\delta_1$. Now, to shift orbits, apply an impulse at $\phi = \phi_0$ and causes an instantaneous change of velocity such that the new energy is $E_2$ and new angular momentum is $\ell_2$. We can then calculate the new orbital parameters $c_2$, $\epsilon_2$, and $\delta_2$. + +Now, as the orbits must join at $\phi_0$, we can see that + +$$\frac{c_1}{1 + \epsilon_1 \cos(\phi_0 - \delta_1)} = \frac{c_2}{1 + \epsilon_2 \cos(\phi_0 - \delta_2)}$$ + +This is tedious, and if possible, we'd like to avoid it. + +Let us assume that we only apply an impulse at the perigee of the initial orbit, and only in the direction of motion. Then, we can use a choice of axis so this occurs at $\phi = 0$, and thus $\phi_0 = 0$ and $\delta_1 = 0$. As this is the perigee of the final orbit, we also set $\delta_2 = 0$. + +Now, we see that + +$$\frac{c_1}{1 + \epsilon_1} = \frac{c_2}{1 + \epsilon_2}$$ + +**Definition**. The *thrust factor* $\gamma$ is a constant such that $v_2 = \gamma v_1$. + +As at the perigee, the angular momentum is $\ell = \mu r v$, and as $r$ is fixed during the impulse, we know that $\ell_2 = \gamma \ell_1$. Furthermore, as $c = \ell / \gamma \mu$, we know that $c_2 = \gamma^2 c_1$. + +Then, we can apply the equation above to see that + +$$\epsilon_2 = \gamma^2 \epsilon_1 + \gamma^2 - 1$$ + +Notably, if $\gamma > 1$ (a thrust in the direction of motion), then $\epsilon_2 > \epsilon_1$ and vice-versa.