Introduction

May 6, 2025

What do we want to know?

Many things actually!

The question of why I leave it for everyone to figure out for themselves.

• How do the (celestial) bodies move?
  • How does the moon move around the Earth?
  • How do planets move around the Sun?
  • How do satellites move around the Earth?
  • How do spacecrafts move around the Sun?
• Can we predict their motion in future?
  • How to predict the motion of a spacecraft around the Earth?
• Can we modify the motion of these bodies?
  • How to change the orbit of a spacecraft around the Earth?

We are infact looking for a quantitative solution for all of these problems.

So What would be a good solution?

• A good solution will be most accurate.
• It will be computationally efficient.
• It will require the least amount of data.
• It will be easy to implement.

More rigorously, we want \(\mathbf{r}(t, \mathbf{r}_0, \mathbf{v}_0)\), where \(\mathbf{r}\) is the position vector of the body at time \(t\), \(\mathbf{r}_0\) is the initial position vector and \(\mathbf{v}_0\) is the initial velocity vector.

Qualitative solution (First baby steps)

Historical Overview

• Ancient civilizations observed the motion of planets among the stars.

• They proposed various models to explain celestial motion.

• By sixteenth century, debates emerged about whether planets actually orbited the Sun or Earth.

• The exact details of how planets moved around the sun, with what precise motion, took more time to uncover.

• Tycho Brahe introduced a groundbreaking idea: To resolve these debates, it would be better to measure the actual positions of planets in the sky accurately. His belief was that precise measurement would provide clarity on planetary motion, more so than philosophical discussions.

• Tycho Brahe spent many years observing and studying the planets from his observatory on the island of Hven, near Copenhagen and he compiled extensive tables of planetary positions.

• After Brahe’s death, Johannes Kepler studied his data and discovered simple yet profound laws of planetary motion.

1.2 Kepler’s Laws

1.2.1 Kepler’s Three Laws:

• First Law:

  • Kepler discovered that each planet moves around the sun in an ellipse.

  • The sun is located at one of the foci of the ellipse.

  • An ellipse is a precise mathematical curve where the sum of the distances from any point on the ellipse to two fixed points (the foci) is constant.

  • This curve can also be described as a foreshortened circle (Figure 1.1).

    Figure 1.1: Kepler’s first law

• Second Law:

  • Kepler observed that planets do not travel at a uniform speed.
  • Planets move faster when they are closer to the sun and slower when they are farther from the sun.
  • More precisely:
    • Consider a planet observed at two successive times (say, a week apart).
    • The radius vector is drawn from the sun to the planet at both positions.
    • The area swept out by the radius vector, bounded by the orbital arc between the two positions (Figure 1.2), is always the same for equal time intervals.
  • This is known as Kepler’s law of equal areas.
  Figure 1.2: Kepler’s second law

• Third Law:

  • This law compares the orbital periods and orbit sizes of different planets.

  • The square of the orbital period (\(T\)) of a planet is proportional to the cube of the semimajor axis (\(a\)) of its orbit: \[ T^{2} \propto a^{3} \]

  • This means that if planets orbited in perfect circles, the time required to complete an orbit would be proportional to the 3/2 power of the radius of the circle.

Reference: Feynman Lectures on Physics

1.2.2 Deriving Kepler’s Laws using Newton’s Laws

• Newton’s laws of motion and gravitation can be used to derive Kepler’s laws.

Kepler’s Laws of Planetary Motion:
1. A planet moves around the sun in an elliptical path with the Sun as one of the Focii.
2. The line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
3. The square of the orbital time period of a planet is proportional to the cube of the semi-major axis of its orbit.

Assumptions:
1. As the Sun, with mass \(M_\text{sun}\), is very massive compared to any object in the solar system, its motion is essentially unaffected by the gravity of the planets.
2. The gravity of the Sun acts along the line joining the Sun and a given planet (Force \(\mathbf{F}\) acts along \(\hat{r}\)). As a result, the motion of the planet is confined to a 2D plane.
3. The mechanical energy \(E\) for Sun-planet system is a conserved quantity.

Derivation:
The equation that describes planetary motion is given by Newton’s law as:

\[ m \mathbf{\ddot{r}} = -G \frac{M_\text{sun} m}{r^2} \hat{r} \]

where \(m\) is the mass of the planet, \(\mathbf{r}\) is the position vector of the planet, \(G\) is the gravitational constant, and \(r\) is the distance between the Sun and the planet.
This problem can be solved in a straightforward fashion in polar coordinates with unit vectors for the radius vector \(\hat{r}\) and the angle about the Sun \(\hat{\theta}\). But for the time being, the equation can be re-expressed in the following form given \(x = r \cos \theta\) and \(y = r \sin \theta\):

\[ m \ddot{x} = -G \frac{M_\text{sun} m}{r^2} \cos \theta \]

\[ m \ddot{y} = -G \frac{M_\text{sun} m}{r^2} \sin \theta \]

We need to find the second derivative of the \(x\) and \(y\) coordinates in terms of the polar coordinates.

\[ \dot{x} = \dot{r} \cos \theta - r \dot{\theta} \sin \theta \]

\[ \ddot{x} = \ddot{r} \cos \theta - 2 \dot{r} \dot{\theta} \sin \theta - r \ddot{\theta} \sin \theta - r \dot{\theta}^2 \cos \theta \]

\[ \dot{y} = \dot{r} \sin \theta + r \dot{\theta} \cos \theta \]

\[ \ddot{y} = \ddot{r} \sin \theta + 2 \dot{r} \dot{\theta} \cos \theta + r \ddot{\theta} \cos \theta - r \dot{\theta}^2 \sin \theta \]

We will now proceed to obtain the orbital equations in polar coordinates.

First, we multiply \(\ddot{x}\) by \(\cos \theta\) and \(\ddot{y}\) by \(\sin \theta\) and add them.

From central equation,

\[ \ddot{x} \cos \theta + \ddot{y} \sin \theta = -G M_\text{sun}(\cos^2 \theta + \sin^2 \theta) \frac{1}{r^2} = -G M_\text{sun} \frac{1}{r^2} \]

From the derivative identities between the Cartesian and polar coordinates,

\[ \ddot{x} \cos \theta + \ddot{y} \sin \theta = \ddot{r} \cos^2 \theta - 2\dot{r} \dot{\theta} \cos \theta \sin \theta - r \dot{\theta}^2 \cos^2 \theta - r \ddot{\theta} \cos \theta \sin \theta + \ddot{r} \sin^2 \theta + 2\dot{r} \dot{\theta} \cos \theta \sin \theta - r \dot{\theta}^2 \sin^2 \theta + r \ddot{\theta} \cos \theta \sin \theta \]

Simplification of the above equation gives,

\[ \ddot{x} \cos \theta + \ddot{y} \sin \theta = \ddot{r} - r \dot{\theta}^2 \]

Using the result from the central equation, we have:

\[ \boxed{ (\ddot{r} - r \dot{\theta}^2) = -G M_\text{sun} \frac{1}{r^2} } \]

Second Law: Equal area in equal time
If we had instead multiplied \(\ddot{x}\) by \(\sin \theta\) and \(\ddot{y}\) by \(\cos \theta\) and subtracted them, we would have obtained the following equation:

\[ r \ddot{\theta} + 2 \dot{r} \dot{\theta} = 0 \]

If we multiply this equation by \(r\),

\[ r^2 \ddot{\theta} + 2 r \dot{r} \dot{\theta} = 0 \]

This is the time derivative of \(r^2 \dot{\theta}\). Thus,

\[ \frac{d}{dt} (r^2 \dot{\theta}) = 0 \]

We know that \(mr^2\dot{\theta}\) is the angular momentum of the planet. Thus, the angular momentum of the planet is conserved. Let it be denoted by \(L\) which is a constant.

\[ \int L \, dt = m \int r^2 \frac{d\theta}{dt} \, dt = m \int_{\theta_i}^{\theta_f} r^2 \, d\theta \]

Now, the integral \(\frac{1}{2} \int_{\theta_i}^{\theta_f} r^2 d\theta\) is the area swept by the radial vector from the Sun to the planet in moving from \(\theta_i\) to \(\theta_f\). However, the result is independent of \(\theta_{i}\) and \(\theta_{f}\), but only depends on \(t\) since the angular momentum \(L\) is constant. Thus,

\[ \boxed{ \frac{1}{2} \int_{\text{any path}} r^2 d\theta = A = \frac{Lt}{2m} } \]

Thus, we have derived Kepler’s second law, the line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

First law: Elliptical orbits, with the Sun at one of the focii
We have,

\[ \ddot{r} - r \dot{\theta}^2 = -G M \frac{1}{r^2} \]

Here, we have put \(M_{\text{sun}} = M\)

Now, the differential equation becomes easy to solve if we make the substitution \(r \rightarrow u^{-1}\). We have,

\[ \begin{aligned} \frac{d r}{d t} & =-u^{-2} \frac{d u}{d t} \\ & =-u^{-2} \frac{d u}{d \theta} \frac{L u^2}{m} \\ & =-\frac{L}{m} \frac{d u}{d \theta} \end{aligned} \]

Further, we have \[ \begin{aligned} \frac{d^2 r}{d t^2} & =-\frac{L}{m} \frac{d}{d t} \frac{d u}{d \theta} \\ & =-\frac{L}{m} \frac{d \theta}{d t} \frac{d}{d \theta} \frac{d u}{d \theta} \\ & =-\left(\frac{L}{m}\right)^2 u^2 \frac{d^2 u}{d \theta^2} \end{aligned} \]

With this identity in hand, our central equation becomes \[ -\left(\frac{L}{m}\right)^2 u^2 \frac{d^2 u}{d \theta^2}-\left(\frac{L}{m}\right)^2 u^3=-G M u^2 \]

which becomes \[ -\frac{d^2 u}{d \theta^2}+\frac{G M m^2}{L^2} = u \]

which has the simple solution \(u=A \cos (\theta+\delta)+\frac{G M m^2}{L^2}\).

We can always define our coordinates so that \(\delta = 0\), and thus we set it to zero for the remainder of the derivation.

In terms of \(r(\theta)\), we have

\[ \begin{aligned} r & =\frac{1}{A \cos \theta+\frac{G M m^2}{L^2}} \\ & =\frac{1}{\frac{G M m^2}{L^2}(e \cos \theta+1)} \end{aligned} \]

Note that in the line above, we made the useful substitution \(e \rightarrow \frac{AL^2}{GMm^2}\).

\(r \sim (1 + e \cos \theta)^{-1}\) is the general form of an ellipse in polar coordinates, with the origin placed at a focus. The parameter \(e\) is called as the eccentricity of the ellipse. Thus, we have derived Kepler’s first law.

Third law: Period squared is proportional to the semi-major axis cubed

From the expression given above for \(r\), we can obtain \(r_{\text{max}}\) and \(r_{\text{min}}\) as

\[ r_{\text{max}} = \frac{1}{\frac{G M m^2}{L^2}(1-e)} , \quad r_{\text{min}} = \frac{1}{\frac{G M m^2}{L^2}(1+e)} \]

Thus the semi-major axis \(a\) is given by

\[ a = \frac{1}{2} (r_{\text{max}} + r_{\text{min}}) = \frac{L^2}{G M m^2} \frac{1}{1-e^2} \]

So,

\[ L^2 = aGMm^2(1-e^2) \]

Now, from the derivation of the second law,

\[ A = \frac{LT}{2m} \]

where A is the total area swept out in one orbital period \(T\). The area of the ellipse is \(\pi a b\), where \(a\) and \(b\) are semi-major axis and semi-minor axis of the ellipse, respectively.
Thus,

\[ \pi a b = \frac{LT}{2m} \]

\[ T = \frac{2 \pi a b m}{L} \]

\[ T^2 = \frac{\pi^2 a^2 b^2 (4 m^2)}{a G M m^2 (1-e^2)} \]

Putting the value of \(b^2 = a^2 (1-e^2)\), we get

\[ T^2 = \frac{\pi^2 a^2 a^2 (1-e^2)(4m^2)}{a GM m^2 (1-e^2)} \]

\[ T^2 = \frac{4 \pi^2}{GM} a^3 \]

Thus,

\[\boxed{ T^2 \propto a^3 } \]

Hence Kepler’s third law has been derived.

Reference: Deriving Kepler’s Laws. Brilliant.org. https://brilliant.org/wiki/deriving-keplers-laws/