Energy Conservation

Section 12.6 Energy Conservation

Two Bodies Interacting with Gravitation.

Since gravitational force is a conservative force, the total energy of the two-body interacting via the gravitational force is conserved. The total energy \(E\) of the two-body system is given by the following.

\begin{equation} E = K + U = \left(\dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2 \right) - G_N\dfrac{m_1m_2}{r}.\tag{12.34} \end{equation}

Now, I will show that the kinetic energy part of this formula can also be written as sum of two parts, the kinetic energy of a fictitious particle of total mass \(M=m_1 + m_2\) moving with the center of mass of the two particles and the kinetic energy of another fictitious particle of mass equal to the reduced mass \(\mu = m_1m_2/M\) of the two particles that is moving with the relative velocity of the two particles.

\begin{equation} E = \left(\dfrac{1}{2}M V_\text{cm}^2 + \dfrac{1}{2}\mu v_\text{rel}^2 \right) - G_N\dfrac{m_1m_2}{r}.\tag{12.35} \end{equation}

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Remark 12.31. Derivation Algebra.

\(\vec R\)\(\vec r\)

\begin{align*} \amp m_1 \vec r_1 + m_2 \vec r_2 = M \vec R\\ \amp \vec r_1 - \vec r_2 = \vec r \end{align*}

\begin{align*} \amp \vec r_1 = \frac{m_2}{M}\,\vec r + \vec R\\ \amp \vec r_2 = -\frac{m_1}{M}\,\vec r + \vec R \end{align*}

\(V\)\(V_\text{cm}\text{.}\)

\begin{align*} \amp \vec v_1 = \frac{m_2}{M}\,\vec v + \vec V \\ \amp \vec v_2 = -\frac{m_1}{M}\,\vec v + \vec V \end{align*}

\begin{align*} \amp v_1^2 = \left(\frac{m_2}{M}\right)^2\,v^2+ V^2 + 2 \frac{m_2}{M} \vec v \cdot \vec V\\ \amp v_2^2 = \left(\frac{m_1}{M}\right)^2\,v^2+ V^2 - 2 \frac{m_1}{M} \vec v \cdot \vec V \end{align*}

\(m_1\)\(m_2\text{,}\)\(1/2\)

\begin{equation*} \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2} \mu v^2 + \frac{1}{2} M V^2. \end{equation*}

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Often we work in a coordinate system whose origin is fixed at the CM. In this system, \(V_\text{cm}=0\text{.}\) This is the CM frame. In this frame, energy of the two particle system needs only the dynamics of the relative position between the two particles!

\begin{equation} E = \dfrac{1}{2}\mu v_\text{rel}^2 - G_N\dfrac{m_1m_2}{r}\ \ (\text{CM Frame})\tag{12.36} \end{equation}

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Subsection 12.6.1 (Calculus) Effective Potential Energy

Since gravitational force is spherically symmetric, the angular momentum is conserved. That makes the motion of the fictitious particle of reduced mass \(\mu\) described by Eq. (12.36) will occur in a plane prpendicular to the angular momentum vector. For planar motion of planets around the Sun, we use polar coordinates. In polar coordinates, the energy of the two-body system in the CM frame, will simply be

\begin{equation} E = \dfrac{1}{2}\mu v_r^2 + \dfrac{1}{2}\mu r^2 \omega^2 - G_N\dfrac{m_1m_2}{r},\tag{12.37} \end{equation}

where \(r \) is the relative distance between the particles, \(v_r = dr/dt\text{,}\) the radial speed, and \(\omega\) speed for the angular variable \(\theta\) of the polar coordinate, i.e., \(\omega = d\theta/dt\text{,}\) in the plane of motion,

\begin{equation*} v_r = \dfrac{dr}{dt},\ \ \text{ and }\ \ \omega = \dfrac{d\theta}{dt}. \end{equation*}

The angular part can be expressed in terms of conserved angular momentum,

\begin{equation} l = \mu \omega r^2 \ \ \longrightarrow\ \ \omega = \dfrac{l}{\mu r^2}.\tag{12.38} \end{equation}

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Remark 12.32. Why is angular momentum = \(\mu \omega r^2\text{?}\)

We work in the CM frame and decorate symbols with prime to indicate the CM frame. We first express the coordinates of the two particles \(\vec r_1^{\prime}\) and \(\vec r_2^{\prime}\) in the CM frame in terms of the relative coordinate \(\vec r = \vec r_1^{\prime}- \vec r_2^{\prime}\text{.}\) From the definition of the CM we have the following for the vector from \(m_1\) to the location of the CM, which is the negative of \(\vec r_1^{\prime}\text{.}\)

\begin{equation*} -\vec r_1^{\prime} = \frac{m_2}{m_1 + m_2} \vec r. \end{equation*}

Subtracting from \(\vec r\) we obtain,

\begin{equation*} \vec r_2^{\prime} = \frac{m_1}{m_1 + m_2} \vec r. \end{equation*}

Let us write the velocities associated with the coordinates \(\vec r\text{,}\) \(\vec r_1^{\prime}\) and \(\vec r_2^{\prime}\) by \(\vec v\text{,}\) \(\vec v_1^{\prime}\) and \(\vec v_2^{\prime}\) respectively. Now, we can carry out the necessary calculations.

\begin{align*} L \amp == \vec r^{\prime}_1 \times m_1\vec v^{\prime}_1 + \vec r^{\prime}_2 \times m_2\vec v^{\prime}_2\\ \amp = \frac{m_1 m_2^2}{(m_1 + m_2)^2} \vec r \times \vec v + \frac{m_1^2 m_2}{(m_1 + m_2)^2} \vec r \times \vec v\\ \amp = \mu \vec r \times v\left[ \frac{m_2}{m_1 + m_2}+ \frac{m_1}{m_1 + m_2}\right] = \vec r \times \mu \vec v. \end{align*}

We will express this formula in polar coordinates. The position and velocity in the polar coordinates will be

\begin{align*} \amp \vec r = r\,\hat u_r \\ \amp \vec v = v_r\hat r + \omega r \hat u_\theta \end{align*}

Therefore, angular momentum of fictitious particle of mass equal to \(\mu\) will be

\begin{equation*} \vec l = \vec r \times (\mu\,\vec v) = \mu \omega r^2\,\hat u_z. \end{equation*}

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After expressing \(\omega\) in terms of \(l\text{,}\) we can group together non-radial-velocity terms.

\begin{equation} E = \dfrac{1}{2}\mu v_r^2 + \left( \dfrac{1}{2} \dfrac{l^2}{\mu r^2} - G_N\dfrac{m_1m_2}{r} \right).\tag{12.39} \end{equation}

The quantity within parenthesis, \(( \cdots )\) is called effective potential energy, and denoted by \(U_\text{eff}\text{.}\) With

\begin{equation} U_\text{eff} = \dfrac{1}{2} \dfrac{l^2}{\mu r^2} - G_N\dfrac{m_1m_2}{r},\tag{12.40} \end{equation}

we write energy more simply as

\begin{equation} E = \dfrac{1}{2}\mu v_r^2 + U_\text{eff} \equiv K_r + U_\text{eff},\tag{12.41} \end{equation}

where \(K_r\) is kinetic energy in the radial motion only. When we write energy this way, the problem looks like the problem of one variable \(r\text{,}\) which can vary in range \([0,\infty)\text{.}\) Of course, in the real world, the planet is moving in a plane with both the radial coordinate and the angular coordinate. Equation (12.41) helps us focus on the radial motion of reduced particle \(\mu\) alone.

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Subsection 12.6.2 Interpreting Effective Potential Energy

How to use effective potential energy in Eq. (12.40) and (12.41)? First of all, we note the \(K_r \ge 0\) since \(\mu\) and square of speed are positive real numbers. Therefore

\begin{equation*} E - U_\text{eff} \ge 0, \text{ for all }r. \end{equation*}

To explore \(U_\text{eff}\) further, let us use a simpler notation.

\begin{equation} U_\text{eff} = \dfrac{1}{2} \dfrac{l^2}{\mu r^2} - G_N\dfrac{m_1m_2}{r} \Leftrightarrow \dfrac{\alpha}{r^2} - \dfrac{\beta}{r},\tag{12.42} \end{equation}

with

\begin{equation*} \alpha = \dfrac{l^2}{2 \mu},\ \ \beta = G_Nm_1m_2. \end{equation*}

Figure 12.33 shows \(U_\text{eff}\text{ vs }r\text{.}\) The figure shows four distinct energy values of interest, \(E=0,\, E_1,\, E_2,\) and \(E_\text{min}=U_0\text{.}\)

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At the minimum of the effective potential, its derivative with respect to \(r\) is zero.

\begin{equation*} \left.\frac{dU_\text{eff}}{dr}\right|_{r=r_0} = 0 \ \implies\ \ r_0 = \frac{l^2}{G_N m_1 m_2 \mu} = \frac{2\alpha}{\beta}. \end{equation*}

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Figure 12.33. Plot of effective potential energy versus radial distance \(r\text{.}\) Four energy levels of interest are indicated on the axis. They are \(E = 0,\ E_1,\ E_2, E_\text{min}\text{.}\) The system cannot have energy less than the minimum of \(U_\text{eff}\) since that would mean \(K_r\lt 0\text{,}\) which is impossible.
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We can see the turning points represented by these energy values by solving \(E = U_\text{eff}\text{,}\) we will have the values of \(r\) for turning points in the motion in relative position, since at those values we will have \(K_r=0\text{,}\) i.e., \(v_\text{r}=0\text{.}\)

\begin{equation} E r^2 + \beta r - \alpha = 0.\tag{12.43} \end{equation}

This equation has following solutions.

\begin{align} \amp r = \dfrac{\alpha}{\beta} \ \ (E=0), \tag{12.44}\\ \amp r = \dfrac{1}{2E} \left(-\beta + \sqrt{ \beta^2 + 4\alpha E } \right)\ \ (E \gt 0), \tag{12.45}\\ \amp r = \dfrac{1}{2|E|} \left(\beta \pm \sqrt{ \beta^2 - 4\alpha |E| } \right)\ \ (E \lt 0), \tag{12.46} \end{align}

where we need to make sure that

\begin{equation*} \beta^2 \ge 4 \alpha |E|. \end{equation*}

The solution for \(E=0\) has one turning point with orbit having the shape of a parabola. The solution for \(E \gt 0\) has also only one turning point with the orbit a hyperbola.

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The two solutions in the case of \(E \lt 0 \) refers to the perihelion and aphelion of a planet in the planet/Sun system.

\begin{align} r_\text{min} \amp = \dfrac{1}{2|E|} \left(\beta - \sqrt{ \beta^2 - 4\alpha|E| } \right)\ \ (E\lt 0,\ \ |E| \lt \beta^2/4\alpha) \tag{12.47}\\ r_\text{max} \amp = \dfrac{1}{2|E|} \left(\beta + \sqrt{ \beta^2 - 4\alpha|E| } \right)\ \ (E\lt 0,\ \ |E| \lt \beta^2/4\alpha) \tag{12.48} \end{align}

Much simpler formulas for these minimum and maximum distances for planetary system is given in the section on Orbit Equation.

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The solutions of orbit equation for these conditions show that the orbit is ellipse with the smallest distance between the masses given by \(r_\text{min} \) and the largest distance \(r_\text{min} \text{.}\) When \((E\lt 0,\ \ |E| = \beta/4\alpha)\text{,}\) there is only one solution with

\begin{equation*} r_0 = \dfrac{2\alpha}{\beta}\ \ (E\lt 0,\ \ |E| = \beta^2/4\alpha), \end{equation*}

which is the radius of the circle of motion of the relative position.

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Example 12.34. Calculating Various Quantities in Earth/Sun Two-Body System.

Find the reduced mass of the two-body system, called the Earth/Sun system consisting of the Earth and the Sun.
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Where is the center of mass of the Earth/Sun system located.
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Find the angular momentum of the Earth about the center of mass of the Earth/Sun system.
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Where in the orbit does the Earth move fastest and where does it move slowest? Why?
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Is the kinetic energy of the Earth with respect to the CM of the Earth/Sun system constant during its flight around the Sun?
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Is the angular momentum of the Earth with respect to the CM of the Earth/Sun system constant during its flight around the Sun?
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Write an energy based on the energy of the Earth/Sun system that is conserved.
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Write an angular momentum based on the angular momentum of the Earth/Sun system that is conserved.
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Solution 1. a

If you use a calculator, you might find that the reduce mass is equal to the mass of the Earth itself. Let us write the reduced mass in terms of the ratio of the small number \(\epsilon \equiv M_E/M_S\) as follows.

\begin{equation*} \mu = \frac{M_E M_S}{M_E + M_S} = M_E \left (\frac{1}{1+\epsilon} \right). \end{equation*}

Now, we can do a Maclaurin series in \(\epsilon\) to obtain.

\begin{equation*} \mu \approx M_E \left (1 - \epsilon \right) = M_E \left (1 - 2.0\times 10^{-6} \right). \end{equation*}

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Solution 2. b

Let us place the origin at the center of the Sun and point the \(x\)-axis towards the Earth. Let \(R\) be the distance of the Earth from the sun, then \(X_{cm}\) will be located at

\begin{equation*} X_{cm} = \frac{M_E}{M_E + M_S} R \approx 3.0\times 10^{-6} R. \end{equation*}

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Solution 3. c

The angular momentum of the Earth will be perpendicular to the the plane of orbit. Let \(v_E\) be the speed of the Earth, then the magnitude of the angular momentum about the CM will be

\begin{equation*} L = M_E v_E (R -X_{cm}) \approx M_E v_E R. \end{equation*}

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Solution 4. d

The Earth will move fastest when it is nearest to the Sun. This is because the force of gravitation is a conservative force and the energy of the Earth is conserved. When the Earth is at the nearest to the Sun, then it has the largest negative potential energy, i.e., the lowest potential energy, which would mean that the Earth has the largest kinetic energy at this instant in the orbit.

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Solution 5. e

No. The kinetic energy of the Earth is not constant in its orbit. The total energy of the Earth is constant but its kinetic energy is not.

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Solution 6. f

The angular momentum of Earth is conserved in its motion since there are no torques on the Earth. The force is central and the central force passes through the CM, which means the torque of this force will have a zero lever arm and hence the angular momentum of Earth cannot change.

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Solution 7. g

The total mechanical energy of either of the two bodies as well as their sum will be conserved.

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Solution 8. h

The angular momentum about the CM of the system for either of the two bodies as well as their sum will be conserved.

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Example 12.35. Anergy and Angular Momentum of Europa Around Jupiter.

Europa is a moon of Jupiter that is most like Earth. It has a mass of \(4.8\times 10^{22}\ \text{kg}\) and moves in an ellipse of eccentricity \(0.009\) with a mean distance of \(670,900\ \text{km}\) from the center of Jupiter whose mass is \(1.9\times10^{27}\ \text{kg}\text{.}\) What are the energy and the angular momentum of Europa? Ignore the motion around the Sun.

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Solution.

Since the mass of Europa is so much smaller than that of Jupiter, we can assume one-particle picture and set \(\mu\) equal to the mass \(m\) of Europa. The data is given for the average distance between Europa and Jupiter which we will equate to \(r_0\) since if Europa moved at the average distance its distance will not change and the orbit would be circular.

\begin{align*} \amp e = 0.009\\ \amp r_0 = 6.7\times 10^8\ \text{m} \end{align*}

We need \(G_NMm\) to find energy \(E\text{.}\) We will calculate the numerical value of this quantity so that we do not clutter the final formulas.

\begin{align*} G_NMm\amp = 6.7\times10^{-11}\frac{\textrm{N.m}^2}{\textrm{kg}^2}\times 1.9\times10^{27}\textrm{kg}\times4.8\times10^{22}\textrm{kg}\\ \amp = 6.1\times 10^{39}\textrm{N.m}^2. \end{align*}

Hence, the energy of Europa will be

\begin{equation*} E = -\frac{G_NMm}{2a} \approx -\frac{G_NMm}{2r_0} = - 4.6\times 10^{30}\ \textrm{J}. \end{equation*}

We use the energy to find the magnitude of the angular momentum.

\begin{equation*} l= b\sqrt{-2\mu E} \approx r_0\sqrt{-2\mu E} = 4.5\times 10^{35}\ \textrm{kg.m}^2/\textrm{s}. \end{equation*}

The direction of the angular momentum is pointed perpendicular to the plane of the orbit.

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