Bound Orbits

Guide

Bound Orbits

Having finished Kepler’s laws, you might ask why we need to study the two-body problem any further at all. Well, the idea is similar to why, after learning Coulomb’s law, we still devote a fair amount of time to learning further electrostatics.

The answer is simply practical application. It is much easier to derive some results regarding the system and use them to make our lives easier when solving problems. For instance, you could, theoretically, now that you know $r(\theta)$ , figure out $r(t)$ without any of the tools we discuss below, but it would be a much more daunting task.

Anyway, we are still only considering the problem of a two-body system, but we’ll primarily work with bound orbits and see what we can do with them. We’ll also consider secondary effects of gravity, that we’ve largely ignored in our body is a point mass model.

As we proved in chapter 3.1, all bound orbits are elliptical. Therefore, a fair amount of the following discussion is going to rely on some knowledge of ellipses. You should have ideally finished reading the Wikipedia page or used an equivalent resource.

Consider an ellipse with semi-major axis $a$ , semi-minor axis $b$ , and eccentricity $e$ . As you may know, the semi-major axis is half the length of the longest radius of the ellipse, while the semi-minor axis is half the length of the shortest radius. Here, radius refers to the distance from the center (not the foci) to the boundary.

I’ll just go through some useful results for ellipses that we will need later. I’ll urge you to go through this before proceeding, as it’s vital to the succeeding discussions.

The semi-minor and semi-major axes are related by

b^2 = a^2 (1 - e^2)

The eccentricity of an ellipse is always less than 1, being 0 for a circle. Now, we can use the equation of the ellipse with the origin at one of the foci:

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

to find the smallest and largest $r$ . This gives the length of the periapsis, which is the point closest to the foci, and apoapsis, the point farthest from the foci:

\begin{align*} r_p &= a (1 - e)\\ r_a &= a (1 + e) \end{align*}

From this, we get the relations

a = \frac{r_p + r_a}{2} \,, \qquad \qquad e = \frac{r_a - r_p}{r_a + r_p}

The semi-latus rectum $p$ of an ellipse is defined as the distance from the focus to the ellipse along a line perpendicular to the major axis. It is given by

p = \frac{b^2}{a} = a (1 - e^2)

From this, eqns. $(3.1.8)$ , $(3.1.10)$ ; we get that the total energy and angular momentum of the orbit in terms of $a$ and $e$ are:

\begin{align*} \tag{3.2.1} \varepsilon &= -\frac{\mu}{2a}\\ \tag{3.2.2} h &= \sqrt{\mu a (1 - e^2)} \end{align*}

From here on, for all subsections except the last one, we consider the case when $m_2 \gg m_1$ , which is the case for, for example, our solar system.

Orbital Velocity

Because we know the value of energy in terms of the parameters now, and the velocity and radius are related by the energy, we can use eqns. $(3.1.7)$ and $(3.2.1)$ to get the vis-viva equation:

\frac{1}{2} v^2 - \frac{\mu}{r} = \varepsilon = -\frac{\mu}{2a}

\tag{3.2.3} \implies \boxed{v^2 = \mu \left( \frac{2}{r} - \frac{1}{a} \right)}

This gives us the velocity of the smaller body at a distance $r$ from the larger body. From here you can see that the velocity is maximum at periapsis and minimum at apoapsis, because it roughly depends on $1/\sqrt{r}$ . The velocity at periapsis is then (using the expression for $r_p$ that we found earlier):

\tag{3.2.4} v_p = \sqrt{\mu \left( \frac{2}{r_p} - \frac{1}{a} \right)} = \sqrt{\frac{\mu}{a} \left( \frac{1+e}{1-e} \right)}

And the velocity at apoapsis is:

\tag{3.2.5} v_a = \sqrt{\mu \left( \frac{2}{r_a} - \frac{1}{a} \right)} = \sqrt{\frac{\mu}{a} \left( \frac{1-e}{1+e} \right)}

You can see that they’re sort of very symmmetric, if we multiply or divide them, one of the term cancels out! This gives us some very nice relations:

v_p v_a = \frac{\mu}{a}\,, \qquad \qquad \frac{v_p}{v_a} = \frac{1+e}{1-e}

As something you should always do when you have a formula with a parameter you can vary, we can ask what the limiting cases of this are. Since $e$ is bound between $0$ and $1$ , let’s see what we can do. $e \to 1$ is not really a nice case, for one the orbit becomes unbound, and everything diverges. On the other hand, we can talk about $e = 0$ precisely!

In the case of a circular orbit, we have $e = 0$ . The body therefore moves with a constant speed $v_c$ , called the circular velocity, given by

\tag{3.2.6} v_c = \sqrt{\frac{\mu}{a}}

For the body to escape the gravitational influence of the larger body and become unbound, the total energy of the system must become more than or equal to 0. We can see this through the conservation of energy. Let the initial velocity be $v$ . Finally, we need the velocity to be just non-zero to find the minimum velocity. Note that far away, the potential energy goes to $0$ . Pairing this up with the fact that $K = mv^2/2$ is going to be $0$ for the minimum velocity, we get that the total energy must be $0$ .

So the escape veloctiy is such that the total energy is $0$ . Using the expression for total energy from Kepler’s Laws, and since initially the seperation between the bodies is $r$ ;

0 = \frac{1}{2}mv_e^2 - \mu\frac{m}{r^2} \implies v_e = \sqrt{\frac{2\mu}{r}}

Thus for the smaller body to escape, the minimum velocity needed is:

\tag{3.2.7} v_e = \sqrt{\frac{2\mu}{r}} = \sqrt{2} v_c

From eqn. $(3.1.13)$ , the time period of the orbit is:

P = 2 \pi \sqrt{\frac{a^3}{\mu}}

Kepler’s Equation

We know what the equation for $r(\theta)$ is, but what about the position as a function of time? It turns out, that’s a fairly more difficult task, and the neatest way to do this is using anomalies, parameters that describe the motion of our lighter body. As I said when discussing the first law of Kepler, we call the angle between the eccentricity vector and the position vector the true anomaly, $\theta$ .

Note that because the position vector from the center of mass (which is effectively the larger body here) $r$ is:

r = \frac{h^2/\mu}{1 + e\cos\theta},

the periapsis occurs at $\theta = 0$ . So the true anomaly, stated in another manner, measures the angle from the periapsis. If we can somehow find the true anomaly as a function of time, our job is essentially done because we already know $r(\theta)$ .

The idea that we use here is to use a fictious circular body (this is closely related to the fact that elliptical motion is a superposition of two independent circular motions). The mean motion $n$ is defined as

\tag{3.2.8} n \overset{\text{def}}{\equiv} \frac{2 \pi}{P}

Let the instant at which the body was at periapsis be $\tau$ . We define the mean anomaly $M$ as the angular distance from the periapsis which a fictitious body would have if it moved in a circle of radius $a$ (the semi-major axis length). At any instant in time $t$ , it is clearly given by:

\tag{3.2.9} M = n (t - \tau)

Mean, Eccentric and True Anomaly

As it turns out, our equations become much neater if we ignore the true anamoly, and insteard work with the eccentric anomaly. We defined the eccentric anamoly $E$ as the eccentric angle of the smaller body in its elliptical orbit (see the figure). In cartesian coordinates with the center of the ellipse being at origin, the equation of the ellipse becomes

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

The eccentric anamoly $E$ in terms of these coordinates is given by

\cos E = \frac{x}{a}\,, \qquad \qquad \sin E = \frac{y}{b}

As a consequence, since $r = x^2 + y^2$ , and $b^2 = a^2(1-e^2)$ we get that the distance of the smaller body from the larger body (foci of the ellipse) is given by

\tag{3.2.10} r = a (1 - \cos E)

Now since $\mathbf{r}(\theta) = r(\cos\theta \hat{\mathbf{{i}}} + \sin\theta \hat{\mathbf{j}})$ , comparing it with expression for $x$ and $y$ that we have, we get that the eccentric anamoly $E$ and true anamoly are related by

\cos \theta = \frac{a}{r} (\cos E - e) \qquad \qquad \sin \theta = \frac{b}{r} \sin E

so that

\tag{3.2.11} \tan \frac{\theta}{2} = \sqrt{\frac{1+e}{1-e}} \tan \frac{E}{2}.

Eccentric Anomaly, The Point P is at (x,y). Reproduced from Katturnen et al., Fundamental Astronomy

Using Kepler’s second law, the total area covered by the smaller body in the shaded region (see the previous figure) is:

A = \pi ab \frac{t - \tau}{P} = \frac{1}{2}ab M

when it is present at the eccentric angle $E$ . Also, since an ellipse is a rescaling of a circle, the area is $b/a$ times the area of $FP'X$ . Thus one obtains from someone geometry that the area of $FPX$ is:

A = \frac{1}{2}ab(E - e\sin E)

We then equate these to find the relation between the mean anamoly $M$ and the eccentric anamoly $E$ as

\tag{3.2.12} M = E - e \sin E

This is called Kepler’s equation. It is a transcendental equation, and we can solve it using numerical methods.

Using an infinite series expansion, we can solve it exactly:

E = M + e \sin M + \frac{e^2}{2} \sin 2M + \frac{e^3}{6} \sin 3M + \ldots

Using this and the relations we found, the true anamoly (if you so want) and distance from focus too can be obtained in terms of infinite series expansion, given by

\begin{align*} \theta &= M + {2e \sin M} + \frac{5e^2}{4} \sin 2M + \frac{e^3}{12} (12 \sin 3M - 3 \sin M) + \cdots\\ \frac{r}{a} &= 1 - e\cos M + \frac{e^2}{2} (1 - \cos 2M) + \frac{3e^3}{8} (\cos M - \cos 3M) + \cdots \end{align*}

And thus we have found $r(t)$ .

Radial Elliptic trajectory

A radial elliptic trajectory is a degenerate case of an elliptical trajectory, where the eccentricity $e$ is equal to 1 and the angular momentum of the system is zero. The trajectory is a straight line, and the two bodies collide with each other at some point in the trajectory. It is still classified as an elliptic trajectory since the total energy of the system is negative.

Binary Systems

Consider a binary system having two bodies of masses $m_1$ and $m_2$ moving in bound orbits. Both the bodies move in ellipses of equal eccentricities but oriented opposite to each other. Properties of orbits of individual bodies will be labelled with a subscript (e.g. $x_1$ and $x_2$ ), and properties of the trajectory of the separation vector, or the orbit of one of the bodies relative to the other will be written without a subscript (e.g. $x$ ).

The semi-major axes $a$ are related by the center of mass relation,

\tag{3.2.13} m_1 a_1 = m_2 a_2 = ma \implies a = a_1 + a_2

Similarily, in the center of mass frame, the speeds $v$ are related by

\tag{3.2.14} m_1 v_1 = m_2 v_2 = m v \implies v = v_1 + v_2

When an elliptical orbit is projected onto a plane, it produces another ellipse. However, the foci of the original ellipse does not project onto the foci of the observed ellipse.

Consider a binary system at a distance $r$ from us, inclined at an angle $i$ , which is the angle between the plane of the orbit and plane of the sky. $\alpha$ and $\tilde{\alpha}$ be the true and observed angular semi-major axes. They are related by the equations

m_1 \alpha_1 = m_2 \alpha_2 = m \alpha \implies \alpha = \alpha_1 + \alpha_2

m_1 \tilde{\alpha}_1 = m_2 \tilde{\alpha}_2 = m \tilde{\alpha} \implies \tilde{\alpha} = \tilde{\alpha_1} + \tilde{\alpha_2}

\tag{3.2.15} \tilde{\alpha} = \alpha \cos i

The true semi-major axis $a$ is given by

a = r \alpha = r \tilde{\alpha} \cos i

Hence Kepler’s third law can be written as

\tag{3.2.16} m_1 + m_2 = \frac{4\pi^2}{G} \frac{a^3}{P^2} = \frac{4\pi^2}{G} \left( \frac{r}{\cos i} \right)^3 \frac{\tilde{\alpha}^3}{P^2}

Usually, for a binary system, only the radial velocity of the bodies can be measured. If the trajectory of the binary system is circular, the radial velocity is given by $v = \frac{2 \pi a}{T}$ . The maximum radial velocity observed is $v_r = v \sin i$

a = a_1 + a_2 = \frac{P}{2 \pi} (v_1 + v_2) = \frac{P}{2 \pi \sin i} (v_{1r} + v_{2r})

\tag{3.2.17} \implies m_1 + m_2 = \frac{P}{2 \pi G} \frac{(v_{1r} + v_{2r})^3}{\sin^3 i}

If only one velocity ( $v_{1r}$ ) is observable, $v_{2r}$ can be obtained using $m_1 v_{1r} = m_2 v_{2r}$ . Putting this into the equation above, we get

\tag{3.2.18} \boxed{ \frac{m_2^3}{(m_1 + m_2)^2} \sin^3 i = \frac{P}{2 \pi G} v_{1r}^3 }

This is the mass function of the binary system. The mass function can be used to determine the mass of the secondary body if the mass of the primary body is known. The mass function can also be used to determine the inclination of the orbit if the masses of both bodies are known. It also sets a lower limit on the mass of $m_2$ :

\tag{3.2.19} m_2 \geq \frac{P}{2 \pi G} v_{1r}^3

Note that for such a binary system, the angular velocities of the two masses about the centre of mass are same. To show this, note that the centripetal force is gravity, so we get that,

M\omega^2 r_1 = \frac{GmM}{R^2}

where $r_1$ is the distance of $M$ from the centre. Since the centre is the centre of mass, using $m_1r_1 = m_2r_2$ , we get that $m/r_1 = (M + m)/R$ which gives us the angular velocity for the body, as:

\tag{3.2.20} \omega = \sqrt{ \frac{G(M + m)}{R^3} }

A similar derivation for the other primary will give you the same angular velocity.

Tidal Forces

In all of our prior discussions we worked with the ideal model of a body being a point object. This is all fine, because we were analysing the motion of such bodies, and it is very much the case that they’re much smaller than their trajectories.

However, there are secondary effects that arise when we consider the body to be an actual, extended body. For example, the tides on earth. In general, Tidal force is the difference in gravitational attraction between different points on a body, causing the body to be pulled unevenly and as a result are being stretched towards the source of attraction.

Phenomenon caused by tidal forces include:

Ocean tides and solid-earth tides
Tidal locking
Breaking apart of celestial bodies and formation of ring systems within the Roche limit
Spaghettification of objects near black holes
Tidal heating of moons and planets

Tidal interaction between a larger galaxy and its smaller companion. (Source: Wikipedia)

Tidal acceleration does not require rotation or orbiting bodies; for example, the body may be freefalling in a straight line under the influence of a gravitational field while still being influenced by (changing) tidal acceleration. The only condition is of course, that it is in a gravitational field. Similar effects arise from other forces, but we don’t really care about them.

The Tidal Effect. Dashed lines show the variation of field.

Let’s try to calculate this apparent effect. Consider a spherical body of radius $R$ , kept at a distance $r$ from a body of mass $M$ . Define

$\mathbf{s}$ as the position vector of mass $m$ on the surface of the sphere relative to the massive body $M$
$\mathbf{R}$ as the position vector of mass $m$ relative to the center of mass of the sphere
$\mathbf{r}$ as the position vector of the sphere relative to the massive body $M$

Since the sphere accelerates towards the massive body $M$ with $\mathbf{a} = -\frac{GM}{r^2} \hat{\mathbf{r}}$ , the force on mass $m$ (as seen by an observer on the body) is given by:

\mathbf{F}_\text{net} = -\frac{GMm}{s^2} \hat{\mathbf{s}} + \frac{GMm}{r^2} \hat{\mathbf{r}} + \sum \mathbf{F}_\text{other}

The last term is of little relevance, it persists in the absence of $M$ while the first two terms arise from the presence of M — the combination of these two additional terms is known as the tidal force $\mathbf{F}_\text{tidal}$ , and the acceleration caused is called the tidal acceleration $\mathbf{a}_\text{tidal}$ .

\tag{3.2.21} \mathbf{a}_\text{tidal} = -GM \left( \frac{\hat{\mathbf{s}}}{s^2} - \frac{\hat{\mathbf{r}}}{r^2} \right)

Since $\mathbf{s} = \mathbf{r} + \mathbf{R}$ ,

\mathbf{a}_\text{tidal} = -GM \left( \frac{\mathbf{r} + \mathbf{R}}{|\mathbf{r} + \mathbf{R}|^3} - \frac{\mathbf{r}}{r^3} \right)

Usually $r \gg R$ , so we can take few approximations (essentially the ones we take to find the field of a dipole), which finally give

\tag{3.2.22} \boxed{ \mathbf{a}_\text{tidal} \approx -\frac{GM}{r^3} \left[ \mathbf{R} - 3\, ( \hat{\mathbf{r}} \cdot \mathbf{R} )\, \hat{\mathbf{r}} \right] }

Consider a cartesian coordinate system with its origin at the center of the sphere, the $x$ -axis pointing towards the massive body, and $y$ -axis oriented such that $\mathbf{R}$ lies in the $xy$ -plane. If the angle $\theta$ is the angle between the $x$ -axis and the vector $\mathbf{R}$ , the tidal acceleration experienced by $m$ is

\tag{3.2.23} \boxed{\mathbf{a}_\text{tidal} = \frac{GMR}{r^3} (2\cos \theta \, \hat{\mathbf{i}} - \sin \theta \, \hat{\mathbf{j}}) }

We see that when $\mathbf{R}$ and $\mathbf{r}$ are parallel, i.e. $m$ lies on the line joining $M$ and the center of the sphere, the tidal acceleration is

\mathbf{a}_\text{tidal} = -2\frac{GM \mathbf{R}}{r^3}

When $\mathbf{R}$ and $\mathbf{r}$ are perpendicular, i.e. $m$ lies on the line perpendicular to the line joining $M$ and the center of the sphere, the tidal acceleration is

\mathbf{a}_\text{tidal} = -\frac{GM \mathbf{R}}{r^3}

The tidal acceleration is maximum when $\mathbf{R}$ and $\mathbf{r}$ are parallel, and minimum when they are perpendicular.

The tidal effect (shown in red) due to the moon and sun on the Earth, and the gravitational effect (shown in blue) due to the moon. S points towards the position of the moon, and O is the centre of the Earth. (Source: Wikipidea)

Roche Limit

You might question that if the oceans did experience tidal forces, why don’t they simply break away? The answer is that Earth’s own gravity stops that from happening. It is held together by its own gravity. Of course, if the tidal acceleration was greater than the gravitational force the body exerts upon itself, the body would disintegrate. In fact, there is a specific distance at which this happens, which depends on the various parameters of the primary body which causes the tidal acceleration, and the secondary body held together by its own gravity (think the Sun and Earth for examples).

The Roche limit or Roche radius is the distance from a massive primary body within which a second body held together only by its own gravity, will disintegrate due to tidal forces. The Roche limit typically applies to a satellite’s disintegrating due to tidal forces induced by its primary, the body around which it orbits.

Let $R_M$ , $\rho_M$ and $M_M$ be the radius, density and mass of the primary (larger) body and $R_m$ , $\rho_m$ and $M_m$ be the radius, density and mass of the satellite. The satellite is at a distance $d$ from the primary. For the satellite to be held together by its own gravity, the tidal force must be less than the gravitational force acting on the satellite. We get the Roche limit when these two interactions are equal.

2 \frac{GM_M R_m m'}{d^3} = \frac{GM_m m'}{R_m^2}

where $m'$ is a mass closest to the primary body, on the satellite’s surface. This gives

\tag{3.2.24} \boxed{ d = R_m \left( \frac{2M_M}{M_m} \right)^{1/3} = R_M \left( \frac{2 \rho_M}{\rho_m} \right)^{1/3}}

If we decrease the distance further, the tidal forces overpower the forces that hold together the body. We in particular care about the force at $m'$ being overpowered because it is the point closest to the primary.

The above expression is for a rigid satellite. For a fluid satellite,

d \approx 2.44\, R_M \left( \frac{\rho_M}{\rho_m} \right)^{1/3}

Consider the satellite to be tidally locked to the primary body, such that the same side of the satellite always faces the primary. Find the Roche limit for such a satellite.

Here along with the tidal forces and the planet’s gravitational force, we also have the centrifugal force acting on the satellite. For the mass $m'$ on the surface of the satellite to remain in equilibrium, we must have

2 \frac{GM_M R_m m'}{d^3} + m' \omega^2 R_m = \frac{GM_m m'}{R_m^2}

where $\omega$ is the angular velocity of the satellite. The angular velocity is given by

\omega = \sqrt{ \frac{GM_M}{d^3} }

Substituting this into the equation, we get

d = R_m \left( \frac{3 M_M}{M_m} \right)^{1/3} = R_M \left( \frac{3 \rho_M}{\rho_m} \right)^{1/3}

Problems

A Sun-orbiting periodic comet is the farthest at

31.5 \mathrm{\,AU}

and the closest at

0.5 \mathrm{\,AU}

. What is the orbital period of this comet? What is the area (in

\mathrm{AU^2/yr}

) swept by the line joining the comet and the Sun?

The semi-major axis of the orbit is

a = \frac{r_p + r_a}{2} = 16 \mathrm{\,AU}

Using Kepler’s third law, we have

P^2 = a^3 = 16^3 = 4096 \implies \boxed{P = 64 \mathrm{\,years}}

The eccentricity and semi-latus rectum can be found using

e = \frac{r_a - r_p}{r_a + r_p} = \frac{31.5 - 0.5}{31.5 + 0.5} = \frac{31}{32}

p = a(1 - e^2) = 16 \left( 1 - \frac{31^2}{32^2} \right) = 16 \left( \frac{63}{1024} \right) = 0.98 \mathrm{\,AU}

Therefore the rate of area swept is

\frac{dA}{dt} = \frac{h}{2} = \frac{\sqrt{\mu p}}{2} = \frac{\sqrt{4 \pi^2 p}}{2} = \pi \sqrt{p} = \pi \cdot \sqrt{0.98} = \boxed{3.1 \mathrm{\, AU^2/yr}}

Estimate the radius of a planet that a man can escape its gravitation by jumping vertically. Assume density of the planet and the Earth are the same.

A man can jump to a height $h \approx 50 \mathrm{\,cm}$ on Earth. This gives the man’s jump velocity to be $v_\text{jump} = \sqrt{2g_\oplus h}$ . Equating it to the escape velocity of the planet, we get

\sqrt{2g_\oplus h} = \sqrt{\frac{2GM}{R}} \implies \sqrt{2\frac{G}{R_\oplus^2} \frac{4 \pi \rho_\oplus R_\oplus^3}{3} h} = \sqrt{\frac{2G}{R} \frac{4 \pi \rho R^3}{3}}

Since $\rho = \rho_\oplus$ , we have

R = \sqrt{R_\oplus h} \approx \boxed{1.79 \times 10^3 \mathrm{\,m}}

A projectile which starts from the surface of the Earth at the sea level is launched with the initial speed of $v_0 = \alpha \sqrt{\frac{GM_\oplus}{R_\oplus}}\,$ ( $\alpha < \sqrt{2}$ ) and the projecting angle (with respect to the local horizon) of $\theta$ . Ignore the air resistance.

a) The orbit of the projectile is an ellipse. Find its semi-major axis $a$ in units of $R_\oplus$ .
b) What is the eccentricity $e$ of this elliptical orbit?
c) Calculate the highest altitude of the projectile with respect to the Earth surface
d) Find the time of flight for the projectile.
e) What is the range of the projectile (surface distance between launching point and falling point)?

a) The total energy of the projectile is

\varepsilon = \frac{1}{2} v_0^2 - \frac{GM_\oplus}{R_\oplus} = -\frac{GM_\oplus}{2a}

Therefore, the semi-major axis is

\boxed{a = \frac{GM_\oplus R_\oplus}{2GM_\oplus - v_0^2 R_\oplus} = \frac{R_\oplus}{2 - \alpha^2}}

b) The angular momentum of the projectile is

h = R_\oplus v_0 \cos \theta = \sqrt{GM_\oplus a (1 - e^2)}

Thus we get the eccentricity as

\begin{align*} e &= \sqrt{1 - \frac{R_\oplus^2 v_0^2 \cos^2 \theta}{GM_\oplus a}}\\ &= \boxed{\sqrt{1 - \frac{\alpha^2 \cos^2 \theta}{2 - \alpha^2}}} \end{align*}

c) The apoapsis distance is

r_a = a(1 + e) = \frac{R_\oplus}{2 - \alpha^2} \left( 1 + \sqrt{1 - \frac{\alpha^2 \cos^2 \theta}{2 - \alpha^2}} \right)

Therefore, the highest altitude of the projectile with respect to the Earth surface is

\boxed{h = R_\oplus \left[ \frac{1}{2 - \alpha^2} \left( 1 + \sqrt{1 - \frac{\alpha^2 \cos^2 \theta}{2 - \alpha^2}} \right) - 1 \right]}

d) To find the time of flight, we find the mean anamoly at the launch and landing points. The eccentric anomaly at the launch point is

\cos E_0 = 1 - \frac{r}{a} = 1 - \frac{R_\oplus}{a} = 1 - (2 - \alpha^2) = \alpha^2 - 1

The mean anamoly at the launch point is

M_0 = E_0 - e \sin E_0 = \cos^{-1}(\alpha^2 - 1) - e \sqrt{2 \alpha^2 - \alpha^4}

The eccentric anomaly at the landing point is simply $E_1 = 2\pi - E_0$ , and the mean anamoly at the landing point is just $M_1 = 2\pi - M_0$ . Therefore, the time of flight is

t = \frac{P}{2 \pi} (M_1 - M_0) = P\left(1 - \frac{M_0}{\pi}\right)

\boxed{t = \sqrt{\frac{R_\oplus^3}{GM_\oplus}} \frac{2}{(2 - \alpha^2)^{3/2}} \left[ \pi - \cos^{-1}(\alpha^2 - 1) + e \sqrt{2 \alpha^2 - \alpha^4} \right]}

e) The get the range of the projectile, we need to find the true anamoly at the launching and landing points. The true anamoly at the launch point is

\theta_0 = 2 \tan^{-1} \left( \sqrt{\frac{1 + e}{1 - e}} \tan \frac{E_0}{2} \right) = 2 \tan^{-1} \left( \sqrt{\frac{1 + e}{1 - e}} \sqrt{\frac{2 - \alpha^2}{\alpha^2}} \right)

The true anamoly at the landing point is $\theta_1 = 2\pi - \theta_0$ . Thus the range of the projectile is

\boxed{R = 2 R_\oplus \left[ \pi - 2\tan^{-1} \left( \sqrt{\frac{1 + e}{1 - e}} \sqrt{\frac{2 - \alpha^2}{\alpha^2}} \right) \right]}

Kepler's Laws Unbound Orbits