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As you might have noticed in the last few units, our principal work is using models to develop equations of motion. Those equations of motion can then be analyzed, integrated, and plotted to understand the behavior of the system. In this unit, we will focus on the equations of motion for a few different systems.

We will set up the equations of motion for a 2D quadratic drag system and show it’s intractable analytically. Here is where our use of [Euler-Cromer integration] can get us out of trouble. We will also develop the analytical solution to the 2D drag case when the drag is linear, which will show us how to solve these problems and compare them things we know.We will then quickly introduce two additional systems that are very common “base models” for more complex systems: (1) the gravitational bound planet system (a proxy for other central force systems) and (2) the simple harmonic oscillator (a common proxy for many oscillatory systems).

Two-Dimensional Quadratic Drag

The drag force in 2D can be written in terms of the velocity vector of the object as:

F⃗drag=−D∣v⃗∣v⃗\vec{F}_{drag} = -D |\vec{v}| \vec{v}

where DD is the drag coefficient and v⃗\vec{v} is the velocity vector. Note that this force is written entirely using velocity, there is no dependence on position. However, the velocity vector is also function of time, v⃗=v⃗(t)\vec{v} = \vec{v}(t).

Define the Coordinate System

To start this analysis, we need to define a coordinate system. Below, we draw the particle at some random time with the velocity vector shown. The axes are typical: xx is horizontal and yy is vertical. The drag force is always opposite to the velocity vector.

Coordinate System

In this coordinate system, the properties of the particle are:

r⃗=xx^+yy^=⟨x,y⟩\vec{r} = x \hat{x} + y \hat{y} = \langle x, y \rangle
v⃗=vxx^+vyy^=⟨vx,vy⟩\vec{v} = v_x \hat{x} + v_y \hat{y} = \langle v_x, v_y \rangle
a⃗=axx^+ayy^=⟨ax,ay⟩\vec{a} = a_x \hat{x} + a_y \hat{y} = \langle a_x, a_y \rangle

where x^\hat{x} and y^\hat{y} are the unit vectors in the xx and yy directions, respectively. And the magnitude of the velocity vector is ∣v⃗∣=vx2+vy2|\vec{v}| = \sqrt{v_x^2 + v_y^2} as you might imagine.

The free body diagram at the point in time shown above is shown below. You see the gravitational force pointing directly downward and the drag force pointing in the opposite direction of the velocity vector. We continue to apply our coordinate system to the forces.

Free Body Diagram

Apply Newton’s Second Law

We now apply Newton’s Second Law to the particle in the chosen coordinate system. The forces acting on the particle are the gravitational force and the drag force.

F⃗net=F⃗gravity+F⃗drag\vec{F}_{net} = \vec{F}_{gravity} + \vec{F}_{drag}

How do we apply the coordinate system to the forces? We focus on the diagram above. We start by writing the sum of the forces. For this, we take the downward component Fgravity,yF_{gravity,y} to have magnitude mgmg, where gg is the magnitude of the acceleration due to gravity.

Fnet,x=Fdrag,xF_{net,x} = F_{drag,x}
Fnet,y=Fdrag,y−Fgravity,yF_{net,y} = F_{drag,y}-F_{gravity,y}

We can now write the forces in terms of the components of the vectors, and we introduce the acceleration vector, a⃗=⟨ax,ay⟩\vec{a} = \langle a_x, a_y \rangle.

max=−D∣v⃗∣vxm a_x = -D |\vec{v}| v_x
may=−D∣v⃗∣vy−mgm a_y = -D |\vec{v}| v_y - mg

Let’s clean this up a little in terms of the components:

x¨=−Dmx˙x˙2+y˙2\ddot{x} = -\frac{D}{m} \dot{x}\sqrt{\dot{x}^2 + \dot{y}^2}
y¨=−Dmy˙x˙2+y˙2−g\ddot{y} = -\frac{D}{m} \dot{y}\sqrt{\dot{x}^2 + \dot{y}^2} - g

We can try to focus on the velocity instead to simplify the equations. Then we integrate those equations to get the position.

v˙x=−Dmvxvx2+vy2\dot{v}_x = -\frac{D}{m} {v_x}\sqrt{{v_x}^2 + {v_y}^2}
v˙y=−Dmvyvx2+vy2−g\dot{v}_y = -\frac{D}{m} {v_y}\sqrt{{v_x}^2 + {v_y}^2} - g

There is generally no simple closed-form analytical solution to these coupled equations.

These are called coupled differential equations because the equations are linked by the terms x˙\dot{x} and y˙\dot{y}. This means that we cannot solve them independently. We need another approach to solve these equations.

Why can’t we solve these equations?

We cannot form a solution because they are coupled and non-linear. Sometimes, we can decouple these equations (as we will see later) and produce partial differentials of the form:

f1(vx)dvx=g1(t)dtf_1(v_x)dv_x = g_1(t)dt
f2(vy)dvy=g2(t)dtf_2(v_y)dv_y = g_2(t)dt

These lead to independent equations of motion. We can use separation of variables to try to solve them. This is not possible in all cases, so functions are still not integrable analytically. But we cannot even form these partials, so an analytical solution is not possible in this case.

Since we cannot integrate these equations by hand, we turn to the Euler-Cromer method to predict the trajectory numerically. The plot below compares a projectile launched with and without quadratic drag; drag shortens the range and steepens the descent.

<Figure size 700x500 with 1 Axes>

Linear Drag in Two-Dimensions

As we saw above, the quadratic drag case is intractable. However, the linear drag case is analytically solvable. The drag force in 2D can be written in terms of the velocity vector of the object as:

F⃗lin=−mγv⃗\vec{F}_{lin} = -m \gamma \vec{v}

where γ\gamma is a proxy for the drag coefficient. The linear drag force is proportional to the velocity vector.

We have the same set up as before and same FBD.

Coordinate System

And thus the same coordinate system. The properties of the particle are the same as above.

Apply Newton’s Second Law

We now apply Newton’s Second Law to the particle in the chosen coordinate system. The forces acting on the particle are the gravitational force and the drag force.

F⃗net=F⃗gravity+F⃗lin\vec{F}_{net} = \vec{F}_{gravity} + \vec{F}_{lin}

Again, the gravitation force magnitude is mgmg, so we write the forces in terms of the components of the vectors:

Fnet,x=Flin,xF_{net,x} = F_{lin,x}
Fnet,y=Flin,y−Fgravity,yF_{net,y} = F_{lin,y}-F_{gravity,y}

In terms of the acceleration vector, a⃗=⟨ax,ay⟩\vec{a} = \langle a_x, a_y \rangle, we have:

max=−mγvxm a_x = -m \gamma v_x
may=−mγvy−mgm a_y = -m \gamma v_y - mg

Notice that vxv_x and vyv_y are the components of the velocity vector; they can be positive, negative, or zero. We can clean this up in terms of the velocity components, and we have two linear, uncoupled differential equations:

v˙x=−γvx\dot{v}_x = -\gamma v_x
v˙y=−γvy−g\dot{v}_y = - \gamma v_y - g

Solve the Equations

We can try to solve these equations by integrating them. We can integrate the first equation to get the velocity in the xx direction as a function of time.

Velocity in the xx direction

v˙x=−γvx\dot{v}_x = -\gamma v_x

We separate the variables and integrate:

dvxvx=−γdt\frac{dv_x}{v_x} = -\gamma dt

We integrate from v0,xv_{0,x} to vxv_x and from 0 to tt:

∫v0,xvxdvxvx=−γ∫0tdt\int_{v_{0,x}}^{v_x} \frac{dv_x}{v_x} = -\gamma \int_{0}^{t} dt
ln⁡(vx)−ln⁡(v0,x)=−γt\ln(v_x) - \ln(v_{0,x}) = -\gamma t
vx(t)=v0,xe−γtv_x(t) = v_{0,x} e^{-\gamma t}

We see an exponential decay in the velocity in the xx direction.

Velocity in the yy direction

Now we can try to do the same for vyv_y:

v˙y=−γvy−g\dot{v}_y = -\gamma v_y - g

We separate the variables and integrate:

dvyvy+gγ=−γdt\frac{dv_y}{v_y + \frac{g}{\gamma}} = -\gamma dt

Note that this integral will be of the form:

∫dxx+a=ln⁡(x+a)+C\int \frac{dx}{x + a} = \ln(x + a) + C

Again, we integrate from v0,yv_{0,y} to vyv_y and from 0 to tt:

∫v0,yvydvyvy+gγ=−γ∫0tdt\int_{v_{0,y}}^{v_y} \frac{dv_y}{v_y + \frac{g}{\gamma}} = -\gamma \int_{0}^{t} dt
ln⁡(vy+gγ)−ln⁡(v0,y+gγ)=−γt\ln(v_y + \frac{g}{\gamma}) - \ln(v_{0,y} + \frac{g}{\gamma}) = -\gamma t
ln(vy+gγv0,y+gγ)=−γtln\left(\dfrac{v_y + \frac{g}{\gamma}}{v_{0,y} + \frac{g}{\gamma}}\right) = -\gamma t

Next we use exponentiation and do a little algebra to solve for vyv_y:

(gγ+vy)=(gγ+v0,y)e−γt\left(\dfrac{g}{\gamma} + v_y\right) = \left(\dfrac{g}{\gamma} + v_{0,y}\right) e^{-\gamma t}
vy(t)=gγ(e−γt−1)+v0,ye−γtv_y(t) = \dfrac{g}{\gamma}\left(e^{-\gamma t} - 1\right) + v_{0,y} e^{-\gamma t}

In the yy direction, the story appears more complex. The plot below shows both components: vxv_x decays exponentially to zero, while vyv_y decays toward the (negative) terminal velocity instead of zero, since gravity keeps acting even after the drag and weight balance.

<Figure size 700x500 with 1 Axes>

Trajectories

One of the main concepts we will discuss the trajectory of a system. We borrow that language and idea from projectile motion -- the location of the particle as a function of time is the trajectory. In this class, we will consider the word trajectory to mean the evolution of any property of the system as a function of time. This connects strongly to the concept of phase space, which we will discuss in the future.

In the prior example, we found the trajectory of the velocity in the xx and yy directions.

vx(t)=v0,xe−γtv_x(t) = v_{0,x} e^{-\gamma t}
vy(t)=gγ(e−γt−1)+v0,ye−γtv_y(t) = \dfrac{g}{\gamma}\left(e^{-\gamma t} - 1\right) + v_{0,y} e^{-\gamma t}

We can expound on that work to find the trajectory of the position of the particle as a function of time. We can integrate the velocity to get the position.

Those integrals are doable, but they can be a little messy. We won’t do them here, but quote the results consistent with the above equations.

x(t)=x0+1γv0,x(1−e−γt)x(t) = x_0 + \dfrac{1}{\gamma}v_{0,x}\left(1 - e^{-\gamma t}\right)
y(t)=y0−gγt+1γ(gγ+v0,y)(1−e−γt)y(t) = y_0 - \dfrac{g}{\gamma}t + \dfrac{1}{\gamma}\left(\dfrac{g}{\gamma} + v_{0,y}\right)\left(1 - e^{-\gamma t}\right)

2D Gravitational Bound System

Consider a massive object (a large star) and a smaller satellite (a moon or small planet). We know that Newton’s Universal Law of Gravitation tells us the force between two objects that interact gravitationally is:

F=Gm1m2r2F = \dfrac{G m_1 m_2}{r^2}

where GG is the gravitational constant, m1m_1 and m2m_2 are the masses of the objects, and rr is the distance between the objects.

But we need to be more clear about the forces and the vector relationships. Consider the figure below with the massive object at the origin and the satellite at some distance rr from the origin. What is the vector r⃗\vec{r} that describes the location of the satellite?

Gravitational Bound System

If we move the sun from the origin a little, we can start to see what r⃗\vec{r} is. The vector r⃗\vec{r} is the vector from the sun to the satellite. See the figure below to see the sketch.

Gravitational Bound System

So if the location of the sun is r⃗sun\vec{r}_{sun} and the Earth is r⃗earth\vec{r}_{earth}, then the vector r⃗\vec{r} is:

r⃗=r⃗earth−r⃗sun\vec{r} = \vec{r}_{earth} - \vec{r}_{sun}

Let’s return to the simplified model with the sun at the origin, and consider the earth at some distance rr from the origin. The force on the earth is:

F⃗grav=−GMsunMearth∣r⃗∣3r⃗\vec{F}_{grav} = -G \dfrac{M_{sun} M_{earth}}{|\vec{r}|^3} \vec{r}

where Msun=2×1030kgM_{sun} = 2\times10^{30} \mathrm{kg} is the mass of the sun and Mearth=6×1024kgM_{earth} = 6 \times 10^{24} \mathrm{kg} is the mass of the earth.

Define the Coordinate System

In the figure below, we show the earth at some distance rr from the origin at an angle ϕ\phi from the xx-axis. This distance is about 1.5×1011  m1.5 \times 10^{11}\;\mathrm{m} or 1  A.U.1\;\mathrm{A.U.} (astronomical unit). While not entirely obvious, the scale of these numbers allow us to assume the Sun is at the origin, and doesn’t move. Although this is not a good assumption for the real solar system, the sun orbits the barycenter of the solar system, which is about 1 solar radii from the center of the sun.

Gravitational Bound System

Let’s use the standard xx and yy axes to write the equations of motion. We can apply Newton’s Second Law to the earth in the chosen coordinate system.

F⃗net=F⃗gravity=ma⃗=m⟨ax,ay⟩\vec{F}_{net} = \vec{F}_{gravity} = m\vec{a} = m\langle a_x, a_y \rangle

So that the forces in the xx and yy directions are:

Fx=−GMsunMearthx(x2+y2)3/2Fy=−GMsunMearthy(x2+y2)3/2F_x = -G \dfrac{M_{sun} M_{earth}x}{(x^2+y^2)^{3/2}} \qquad F_y = -G \dfrac{M_{sun} M_{earth}y}{(x^2+y^2)^{3/2}}

and thus the acceleration of the Earth in the xx and yy directions are:

ax=−GMsunx(x2+y2)3/2ay=−GMsuny(x2+y2)3/2a_x = -G \dfrac{M_{sun} x}{(x^2+y^2)^{3/2}} \qquad a_y = -G \dfrac{M_{sun} y}{(x^2+y^2)^{3/2}}

This gives us a set of coupled differential equations.

x¨=−GMsunx(x2+y2)3/2y¨=−GMsuny(x2+y2)3/2\ddot{x} = -G \dfrac{M_{sun} x}{(x^2+y^2)^{3/2}} \qquad \ddot{y} = -G \dfrac{M_{sun} y}{(x^2+y^2)^{3/2}}

How do we then get the trajectories?

We can’t solve these equations without more information. We need to know the initial conditions of the system (the initial position and velocity of the Earth). We can then integrate these equations to get the position of the Earth as a function of time.

We have three potential ways to solve these EOMs:

  1. Direct Integration: We can integrate the equations of motion directly. This is possible in some cases where the equations are simple enough; think about the falling ball without air resistance, or the linear 1D drag case.

  2. Decouple and Solve: We try to solve the coupled differential equations by decoupling them. This is possible in some cases, but not all. We can frequently decouple the equations by writing them in terms of the velocity, or by making a change of position variables.

  3. Numerical Integration: We use numerical methods to predict the motion in small time steps. This is the most common method for solving complex systems.

Let’s see numerical integration in action. Starting the Earth at 1 A.U. with a speed close to (but not exactly) circular orbital speed produces the elliptical orbit shown below.

<Figure size 600x600 with 1 Axes>

The Simple Harmonic Oscillator (SHO)

In 1D, the simple harmonic oscillator is a system where the force is proportional to the displacement from the equilibrium position. The force is given by:

F=−ksF = -ks

where kk is the spring constant and ss is the displacement from the equilibrium position, x−L0x-L_0. The quantity L0L_0 is the relaxed length of the spring. The figure below shows the typical horizontal spring system.

SHO

We can typically choose to measure the displacement from the equilibrium position, and write the force instead as:

F=max=mv˙x=mx¨=−kxF = m a_x = m \dot{v}_x = m\ddot{x} = -kx

So the equation of motion that we will try to solve is:

x¨=−kmx\ddot{x} = -\dfrac{k}{m}x

How do we solve this?

d2xdt2=−kmx=−ω2x\dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x = - \omega^2 x

where ω=km\omega = \sqrt{\dfrac{k}{m}} is the natural oscillation frequency of the system.

It might seem strange, but let’s try the following potential solution to the differential equation:

x(t)=Ceiωtx(t) = C e^{i\omega t}

where CC is a constant. We can take the second derivative of x(t)x(t) with respect to time to see if it satisfies the differential equation.

x˙(t)=iωCeiωt\dot{x}(t) = i\omega C e^{i\omega t}
x¨(t)=−ω2Ceiωt=−ω2x(t)\ddot{x}(t) = -\omega^2 C e^{i\omega t} = -\omega^2 x(t)

This is a solution to the differential equation as long as CC is a constant, but it’s a complex one (C=a+ibC=a+ib). We can also write the solution in terms of the cosine and sine functions because the exponential function can be written in terms of these functions.

eiωt=cos⁡(ωt)+isin⁡(ωt)e^{i\omega t} = \cos(\omega t) + i\sin(\omega t)

Thus, another general solution to this EOM that we can write is:

x(t)=Acos⁡(ωt)+Bsin⁡(ωt)x(t) = A \cos(\omega t) + B \sin(\omega t)

where AA and BB are constants that depend on the initial conditions of the system. Let’s see how that works:

x˙(t)=−Aωsin⁡(ωt)+Bωcos⁡(ωt)\dot{x}(t) = -A \omega \sin(\omega t) + B \omega \cos(\omega t)
x¨(t)=−Aω2cos⁡(ωt)−Bω2sin⁡(ωt)=−ω2x(t)\ddot{x}(t) = -A \omega^2 \cos(\omega t) - B \omega^2 \sin(\omega t) = -\omega^2 x(t)

Another form that works is:

x(t)=Dcos⁡(ωt+ϕ)x(t) = D \cos(\omega t + \phi)

where DD is the amplitude of the oscillation and ϕ\phi is the phase of the oscillation. Let’s check that again:

x˙(t)=−Dωsin⁡(ωt+ϕ)\dot{x}(t) = -D \omega \sin(\omega t + \phi)
x¨(t)=−Dω2cos⁡(ωt+ϕ)=−ω2x(t)\ddot{x}(t) = -D \omega^2 \cos(\omega t + \phi) = -\omega^2 x(t)

So we have several forms of the general solution to the simple harmonic oscillator. We can use these solutions to understand the behavior of the system. We can also use these solutions to understand the behavior of more complex systems that can be approximated by the simple harmonic oscillator.

One critical aspect of these solutions is that they have 2 free parameters, AA and BB, or DD and ϕ\phi. These parameters are determined by the initial conditions of the system. There are N free parameters in the general solution to an Nth order differential equation.