Two-Body Dynamics#

Prepared by: Emmanuel Airiofolo, Joost Hubbard, Ceyda Alan and Angadh Nanjangud

In this lecture we cover the following topics:

  1. Gravitational Force and Newton’s Second Law

  2. Two-Body Relative Dynamics

  3. The Specific Angular Momentum

  4. Consequences of Specific Angular Momentum = constant

  5. Kepler’s Second Law

  6. Polar Coordinates

  7. Conservation of Specific Orbital Energy: E

  8. Admissible Orbital Radius

Gravitational Force and Newton’s Second Law#

  • The gravitational force between two masses \(m_1\) and \(m_2\) is given by:

Gravitational Force#

(20)#\[\begin{split}\begin{aligned} {\bf{F}}_{12} &= \frac{G m_1 m_2}{r^2} \hat{\bf{r}} = \frac{G m_1 m_2}{r^3} {\bf{r}} \\ {\bf{F}}_{21} &= -\frac{G m_1 m_2}{r^2} \hat{\bf{r}} = -\frac{G m_1 m_2}{r^3} {\bf{r}} \end{aligned}\end{split}\]

where:

\(G = 6.672 \times 10^{-11} \ \text{kg}^{-1} \ \text{m}^3 \ \text{s}^{-2}\) (Universal gravitational constant) and \(\hat{\bf{r}} = \frac{{\bf{r}}}{|\bf{r}|}\) is the unit vector.

Figure 6

  • \({\bf{F}}_{12}\) represents the force on \(m_1\) exerted by \(m_2\)’s gravitational attraction.

  • \({\bf{F}}_{21}\) represents the force on \(m_2\) exerted by \(m_1\)’s gravitational attraction.

Center of Mass ( G )#

The center of mass \({\bf{r}}_G\) of the system is given by:

(21)#\[{\bf{r}}_G = \frac{m_1 {\bf{r}}_1 + m_2 {\bf{r}}_2}{m_1 + m_2}\]

Newton’s Second Law (F = ma)

(22)#\[m_1 \ddot{\bf{r}}_1 = {\bf{F}}_{12}\]
(23)#\[m_2 \ddot{\bf{r}}_2 = {\bf{F}}_{21}\]

Substituting (133) into equations (146) + (147):

(24)#\[m_1 \ddot{\bf{r}}_1 + m_2 \ddot{\bf{r}}_2 = {\bf{F}}_{12} + {\bf{F}}_{21} = \frac{G m_1 m_2 {\bf{r}}}{r^3} - \frac{G m_1 m_2 {\bf{r}}}{r^3} = 0 \]

Thus:

(25)#\[\ddot{\bf{r}}_G = 0, \quad \dot{\bf{r}}_G = \text{constant}\]

This implies that \(G\) can be chosen as the origin of the inertial frame.

Two-Body Relative Dynamics#

Newton’s Second Law#

(26)#\[m_2 \ddot{\bf{r}}_2 = {\bf{F}}_{21} = - \frac{G m_1 m_2 {\bf{r}}}{r^3} \rightarrow \ddot{\bf{r}}_2 = - \frac{G m_1 {\bf{r}}}{r^3}\]
(27)#\[m_1 \ddot{\bf{r}}_1 = {\bf{F}}_{12} = \frac{G m_1 m_2 {\bf{r}}}{r^3} \rightarrow \ddot{\bf{r}}_1 = \frac{G m_2 {\bf{r}}}{r^3}\]

(170) - (171):

(28)#\[\ddot{\bf{r}} = \ddot{\bf{r}}_2 - \ddot{\bf{r}}_1 = - \frac{G m_1 {\bf{r}}}{r^3} - \frac{G m_2 {\bf{r}}}{r^3} = - \frac{G (m_1 + m_2) {\bf{r}}}{r^3}\]
(29)#\[\ddot{\bf{r}} = - \frac{G (m_1 + m_2)}{r^3} {\bf{r}}\]
(30)#\[\ddot{\bf{r}} = - \frac{\mu {\bf{r}}}{r^3}\]

where \(\mu\) is the gravitational parameter.

Implications of the Equation of Motion If \(m_1 \gg m_2\)

e.g., \(m_1\) is \(m_\oplus = 5.974 \times 10^{24} \ \text{kg}\) and \(m_2\) is a spacecraft, \(m_2 \approx 1000 \ \text{kg}\)

(31)#\[{\bf{r}}_G \approx {\bf{r}}_1\]
(32)#\[\mu \approx G m_1\]

This is referred to as a restricted two-body problem.

Two-Body Relative Dynamics#

\[{\bf{r}} (t_0) = \bf{r_0}\]
(33)#\[\bf{\ddot r} = -\frac{\mu}{r^3}\bf{r}\]

  • System of 3 scalar second order differential equations

  • It can be reduced to a system of 6 first order equations

(34)#\[\bf{\dot r} = \bf{v}\]
(35)#\[\bf{\dot v} = -\frac{\mu}{r^3}\bf{r}\]

  • To solve this system we need 6 initial conditions with \(\bf{r_0}\) and \(\bf{v_0}\) as known intial values

(36)#\[{\bf{r}} (t_0) = \bf{r_0}\]
(37)#\[{\bf{v}} (t_0) = \bf{v_0}\]

  • The system admits at maximum 6 independent integrals of motion

An Integral of Motion is a function \(f({\bf{r}} , {\bf{v}} , t )\) that is constant for all times, t

(38)#\[f({\bf{r}} , {\bf{v}} , t) = constant\]

  • The value of the constant is determined by the intial conditions

  • An integral of motion reduces the degree of freedom of our problem

The Specific Angular Momentum#

  • The relative angular momentum of \(m_2\) with respect \(m_1\) is

(39)#\[{\bf H}_{2/1} = {\bf{r}}\times( m_2 {\bf{v}}) = {\bf{r}}\times( m_2 {\bf{\dot r}})\]

  • per unit mass gives the specific angular momentum

(40)#\[{\bf{h}}=\frac{{\bf H}_{2/1}} {m_2} = \bf{r}\times\bf{\dot r}\]

  • the time derivative of \(\bf{h}\) is

(41)#\[\frac{d\bf{h}}{dt} =\frac{d}{dt}(\bf{r}\times\bf{\dot r}) = \bf{\dot r}\times\bf{\dot r} + \bf{r}\times\bf{\ddot r} = \bf{r}\times\bf{\ddot r}\]

  • From the dynamics \(\bf{\ddot r}=-\frac{\mu}{r^3}\bf{r}\), thus;

(42)#\[\frac{d\bf{h}}{dt}=\bf{r}\times\bf{\ddot r}=\bf{r}\times-\frac{\mu}{r^3}\bf{r}=0\]
(43)#\[\frac{d\bf{h}}{dt}=0\]
(44)#\[\bf{h} = constant\]

This result has the following implications:

  • \({\bf{h}}({\bf{r}} , {\bf{\dot r}} , t)\) = constant.

  • \(\bf{h}\) is an integral of motion.

  • All 3 scalar quantities are constants: \(h_x({\bf{r}} , {\bf{\dot r}} , t)\) = constant, \(h_y({\bf{r}} , {\bf{\dot r}} , t)\) = constant, \(h_z({\bf{r}} , {\bf{\dot r}} , t)\) = constant.

Consequences of Specific Angular Momentum = constant#

(45)#\[\bf{h} = \bf{r}\times\bf{v}\]

hence:

(46)#\[\bf{h}\perp\bf{r}\times\bf{v} \]

Direction

Figure 1

  • \(\bf{r}\) and \(\bf{v}\) must at all times lie in a plane perpendicular to \(\bf{h}\)

  • The motion is planar

Magnitude

Figure 2

(47)#\[\bf{h} = \bf{r}\times\bf{v} = \bf{r}\times(v_r\bf{e_r} + v_\theta\bf{e_\theta})\]
(48)#\[=\bf{r}\times(v_r\bf{e_r}) + \bf{r}\times(v_\theta\bf{e_\theta})\]
(49)#\[=rv_\theta{\bf{\hat{h}}} ⇒ rv_\theta = constant\]
(50)#\[dA = \frac{1}{2}rvdt\sin(\phi)\]
(51)#\[\frac{dA}{dt} = \frac{1}{2}rv\sin(\phi) = \frac{1}{2}rv_\theta \]
(53)#\[⇒ \frac{dA}{dt}=\frac{1}{2}h = constant\]

Kepler’s Second Law: Areal velocity is constant

Kepler’s Second Law#

(53)#\[⇒ \frac{dA}{dt}=\frac{1}{2}h = constant\]

Polar Coordinates#

If the motion is planar we can use polar coordinates to fully describe it.

Figure 3

  • In polar coordinates \(\bf{r} = r\bf{e_r}\)

(54)#\[\frac{d\bf{r}}{dt} = \bf{v} = {\dot r}\bf{e_r} + {r}\bf{e_\theta}\dot \theta \]
(55)#\[\bf{h}=\bf{r}\times\bf{v}=r\bf{e_r}\times({\dot r}\bf{e_r} + {r}\bf{e_\theta}\dot \theta)\]
(56)#\[⇒r^2\dot \theta{\bf{k}} = rv_\theta {\bf{k}} = constant\]
(57)#\[\frac{dA}{dt}=\frac{1}{2}h=\frac{1}{2}r^2\dot \theta\]
(58)#\[{\frac{d^2\bf{r}}{dt}} = {\bf{a}} = ({\ddot r} - r\dot\theta^2) {\bf{e_r}} + (r\ddot \theta+2\dot r\dot \theta) {\bf{e_\theta}}\]
(59)#\[\frac{-\mu}{r^2}\bf{e_r}\]
(60)#\[{\ddot r} - r\dot \theta^2 = \frac{-\mu}{r^2}\]
(61)#\[r\ddot \theta + 2{\dot r}\dot \theta = 0 ↔ \frac{d}{dt}(r^2\dot \theta)=2r{\dot r}\dot \theta+r^2\ddot \theta=r(r\ddot \theta+2{\dot r}\dot \theta) =0\]

  • (61) is equivalent to \(\bf{h}\) = constant which is a scalar

Conservation of Specific Orbital Energy: \(E\)#

Take the two-body dynamics and the dot product with \(\bf{\dot r}\);

(62)#\[\bf{\dot r}\cdot\bf{\ddot r}=\frac{-\mu}{r^2}\bf{\hat{r}}\cdot\bf{\dot r}\]

  1. Note that:

(63)#\[\frac{d}{dt}(\frac{1}{2}v^2)=\frac{1}{2}\frac{d}{dt}(\bf{\dot r\cdot\dot r})\]
(64)#\[⇒\frac{1}{2}{\bf{\ddot r}} {\cdot\bf{\dot r}} + {\frac{1}{2}\bf{\dot r}}{\cdot\bf{\ddot r}} = {\bf{\dot r}}{\cdot\bf{\ddot r}}\]

thus;

(65)#\[⇒({\bf{\dot r}}\cdot{\bf{\ddot r}}) = \frac{d}{dt}(\frac{1}{2}v^2)\]

Notice that this is the derivative of the kinetic energy

  • In polar coordinates \(v^2={v_r}^2+{v_\theta}^2={\dot r}^2+r^2\dot \theta^2\)

  1. Note that:

(66)#\[\frac{d}{dt}(\frac{\mu}{r})=\frac{\partial}{\partial r}\frac{\mu}{r}\frac{dr}{dt} = -\frac{\mu}{r^2}{\dot r}\]

thus

(67)#\[-\frac{\mu}{r^2}{\bf{\hat{r}}}{\cdot\bf{\dot r}} = \frac{\mu}{r^2}\dot r = \frac{d}{dt}(\frac{\mu}{r})\]

Notice that this is the derivative of the potential of the gravotational force

Combining the Derivatives of the Kinetic Energy (65) and Potential of Gravitationl Force (67), we get the Conservation of Specific Orbital Energy (69).

(68)#\[\frac{d}{dt}(\frac{1}{2}v^2-\frac{\mu}{r})=0\]
(69)#\[\frac{1}{2}v^2-\frac{\mu}{r} = constant\]

Notice that this is the conservation of specific orbital energy

Admissible Orbital Radius#

(70)#\[E=\frac{1}{2}v^2-\frac{\mu}{r}=\frac{1}{2}{\dot r}^2+\frac{1}{2}r^2\dot \theta-\frac{\mu}{r}=\frac{1}{2}{\dot r}^2+\frac{1}{2}\frac{h^2}{r^2}-\frac{\mu}{r}\]

  • \(\frac{1}{2}\frac{h^2}{r^2}-\frac{\mu}{r}\) is the effective potential \(\phi(r)\)

  • \(\frac{1}{2}\frac{h^2}{r^2}\) is the potential of the contributed force

  • \(\frac{\mu}{r}\) is the potential gravitational force

Figure 4

The motion is only possible for those values of \(r\) such that \(E\geq\phi_{eff}\)

Figure 5

  • For \(E<0\) the motion occurs between \(r_{min}-r_{max}\)

  • For \(E\geq0\) the motion is unbounded

  • For \(E=min\phi(r), r_{min}=r_{max}\) hence we have a circular orbit

Figure 6

Given initial position \(\bf{r_0}\) and velocity \(\bf{v_0}\) we can

  • Compute

(71)#\[\bf{h}=\bf{h_0}=constant\]

orbital plane

\(r_0^2\dot \theta_0^2\rightarrow\) Areal velocity

  • Compute

(72)#\[E_0=\frac{1}{2}v_0^2 - \frac{\mu}{r_0}\]

if \(E<0\) bounded motion and \(E\geq0\) unbounded motion

  • Compute

(73)#\[r_{min/max}\]

when

(74)#\[\phi_{eff}=E=E_0\]