Lecture 11 - Gravity and Kepler’s Laws

Lecture 11 - Gravity and Kepler’s Laws#

Important

The content on this page is not examinable.

Gravity#

In between determining his three laws of motions plus (probably) inventing calculus, Newton also found time to look at the solar system and ponder on why the planets stay in their orbit given the lack of any visible centipetal force. He postulated that

“Every particle of matter in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.”

which can be written mathematically as

\[\begin{align*} F_g &= \frac{Gm_1m_2}{r^2} \end{align*}\]

where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the two bodies attracting one another and \(r\) is the distance between them.

It turns out that having two or more bodies is unnecessary, although Newton would not have been able to observe gravity from a single mass. All mass creates a gravitational field and we can use a test mass \(\delta m\) to measure the gravitational force acting on it from some larger mass \(M\). Once again we make use of Newton’s Second Law to state

\[\begin{align*} \delta m\, \textbf{a} &= \textbf{F}_g\\ \therefore \textbf{a} &= -\frac{GM}{r^2}\,\hat{\textbf{r}} \end{align*}\]

where the minus sign is necessary to indicate that gravity is acting in the opposite direction to the radius vector that points from the large mass to the test mass.

Example - gravitational force between particle and a bar#

We will extend this theory further to consider the gravitational force between a particle and a more extended object, in this case a thin bar of uniform density as shown in fig:C11_bar.

A schematic of a uniform bar of mass $M$ and length $L$ next to a small mass $m$ some distance $h$ away. The bar is split into small mass elements and the force of gravity is calculated between these elements and the small mass, over the entire length of the bar.

A schematic of a uniform bar of mass \(M\) and length \(L\) next to a small mass \(m\) some distance \(h\) away. The bar is split into small mass elements and the force of gravity is calculated between these elements and the small mass, over the entire length of the bar.#

As the gravitational force depends on distance and different parts of the bar are different distances away from the particle we need to split our bar up into small slices of thickness \(\mathrm{d}x\) and mass \(\mathrm{d}m\), and consider the gravitational force on each of these masses. The force acting on a thin slice \(\mathrm{d}F\) is

\[\begin{align*} \mathrm{d}F &= -\frac{Gm\,\mathrm{d}m}{x^2} \end{align*}\]

and we find the total force by integrating over the entire bar, taking care to use the correct limits. In order to integrate over the length of the bar we need to change our integral variable which requires us to make use of the fact that the bar has a uniform length density \(\lambda\) and so \(\mathrm{d}m = \lambda \,\mathrm{d}x\). Therefore

\[\begin{align*} F_g &= -Gm\lambda\int_h^{h+L} \frac{\mathrm{d}x}{x^2}\\ &= -\frac{GmM}{L}\left[-\frac{1}{x}\right]_h^{h+L}\\ &= \frac{GmM}{h(h+L)} \end{align*}\]

You can perform some sense checks here. If we squash the bar sufficiently that the length tends towards zero our equation should reduce to the simple case of gravitational force between two point masses. A similar result should happen if you make the distance between the bar and mass much much larger than the size of the bar.

Kepler’s Laws#

Before we jump into the three Laws it’s worth having a reminder about 2D polar coordinate systems and ellipses, as they are important pieces of information needed to derive and understand some of the Laws.

2D Polar Coordinates#

When dealing with circular and elliptical systems it is often more convenient to work with 2D polar coordinates, i.e. \((r,\theta)\) rather than \((x,y)\).

If we consider a planetary system orbiting a single star then we first define the origin as the centre of the star such that the \(r\) coordinate becomes the distance between the star and planet. Our choice of \(\theta=0\) is entirely arbitrary as we will be looking either small angular changes or complete orbits.

Having defined our coordinate system we recall that the velocity of the planet over a small section of the orbit can be resolved into radial and tangential components, i.e.

\[\begin{align*} v_r &= \dfrac{\mathrm{d}r}{\mathrm{d}t}\\ v_t &= r\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\\ &= r\omega \end{align*}\]

Ellipses#

The general equation for an ellipse (in Cartesian form) with the longest length aligned along the \(x\)-axis is given by:

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\]

which takes the general form in 2D polar coordinates:

\[\dfrac{a(1-e^2)}{r}=1+e\cos\theta\]

where \(a\) and \(b\) are the semi-major and semi-minor axes respectively and the eccentricity of the ellipse \(e\) describes how the shape of the orbit deviates from a circular path (\(e=0\) describes a perfect circle, and \(0\leq e \leq 1)\).

First Law#

There are different methods of deriving Kepler’s First Law but here I will present a proof based on conservation of energy.

A body of mass $m$ orbiting on an elliptical path around a larger mass $M$. The distance between the two masses changes throughout the orbit. The velocity of the small mass is resolved into two components, one pointing radially and the other perpendicular to this.

A body of mass \(m\) orbiting on an elliptical path around a larger mass \(M\). The distance between the two masses changes throughout the orbit. The velocity of the small mass is resolved into two components, one pointing radially and the other perpendicular to this.#

For an orbiting body of mass \(m\) around a larger body of mass \(M\) the total energy \(E\) of the system is given by

()#\[\begin{split}E &= \dfrac{mv^2}{2}-\dfrac{GMm}{r}\nonumber\\ &=\dfrac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)-\dfrac{GMm}{r}\end{split}\]

where I have used the dot notation for first derivative with respect to time, and split the velocity into radial and tangential components.

The angular momentum \(L\) of the system is given by

()#\[\begin{split}L &= mv_t r \nonumber \\ &= mr^2\omega \nonumber\\&= mr^2\dot{\theta}\end{split}\]

Next we make a subsitution that isn’t obvious at first glance but makes our subsequent steps easier. We define \(\rho = \dfrac{1}{r}\) such that equation () can be rewritten as

()#\[\begin{split} \dot{\theta} &= \dfrac{L\rho^2}{m}\nonumber \\ \therefore \theta &= \int\dfrac{L\rho^2}{m}\mathrm{d}t \nonumber\\ &= \int \dfrac{L\rho^2}{m}\dfrac{\mathrm{d}t}{\mathrm{d}\rho}\mathrm{d}\rho\end{split}\]

However we can manipulate our definition for \(\rho\) to find \(\dot{r}=-\dfrac{1}{\rho^2}\dfrac{\mathrm{d}\rho}{\mathrm{d}t}\) such that equation () becomes

()#\[ \theta = -\int\dfrac{L}{m\dot{r}}\mathrm{d}\rho \]

We now turn our attention back to equation () which can be rearranged into

()#\[\begin{split}\dot{r}^2 &= \dfrac{2E}{m}+2GM\rho-\dfrac{L^2}{m^2}\rho^2 \nonumber\\ \therefore \dot{r} &= \dfrac{L}{m}\left[\dfrac{e^2}{r_0^2}-\left(\rho-\dfrac{1}{r_0}\right)^2\right]^{\frac{1}{2}}\end{split}\]

This last step requires a bit of algebraic manipulation as well as the (again not so obvious) substitutions

\[\begin{align*} r_0 &= \dfrac{L^2}{GMm^2}\\ e^2 &= 1+\dfrac{2Er_0}{GMm} \end{align*}\]

Finally we can substitute equation () into equation () to find

\[\begin{align*} \theta &= -\int\dfrac{\mathrm{d}\rho}{\sqrt{\left(\dfrac{e}{r_0}\right)^2 - \left(\rho-\dfrac{1}{r_0} \right)^2 }}\\ &= \cos^{-1}\left(\dfrac{\rho-\frac{1}{r_0}}{\frac{e}{r_0}}\right) \end{align*}\]

which can be rearranged into the general form of an eclipse, i.e.

()#\[r = \dfrac{r_0}{1+e\cos\theta}\]

Second Law#

An elliptical path of an object orbiting a large mass. The arc length between two points in the orbit reduces to the straight line distance between the points when the angle between the points $\Delta \theta$ is sufficiently small.

An elliptical path of an object orbiting a large mass. The arc length between two points in the orbit reduces to the straight line distance between the points when the angle between the points \(\Delta \theta\) is sufficiently small.#

Consider fig:C11_Kepler2 above in which an orbiting body moves from position \(A\) to \(B\) in a small time interval \(\Delta t\).

The area swept out in this time interval is approximately equal to the area of the triangle \(ABS\) which, for small arc lengths \(\Delta s\), reduces to the area of the triangle \(BCS\).

The base of this triangle is \(r \Delta\theta\) and height is \(r\) which means that the small area swept out, \(\Delta A\) is

\[\Delta A = \dfrac{1}{2}r^2 \Delta\theta\]

and so the rate of area being swept out is

()#\[\begin{split}\dfrac{\Delta A}{\Delta t} &= \dfrac{1}{2}r^2\dfrac{\Delta\theta}{\Delta t}\nonumber\\ &= \dfrac{1}{2}r^2\omega\nonumber\\ &= \dfrac{L}{2m}\end{split}\]

where the angular momentum \(L=mr^2\omega\) is constant as there is no net torque acting on the system.

Third Law#

As the area of an ellipse is \(\pi ab\) and the rate of sweeping out area is \(\dfrac{L}{2m}\) from equation () we can show that the time taken for a complete orbit \(T\) is the area divided by the rate of area swept, i.e.

()#\[\begin{split}T&=\dfrac{\pi ab}{\frac{L}{2m}}\nonumber \\ &= \dfrac{2m\pi ab}{L}\end{split}\]

We can also rearrange equation () into the form \(\dfrac{L^2}{GMm^2}=a(1-e^2)\) and from geometry we know that

\[\begin{align*} b^2 &= a^2 - e^2a^2\\ &= a^2(1-e^2) \end{align*}\]

So by squaring equation () and combining these equations we find

\[\begin{align*} T^2 &= \dfrac{\left(2m\pi ab \right)^2}{L^2}\\ &= \dfrac{\left(2m\pi ab\right)^2}{GMm^2a(1-e^2)}\\ &= \dfrac{\left(2m\pi ab\right)^2}{GMm^2a(\frac{b^2}{a^2})}\\ &= \dfrac{4\pi^2a^3}{GM} \end{align*}\]

In other words, the orbital period squared is proportional to the semi-major axis cubed, with a constant of proportionality that only depends on the mass of the star and not on the eccentricity of the orbit.

Lecture Questions#

Question 1

Calculate the acceleration of free fall on the Moon and compare it with the acceleration of free fall on the Earth.

A common way to interpret “compare” in physics is taking a ratio of one value to another.

We take \(M_{\text{moon}}=7.35\times10^22\text{ kg}\) and \(R_{\text{moon}}=1.74\times10^{6}\text{ m}\). Therefore

\[\begin{align*} g_{\text{moon}}&= \frac{GM_{\text{moon}}}{R_{\text{moon}}^2}\\ &= 1.62\text{ m s}^{-2} \end{align*}\]

and so \(\dfrac{g_{\text{moon}}}{g_{\text{Earth}}} = \dfrac{1.62}{9.81} \approx{1}{6}\)

Question 2

The masses of the Sun, Earth and Moon are \(1.99\times10^{30}\text{ kg}\), \(5.98\times10^{24}\text{ kg}\) and \(7.35\times10^{22}\text{ kg}\) respectively. Assume that the location of the Moon is such that the angle subtended by the lines from the Moon to the Sun and from the Moon to the Earth is \(45.0^{\circ}\). What is the net force on the Moon due to the gravitational forces of the Sun and Earth?

Take the Moon-Sun distance to be \(1.50\times10^{11}\text{ m}\) and the Moon-Earth distance to be \(3.84\times10^{8}\text{ m}\).

Draw a diagram and carefully choose where you want your axes to lie in order to simplify the system.

We align our axes such that the Moon-Sun line lies along the \(x\) axis, and that the Moon-Earth-Sun all lie in the \((x,y)\) plane. The force due to the Sun on the Moon is

\[\begin{align*} \textbf{F}_{SM} &= \frac{GM_SM_M}{R_{SM}^2}\hat{\textbf{i}} \end{align*}\]

and the force between the Earth and Moon is

\[\begin{align*} \textbf{F}_{EM} &= \frac{GM_EM_M}{\sqrt{2}R_{EM}^2}\hat{\textbf{i}} + \frac{GM_EM_M}{\sqrt{2}R_{EM}^2}\hat{\textbf{j}} \end{align*}\]

The total force is the sum of these two forces

\[\begin{align*} \textbf{F}_T &= \textbf{F}_{SM} + \textbf{F}_{EM}\\ &= GM_M\left[\left(\frac{M_S}{R_{SM}^2}+\frac{M_E}{\sqrt{2}R_{EM}^2}\right) \hat{\textbf{i}}+ \frac{M_E}{\sqrt{2}R_{EM}^2}\hat{\textbf{j}}\right]\\ &= \left(5.74\times10^{20}\hat{\textbf{i}}+1.41\times10^{20}\hat{\textbf{j}}\right) \text{ N} \end{align*}\]

The magnitude of the force is \(|F_T| = \sqrt{(5.74\times10^{20})^2+(1.41\times10^{20})^2} = 5.91\times10^{20}\text{ N}\) and points at an angle \(\theta = \tan^{-1}\left(\dfrac{1.41\times10^{20}}{5.74\times10^{20}}\right) = 13.8^{\circ}\) with respect to the \(x\) axis (i.e. the Sun-Moon line).

Question 3

Both Venus and Earth have approximately circular orbits around the Sun. The period of the orbital motion of Venus is \(0.165\) Earth year. What is the orbital radius of Venus expressed in astronomical units?

This is a fairly straightforward case of using Kepler’s Third Law (twice).

From Kepler’s Third law, where the semi-major axis of a circular orbit is just the radius of the orbit, and so

\[\begin{align*} T_E^2 = \frac{4\pi^2}{GM_S}r_E^3 \qquad &\text{and} \qquad T_V^2 = \frac{4\pi^2}{GM_S}r_V^3\\ \therefore r_V &= \left(\frac{T_V}{T_E}\right)^{\frac{2}{3}}r_E\\ &= 0.72r_E \end{align*}\]

Question 4

A communications satellite is in a circular orbit around the Earth, in the equatorial plane. The period of the orbit is such that the satellite `hovers’ in a fixed position relative to the rotating Earth. What is the radius of such a “geosynchronous” or “geostationary” orbit?

Geostationary means the satellite must have the same period as the surface of the Earth it is fixed above, which you can use in Kepler’s Third Law.

A geosynchronous satellite has an orbital period of 1 day, so we use Kepler’s Third Law as

\[\begin{align*} T^2 &= \frac{4\pi^2}{GM_E}r^3\\ \therefore r &= \left(\frac{GM_ET^2}{4\pi^2}\right)^{\frac{1}{3}}\\ &= 4.23\times10^7\text{ m} \approx 6.6R_{\text{Earth}} \end{align*}\]
Contents