We won't be sending any people (or animals) with our rocket, so we won't have to worry about dying, but hitting our target would be prefarable.
Last post we decided that we want to try and land on Planet1, the fourth planet from our star. Turns out, though, that choosing the planet and actually landing are two very different things. Planets have this tendancy to not stand still, which means that is you simply aim at one of them, it'll be long gone when you finally reach the point you aimed at.
If we recall our first simulation of the rocket launch, we launched our rocket without any consideration of where we wanted to go, simply launching straight up in the x-direction at the start of our simulation. This time, we have a specific goal in mind, and we must edit our simulation to suit this goal. This means
- Generalizing our starting coodinates so that we can launch from anywhere on the planet, thereby being able to launch in any direction needed.
- Finding the best time to launch. We may not want to launch at t=0, as this may not be the best placement of our planets. Instead, we might want to let our simulation run until we get a better placement.
We should keep every planet in mind, as their gravity affects our rocket, but primarely we will look at our home planet and our destination planet for now.
Let's begin by generalizing our starting coodinates.
We change our initial position on the ground to a more flexible version that easily let's us change our lauch site. We could technically launch from anywhere on the planet, but as all our planets orbit the star in the xy-plane, we will launch our rocket in this same plane, thus our rocket may only launch from the equator (the part of the planet in the xy-plane).
Launching from the equator also gives us the edvantage of the greatest rotational velocity (the rockets tangential velocity, as this is the linear speed of the rocket, along the circular path that is the planet), as this is where the radius is the largest. You can imagine tracing the lines of a globus around the globe. If you try to use the same speed everywhere, you'll notice that the further from the equator, the faster you'll complete a round (really, try it). The planet needs to keep up with it self, sort of, thus the greatest tangential velocity is found at the largest radius - the equator!
If we 'slice' our planet so that only the thin line that is the equator remains, we have a circle, and we want to be able to launch from any point on this circle. You may recall that that we mentioned coordinates specifically suited for circular shapes? That's right, polar coordinates! Take a look here if you can't recall what they are. We convert our cartesian coordinates \(\vec{r}_0 = [x,y]\) to polar coordinates giving our start position as \(\vec{r}_0 = [r_p \cos{\theta}, r_p \sin{\theta}]\), where \(r_p\) is the radius of our planet. Thus we can simply insert any point around the equator we think is best to launch from in order to reach our destination!
When we change coordinate systems, we also need to change our initial veocity, also taking into account the tangential velocity, vt, from the planet's rotation around itself. Remembering that velocity is the derivative of the position, we get \(\vec{v}_0 = [-v_0 \sin{\theta} , v_0 \cos{\theta}]\).
One way of launching our rocket would be by utilizing these flexible initial conditions and simulating the trajectory of our rocket when launch from different places on our planet. We can then check if we are close enough to be caught in the destination planet's orbit, or if we accidentally flung our rocket into the deep recesses of outer space. As simulating a launch on a computer is free, we can check for several angles until we find a suitable position to launch. It will be important to remember that our trajectory will be curved because of the tangential velocity. We will also have to take into account the movement of the destination planet during the travel time. This isn't the most scientific way of launching a rocket, but after a few tries it might work!
We'd prefer to have some idea of where and when to launch, however, which is why we want to look into something called a Hohmann transfer when deciding the best time to launch.
We thought for a moment that after launching we would enter something called a parking orbit, where we just keep spinning (or falling, actually) in orbit around our planet. However, as our goal has been to reach escape velocity, we saw that we would not enter any parking orbit (we've escaped our planet's orbit), and so we can't use a Hohmann transfer exactly as planned, but it's still a good starting point when finding the best time to launch! We have therefore included a part about the general concepts behind the transfer.
When using a Hohmann transfer, our goal is to move from our planet's orbit to that of a neighboring planet. There are different ways of doing this, all with different pros and cons. A Hohmann transfer is not the quickest way, but it is very fuel efficient, and our basement space agency is on a budget. So how does it work? We use a transfer orbit to move between orbits!
This next part is a little mathematical, but basically we first find the time our trip should take, then how the planets should be positioned in relation to each other for launch, and when this position happens for our home and destination planet.
Let's puzzle together the information we have in order to describe this new orbit.
We already know that the maximum width of the elliptical orbit is given by \(2a\), where \(a\) is the semi-major axis. That gives us the semi-major axis of the transfer orbit: \(a_t = \frac{1}{2}(r_1 + r_2)\), where \(r_1\) and \(r_2\) are the radii of the orbits of our planet and the destination planet.
We can then find the period, Pt, of the transfer orbit from Kepler's third law:
\(P_t^2 = a^3 \Longleftrightarrow P_t = \sqrt{a^3}\)
So how much time will our trip take? As the return trip is a problem for another day, we are only interested in the first half of our transfer orbit, and the total travel time will be \(T = \frac{1}{2}P_t\).
Now that we know where we want to go, and how long this will take, how do we know when to launch?
We know how much time (\(T \)) the trip will take and the period (\(P_2\)) of the destination planet . We want to find the angle \(\theta_{launch}\) between the position vectors \(\vec{r}_1\) and \(\vec{r}_2\) of our planets at the time of launch.
We use this relation and solve for \(\theta\):
\(\frac{T}{P_2} = \frac{180^{\circ} - \theta_{launch}}{360^{\circ}} \Longleftrightarrow \theta_{launch} = 180^{\circ} - \frac{360^{\circ} T}{P_2} \)
Now that we know the correct angle for launch, we wish to know when this is the angle between the position vectors, as the angle between the planets are constantly changing. We use the familiar equation for the angle between vectors:
\(\cos{\theta} = \frac{\vec{r}_1 \cdot \vec{r}_2}{|\vec{r}_1| \cdot |\vec{r}_2|}\)
and find the point in time where the angle \(\theta\) between the planets matches our optimal launch angle, or \(\cos{\theta} = \frac{\vec{r}_1 \cdot \vec{r}_2}{|\vec{r}_1| \cdot |\vec{r}_2|} = \cos{\theta_{launch}} \). As we know all our planets, we can find when this expression is true, and this will be our launch window.
If we were in a parking orbit we would be circling around the planet at a speed of \(v_1 =\sqrt{ \frac{GM_*}{r}}\), where \(M_* \) is the mass of the star and \(r\) is the radius of our rocket's orbit around our planet. We would need to give our rocket a speed boost in order to escape the planet's orbit and enter the transfer orbit: \(v_1 + \Delta v_1 = v_p\), where \(v_p\) is the orbital speed at perihel (the point in an elliptical orbit closest to the star, which is where we want to be at the time of launch) in the transfer orbit. As we want this speed boost to be tangential to the transfer orbit, the rocket needs to be aligned with our home planet at the time of exit. As a launch window leaves us time to make some final adjustments, we only need to wait until the rocket is at a tangential position to the transfer orbit when it's in parking orbit around the planet.
However, this only gives us an estimate as we are not in a parking orbit, and we therefore write our program in a way that makes it possible for us to adjust the time of launch. We should also keep in mind that the Hohmann transfer does not take the gravitational forces from our planet and the sun into account, so we should not trust it entirely. There will be some trial and error, but with this method we do at least have a starting point for a good time to launch.
With a method to generalize our launch position, and a rough estimate of when we should launch, we are ready to (simulate) leaving home. On our way out, we want to change our frame of reference to one that puts the star in the origin. We do this by adding the position of our home planet at the time of exit to the the position of the rocket relative to the planet.
This raises the question of weather we need to the planet's movement after launch into account?
As we can see, we know the position vector from the sun to the planet at launch, and the position vector from the planet at launch to the rocket. We can therefore ignore the planet's movement after launch, as this is not part of the vector sum we need to calculate in order to change our frame of reference.
The planet has moved on, and so must we.
We do need to add the planet's velocity to the velocity of the rocket. At the time of launch, when we are in the planets system of frame, we are moving with the planet's frame (the planet is technically not moving when it is the center of the frame of reference). However, as we change our frame of reference, we have to add our planets velocity. Newton's first law tells us that we are either at rest, or moving at constant speed. As we have not exerted any forces in the direction of the movement of the planet, we know that we are not at rest, and therefore our speed must remain constant in this specific direction.
We now have a position vector between the star and the rocket, which will be of great benefit as we soar through space and leave our moving planet behind us. (It's very confusing when the origin in your coordinate system refuses to stay in place...)
Two last minute thoughts
When we (hopefully) get to our destination planet we need to know we can keep track of where we are compared to the planet. Navigating a vast space is one thing, actually orbiting and landing on a planet is a whole other thing. Luckily for us we know that we can find the distance, l, to a planet from
\(l=|\vec{r}|\sqrt{\frac{M_p}{kM_s}}\)
where \(|\vec{r}|\) is the distance from the spacecraft to the star, \(M_p\) is the planet's mass, \(M_s\) is the mass of our star, and k is a scaling factor depending on how much larger the gravitational force from the star is on the spacecraft.
How on (not) Earth do we know this?
As usual we put our trust in Newton and his laws. By now we should recall that the gravitational force is given by
\(\vec{F}_g=G\frac{M_pM_s}{r^2}\hat{r}\)
where, r is the distance between two masses and G remains the gravitational constant.
In our case we have two forces working on our spacecraft, the force \(F_p\) from the destination planet, and the force \(F_s\) from the star (and technically some from other planets, but compared to the gravity of the planet when we are considerably close to it, and the huge gravity of the star, the gravity of other planets can be ignored for simplicity). If we now assume our spacecraft is not moving, maybe you know what this means for our forces? (hint: Newton's 2nd and 3rd laws)
Right, if there is no movement, the total force must be zero, meaning the forces pulling from opposite sides on the spacecraft must 'cancel each other out'!
\(F_p-F_s = 0 \longleftrightarrow F_p = F_s\)
We write out the force on the spacecraft form the planet and the force on the spacecraft from the star using Newton's gravitational law:
\(G\frac{M_pm}{l^2}\hat{r}=kG\frac{M_sm}{r^2}\hat{r}\)
We've added the scale factor, which is how many times greater the gravitation from the planet is on the spacecraft, compared to the gravitational force from the star.
We are looking for the distance from the spacecraft to the planet, and write this as l. The mass of the destination planet is written as m, and the masses of the planet and star as \(M_p\) and \(M_s\) respectively. If we cross out the common factors, we're left with:
\(\frac{M_p}{l^2}\hat{r}=\frac{M_s}{r^2}\hat{r}\)
Rearranging this with respect to the distance l, gives:
\(l=\sqrt{\frac{M_p}{kM_s}}|\vec{r}|\)
Exactly like we wanted!
Another sort of important thing to know when we're navigating up there, is when we should be able to see our destination planet. It would certainly help if we knew when it should appear on a photo in order to make sure we're heading in the right direction, and not just cruising away into the vast darkness.
Luckily, our spacecraft is equipped with a camera. In our case, we can picture our planet when it makes up one pixel on a picture taken, we then say that the planet is resolved. Not the fanciest picture, but at least we'll know it's there.
Imagine you're standing a distance, L, away from an object you want to picture. The width of what your camera pictures at this distance is then the field of view, F, and is measured in degrees.
If we have a look at the figure to the left, we recognize a familiar shape- the triangle. We are curious about how far away we need to be in order to resolve our planet, in other words, we need to find the distance L.
Using a bit of trigonometry, we can say that
\(\tan{\frac{\theta}{2}}=\frac{opposite}{adjacent}=\frac{h}{L}\)
where we set the opposite length as h, the adjacent length as L (what we're looking for) and we say that the angle of the top triangle is half the angle of the full field of view. This angle can be written as the field of view, meaning we can rewrite:
\(\tan{\frac{F}{2}}=\frac{h}{L}\)
and (from the )
\(\cos{\frac{F}{2}}=\frac{F}{P}\)
If we set the height to the radius of the planet, we get that \(h = R\). Our angle is then very, very small!
This means that we can get a little naugty and simply say that for a small angle, \(\theta<<\), we have that \(\tan{\theta}\approx\theta\), and \(\cos{\theta}\approx\theta\) meaning we can simplify the expression and put in h:
\(\frac{R}{L}\approx\frac{F}{P}\leftrightarrow L \approx \frac{RP}{F}\)
We need our spacecraft to be either at or closer than this distance L:
\(L \lessapprox \frac{RP}{F}\)
Okay! That should help us when we're navigating. I believe it's been mentioned before but space is really really big. We need all the resources we can get, no equation is too little!
We now have a possible placement for our planets when we want to launch, and because we know when the planets are at these positions, we know approximately when we want to launch. We can launch from anywhere (on the equator) of our planet, meaning that we can attempt a few simulations in order to attempt to hit our destination planet (perhaps we will find a better method for this when we attempt these simulations in the future, but for now the fact that we can change when and where we launch, is what matters). Once we are up there, we know how we can find the distance to our planet and when we should be able to see it, so we can check if we are completely off course
We won't take off just yet, we still need to be able to see where we're going a little more thourougly, but we are definitly getting closer!