Nothing new here, just wanted to capture these proofs into a single location for easy reference. I've tried to arrange them into a form that is easy to understand and follow.

\(\require{cancel}\)Proof that a Newtonian force between two masses will produce an elliptical orbit - this is a very textbook approach using polar coordinates, I'm just capturing it here for reference. There are many other approaches, and this has likely been done millions of times now.

Remember that Newton's Law is:

\[ \begin{align} F &= G \frac{Mm}{r^2} \\ &\therefore \nonumber \\ F &= m a \end{align} \]

We define a unit vector in the radial direction \( \hat{r} \) along the angle \( \theta \):

\[ \hat{r} = \hat{x} \cos \theta + \hat{y} \sin \theta \] Therefore \( \hat{r} \) changes as per angle \(\theta\) perpendicular to \( \hat{r} \), giving us the tangential unit vector \( \hat{\theta} \)

\[ \frac{\mathrm{d}\hat{r}}{\mathrm{d}\theta} = \hat{x} ( - \sin \theta ) + \hat{y} \cos \theta = \hat{\theta} \] And the derivative of \( \hat{\theta} \) is another 90° rotation, giving us \( -\hat{r} \):

\[ \frac{\mathrm{d}\hat{\theta}}{\mathrm{d}\theta} = \hat{x} ( - \cos \theta ) + \hat{y} ( - \sin \theta ) = -\hat{r} \] Find Acceleration \( \vec{a} \) in polar coordinates, first we find velocity \( \vec{v} \) as the change in radial vector over time:

\[ \vec{v} = \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \left( r \hat{r} \right) \] Applying the product rule and the chain rule we get our expression for how fast you are moving radially (\(\dot{r}\) in \(\hat{r}\)), and how fast you are moving in the tangential direction (\(r \dot{\theta}\) in \(\hat{\theta}\)):

\[ \vec{v} = \frac{\mathrm{d}r}{\mathrm{d}t} \hat{r} + r \frac{\mathrm{d}\theta}{\mathrm{d}t} \frac{\mathrm{d}\hat{r}}{\mathrm{d}\theta} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} \] Acceleration is the derivative of velocity:

\[ \vec{a} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \left( \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} \right) \] \[ \vec{a} = \color{Red}{\ddot{r} \hat{r}} + \color{ForestGreen}{\dot{r} \dot{\theta} \hat{\theta} + \dot{r} \dot{\theta} \hat{\theta} + r \ddot{\theta} \hat{\theta}} + \color{Red}{r \dot{\theta} (-\hat{r}) \dot{\theta}} \] Grouping terms in \( \hat{r} \; \text{and} \; \hat{\theta} \) we get:

\[ \vec{a} = \color{Red}{\hat{r} \left( \ddot{r} - r \hat{\theta}^2 \right)} + \color{ForestGreen}{ \hat{\theta} \left( 2 \dot{r} \dot{\theta} + r \ddot{\theta} \right) } \label{refa} \] From Newton

\[ \vec{F} = m \vec{a} \] We only need the radial term from \(\eqref{refa}\) since the Force is only in the radial direction, substitute along with Law of Gravity:

\[ - \frac{GM\cancel{m}}{r^2} \cancel{\hat{r}} = \cancel{m} \cancel{\hat{r}} ( \ddot{r} - r \dot{\theta}^2 ) \] The radial unit vector \( \hat{r} \) and mass term \( m \) cancel out, giving us this equation that we must show is equal:

\[ - \frac{GM}{r^2} = \ddot{r} - r \dot{\theta}^2 \label{refb} \] The angular momentum is a constant because there are no torques acting on our system, the only Force is gravity acting radially inwards so we can define \( \dot{\theta} \):

\[ L = m r^2 \dot{\theta} \; \therefore \; \dot{\theta} = \frac{L}{m r^2} \label{eqL} \] And substitute this into \eqref{refb}:

\[ - \frac{GM}{r^2} \stackrel{?}{=} \ddot{r} - \frac{L^2}{m^2 r^3} \label{refc} \] Next we need to show that the second time derivative of this equation of an ellipse works in our force equation:

\[ r = \frac{a(1 - \epsilon^2)}{1 + \epsilon \cos \theta} \] First derivative [Wolfram|Alpha]:

\[ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}r}{\mathrm{d}\theta} \frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{-a(1-\epsilon^2) ( - \epsilon \sin \theta )}{(1+\epsilon \cos \theta)^2} \dot{\theta} \] Substitute for \( \dot{\theta} \) and again for \( r \)

\[ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\cancel{a(1-\epsilon^2)} ( \epsilon \sin \theta )}{\cancel{(1+\epsilon \cos \theta)^2}} \frac{L \cancel{(1 + \epsilon \cos \theta)^2}}{m a(1-\epsilon^2) \cancel{a(1-\epsilon^2)}} \] \[ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{L \epsilon \sin \theta}{m a (1-\epsilon^2)} \] Second derivative:

\[ \frac{\mathrm{d}^2 r}{\mathrm{d} t^2} = \frac{\mathrm{d}}{\mathrm{d}\theta} \left( \frac{\mathrm{d} r}{\mathrm{d} t} \right) \left( \frac{\mathrm{d}\theta}{\mathrm{d} t} \right) = \frac{L \epsilon \cos \theta}{m a (1-\epsilon^2)} \frac{L}{m r^2} \] Which we can substitute back into our force equation \(\eqref{refc}\), giving us:

\[ - \frac{GM}{\cancel{r^2}} \cancel{\color{Red}{r^2}} \color{Red}{m^2} \stackrel{?}{=} \left( \frac{L^2 \epsilon \cos \theta}{\cancel{m^2} a(1-\epsilon^2) \cancel{r^2}} - \frac{L^2}{\cancel{m^2 r^2} r} \right) \cancel{\color{Red}{{r^2}{m^2}}} \] Cancel out the mass and radius terms and substitute for \( r \) again:

\[ -GMm^2 \stackrel{?}{=} \frac{L^2 (\epsilon \cos \theta)}{a(1-\epsilon^2)} - \frac{L^2(1+\epsilon \cos \theta)}{a(1-\epsilon^2)} \] Which will now simplify further and we can solve for \( L \):

\[ L^2 = GMm^2 \; a(1-\epsilon^2) \] Both sides are constants being equal with no dependence on any term, but defined by the ellipse of semi-major axis \( a \) and eccentricity \( \epsilon \).

This shows that a Newtonian force (proportional to the product of the masses and inversely proportional to the square of their distances) acting mutually between two masses will produce an elliptical orbit.

Next we want to show that a planet in orbit would sweep out equal areas over equal time.

We can find the angular momentum:

\[ \vec{L} = \vec{r} \times \vec{p} \] We only need the perpendicular component, so we can drop the vectors: \[ L = r p_\perp = r (m v_\perp) \] And \( v_\perp = r \omega \):

\[ L = r m (r \omega) \] Therefore, as we also found in the First Law \( \eqref{eqL} \):

\[ L = m r^2 \omega \] and we find that:

\[ r^2 \omega = \frac{L}{m} \label{eqr2} \]

Now we need to show that for the integral:

\[ A = \int \frac{1}{2} r^2 \mathrm{d}\theta \] that \( r^2 \) is a constant. The derivative with respect to \(\theta\) is:

\[ \frac{\mathrm{d}A}{\mathrm{d}\theta} = \frac{1}{2} r^2 \] and with respect to time (\(t\)):

\[ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}\theta} \frac{\mathrm{d}\theta}{\mathrm{d}r} = \frac{1}{2} r^2 \frac{\mathrm{d}\theta}{\mathrm{d}t} \] but \( \frac{\mathrm{d}\theta}{\mathrm{d}t} \) is just \( \omega \): \[ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2} r^2 \omega \] And we can replace \( r^2 \omega \) from \( \eqref{eqr2} \):

\[ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2} \frac{L}{m} \] Which is a constant.

Finally, we want to show that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Since we know that centripetal force is \( m \frac{v^2}{R} \) we can consider what Force is needed:

\[ \frac{GMm}{R^2} = m \frac{v^2}{R} \] For a nearly circular orbit we can substitute \( v^2 = \frac{4 \pi^2 R^2}{T^2} \)

\[ \frac{G M \cancel{m}}{R^2} = \frac{\cancel{m} 4 \pi^2 R \cancel{R}}{\cancel{R} T^2} \] And we find:

\[ \frac{T^2}{R^3} = \frac{4 \pi^2}{G M} \] Therefore this ratio depends only on M and our ratio would be constant for masses in orbit around it.

## Kepler's First Law: Ellipses

\(\require{cancel}\)Proof that a Newtonian force between two masses will produce an elliptical orbit - this is a very textbook approach using polar coordinates, I'm just capturing it here for reference. There are many other approaches, and this has likely been done millions of times now.

Remember that Newton's Law is:

\[ \begin{align} F &= G \frac{Mm}{r^2} \\ &\therefore \nonumber \\ F &= m a \end{align} \]

We define a unit vector in the radial direction \( \hat{r} \) along the angle \( \theta \):

\[ \hat{r} = \hat{x} \cos \theta + \hat{y} \sin \theta \] Therefore \( \hat{r} \) changes as per angle \(\theta\) perpendicular to \( \hat{r} \), giving us the tangential unit vector \( \hat{\theta} \)

\[ \frac{\mathrm{d}\hat{r}}{\mathrm{d}\theta} = \hat{x} ( - \sin \theta ) + \hat{y} \cos \theta = \hat{\theta} \] And the derivative of \( \hat{\theta} \) is another 90° rotation, giving us \( -\hat{r} \):

\[ \frac{\mathrm{d}\hat{\theta}}{\mathrm{d}\theta} = \hat{x} ( - \cos \theta ) + \hat{y} ( - \sin \theta ) = -\hat{r} \] Find Acceleration \( \vec{a} \) in polar coordinates, first we find velocity \( \vec{v} \) as the change in radial vector over time:

\[ \vec{v} = \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \left( r \hat{r} \right) \] Applying the product rule and the chain rule we get our expression for how fast you are moving radially (\(\dot{r}\) in \(\hat{r}\)), and how fast you are moving in the tangential direction (\(r \dot{\theta}\) in \(\hat{\theta}\)):

\[ \vec{v} = \frac{\mathrm{d}r}{\mathrm{d}t} \hat{r} + r \frac{\mathrm{d}\theta}{\mathrm{d}t} \frac{\mathrm{d}\hat{r}}{\mathrm{d}\theta} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} \] Acceleration is the derivative of velocity:

\[ \vec{a} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \left( \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} \right) \] \[ \vec{a} = \color{Red}{\ddot{r} \hat{r}} + \color{ForestGreen}{\dot{r} \dot{\theta} \hat{\theta} + \dot{r} \dot{\theta} \hat{\theta} + r \ddot{\theta} \hat{\theta}} + \color{Red}{r \dot{\theta} (-\hat{r}) \dot{\theta}} \] Grouping terms in \( \hat{r} \; \text{and} \; \hat{\theta} \) we get:

\[ \vec{a} = \color{Red}{\hat{r} \left( \ddot{r} - r \hat{\theta}^2 \right)} + \color{ForestGreen}{ \hat{\theta} \left( 2 \dot{r} \dot{\theta} + r \ddot{\theta} \right) } \label{refa} \] From Newton

\[ \vec{F} = m \vec{a} \] We only need the radial term from \(\eqref{refa}\) since the Force is only in the radial direction, substitute along with Law of Gravity:

\[ - \frac{GM\cancel{m}}{r^2} \cancel{\hat{r}} = \cancel{m} \cancel{\hat{r}} ( \ddot{r} - r \dot{\theta}^2 ) \] The radial unit vector \( \hat{r} \) and mass term \( m \) cancel out, giving us this equation that we must show is equal:

\[ - \frac{GM}{r^2} = \ddot{r} - r \dot{\theta}^2 \label{refb} \] The angular momentum is a constant because there are no torques acting on our system, the only Force is gravity acting radially inwards so we can define \( \dot{\theta} \):

\[ L = m r^2 \dot{\theta} \; \therefore \; \dot{\theta} = \frac{L}{m r^2} \label{eqL} \] And substitute this into \eqref{refb}:

\[ - \frac{GM}{r^2} \stackrel{?}{=} \ddot{r} - \frac{L^2}{m^2 r^3} \label{refc} \] Next we need to show that the second time derivative of this equation of an ellipse works in our force equation:

\[ r = \frac{a(1 - \epsilon^2)}{1 + \epsilon \cos \theta} \] First derivative [Wolfram|Alpha]:

\[ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}r}{\mathrm{d}\theta} \frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{-a(1-\epsilon^2) ( - \epsilon \sin \theta )}{(1+\epsilon \cos \theta)^2} \dot{\theta} \] Substitute for \( \dot{\theta} \) and again for \( r \)

\[ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\cancel{a(1-\epsilon^2)} ( \epsilon \sin \theta )}{\cancel{(1+\epsilon \cos \theta)^2}} \frac{L \cancel{(1 + \epsilon \cos \theta)^2}}{m a(1-\epsilon^2) \cancel{a(1-\epsilon^2)}} \] \[ \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{L \epsilon \sin \theta}{m a (1-\epsilon^2)} \] Second derivative:

\[ \frac{\mathrm{d}^2 r}{\mathrm{d} t^2} = \frac{\mathrm{d}}{\mathrm{d}\theta} \left( \frac{\mathrm{d} r}{\mathrm{d} t} \right) \left( \frac{\mathrm{d}\theta}{\mathrm{d} t} \right) = \frac{L \epsilon \cos \theta}{m a (1-\epsilon^2)} \frac{L}{m r^2} \] Which we can substitute back into our force equation \(\eqref{refc}\), giving us:

\[ - \frac{GM}{\cancel{r^2}} \cancel{\color{Red}{r^2}} \color{Red}{m^2} \stackrel{?}{=} \left( \frac{L^2 \epsilon \cos \theta}{\cancel{m^2} a(1-\epsilon^2) \cancel{r^2}} - \frac{L^2}{\cancel{m^2 r^2} r} \right) \cancel{\color{Red}{{r^2}{m^2}}} \] Cancel out the mass and radius terms and substitute for \( r \) again:

\[ -GMm^2 \stackrel{?}{=} \frac{L^2 (\epsilon \cos \theta)}{a(1-\epsilon^2)} - \frac{L^2(1+\epsilon \cos \theta)}{a(1-\epsilon^2)} \] Which will now simplify further and we can solve for \( L \):

\[ L^2 = GMm^2 \; a(1-\epsilon^2) \] Both sides are constants being equal with no dependence on any term, but defined by the ellipse of semi-major axis \( a \) and eccentricity \( \epsilon \).

This shows that a Newtonian force (proportional to the product of the masses and inversely proportional to the square of their distances) acting mutually between two masses will produce an elliptical orbit.

## Kepler's Second Law: Equal Areas

Next we want to show that a planet in orbit would sweep out equal areas over equal time.

We can find the angular momentum:

\[ \vec{L} = \vec{r} \times \vec{p} \] We only need the perpendicular component, so we can drop the vectors: \[ L = r p_\perp = r (m v_\perp) \] And \( v_\perp = r \omega \):

\[ L = r m (r \omega) \] Therefore, as we also found in the First Law \( \eqref{eqL} \):

\[ L = m r^2 \omega \] and we find that:

\[ r^2 \omega = \frac{L}{m} \label{eqr2} \]

Now we need to show that for the integral:

\[ A = \int \frac{1}{2} r^2 \mathrm{d}\theta \] that \( r^2 \) is a constant. The derivative with respect to \(\theta\) is:

\[ \frac{\mathrm{d}A}{\mathrm{d}\theta} = \frac{1}{2} r^2 \] and with respect to time (\(t\)):

\[ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}\theta} \frac{\mathrm{d}\theta}{\mathrm{d}r} = \frac{1}{2} r^2 \frac{\mathrm{d}\theta}{\mathrm{d}t} \] but \( \frac{\mathrm{d}\theta}{\mathrm{d}t} \) is just \( \omega \): \[ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2} r^2 \omega \] And we can replace \( r^2 \omega \) from \( \eqref{eqr2} \):

\[ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2} \frac{L}{m} \] Which is a constant.

## Kepler's Third Law: Harmonies

Finally, we want to show that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Since we know that centripetal force is \( m \frac{v^2}{R} \) we can consider what Force is needed:

\[ \frac{GMm}{R^2} = m \frac{v^2}{R} \] For a nearly circular orbit we can substitute \( v^2 = \frac{4 \pi^2 R^2}{T^2} \)

\[ \frac{G M \cancel{m}}{R^2} = \frac{\cancel{m} 4 \pi^2 R \cancel{R}}{\cancel{R} T^2} \] And we find:

\[ \frac{T^2}{R^3} = \frac{4 \pi^2}{G M} \] Therefore this ratio depends only on M and our ratio would be constant for masses in orbit around it.

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