Following our text, Fundamentals of Astrodynamics by Bate, Mueller, and White, we start with Kepler’s Laws of Planetary Motion, which are general-izations derived from the planetary position data of Tycho Brahe. According to our text, Kepler published the first two. Ordinary Differential Equations-Lecture Notes Eugen J. Ionascu c Draft date April 25, 2006.
Discussion
constant acceleration
For the sake of accuracy, this section should be entitled 'One dimensional equations of motion for constant acceleration'. Given that such a title would be a stylistic nightmare, let me begin this section with the following qualification. These equations of motion are valid only when acceleration is constant and motion is constrained to a straight line.
Given that we live in a three dimensional universe in which the only constant is change, you may be tempted to dismiss this section outright. It would be correct to say that no object has ever traveled in a straight line with a constant acceleration anywhere in the universe at any time — not today, not yesterday, not tomorrow, not five billion years ago, not thirty billion years in the future, never. This I can say with absolute metaphysical certainty.
So what good is this section then? Well, in many instances, it is useful to assume that an object did or will travel along a path that is essentially straight and with an acceleration that is nearly constant; that is, any deviation from the ideal motion can be essentially ignored. Motion along a curved path may be considered effectively one-dimensional if there is only one degree of freedom for the objects involved. A road might twist and turn and explore all sorts of directions, but the cars driving on it have only one degree of freedom — the freedom to drive in one direction or the opposite direction. (You can't drive diagonally on a road and hope to stay on it for long.) In this regard, it is not unlike motion restricted to a straight line. Approximating real situations with models based on ideal situations is not considered cheating. This is the way things get done in physics. It is such a useful technique that we will use it over and over again.
Our goal in this section then, is to derive new equations that can be used to describe the motion of an object in terms of its three kinematic variables: velocity (v), position (s), and time (t). There are three ways to pair them up: velocity-time, position-time, and velocity-position. In this order, they are also often called the first, second, and third equations of motion, but there is no compelling reason to learn these names.
Since we are dealing with motion in a straight line, direction will be indicated by sign — positive quantities point one way, while negative quantities point the opposite way. Determining which direction is positive and which is negative is entirely arbitrary. The laws of physics are isotropic; that is, they are independent of the orientation of the coordinate system. Some problems are easier to understand and solve, however, when one direction is chosen positive over another. As long as you are consistent within a problem, it doesn't matter.
Basic Astrodynamics Formulas Free
velocity-time
The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant. If velocity increases by a certain amount in a certain time, it should increase by twice that amount in twice the time. If an object already started with a certain velocity, then its new velocity would be the old velocity plus this change. You ought to be able to see the equation in your mind's eye already.
This is the easiest of the three equations to derive using algebra. Start from the definition of acceleration.
| a = | ∆v |
| ∆t |
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Expand ∆v to v − v0 and condense ∆t to t.
| a = | v − v0 |
| t |
Then solve for v as a function of t.
v = v0 + at[1]
This is the first equation of motion. It's written like a polynomial — a constant term (v0) followed by a first order term (at). Since the highest order is 1, it's more correct to call it a linear function.
The symbol v0 [vee nought] is called the initial velocity or the velocity a time t = 0. It is often thought of as the 'first velocity' but this is a rather naive way to describe it. A better definition would be to say that an initial velocity is the velocity that a moving object has when it first becomes important in a problem. Say a meteor was spotted deep in space and the problem was to determine its trajectory, then the initial velocity would likely be the velocity it had when it was first observed. But if the problem was about this same meteor burning up on reentry, then the initial velocity likely be the velocity it had when it entered Earth's atmosphere. The answer to 'What's the initial velocity?' is 'It depends'. This turns out to be the answer to a lot of questions.
The symbol v is the velocity some time t after the initial velocity. It is often called the final velocity but this does not make it an object's 'last velocity'. Take the case of the meteor. What velocity is represented by the symbol v? If you've been paying attention, then you should have anticipated the answer. It depends. It could be the velocity the meteor has as it passes by the moon, as it enters the Earth's atmosphere, or as it strikes the Earth's surface. It could also be the meteorite's velocity as it sits in the bottom of a crater. (In that case v = 0 m/s.) Are any of these the final velocity? Who knows. Someone could extract the meteorite from its hole in the ground and drive away with it. Is this relevant? Probably not, but it depends. There's no rule for this kind of thing. You have to parse the text of a problem for physical quantities and then assign meaning to mathematical symbols.
The last part of this equation at is the change in the velocity from the initial value. Recall that a is the rate of change of velocity and that t is the time after some initial event. Rate times time is change. Given an object accelerating at 10 m/s2, after 5 s it would be moving 50 m/s faster. If it started with a velocity of 15 m/s, then its velocity after 5 s would be…
15 m/s + 50 m/s = 65 m/s
position-time
The displacement of a moving object is directly proportional to both velocity and time. Move faster. Go farther. Move longer (as in longer time). Go farther. Acceleration compounds this simple situation since velocity is now also directly proportional to time. Try saying this in words and it sounds ridiculous. 'Displacement is directly proportional to time and directly proportional to velocity, which is directly proportional to time.' Time is a factor twice, making displacement proportional to the square of time. A car accelerating for two seconds would cover four times the distance of a car accelerating for only one second (22 = 4). A car accelerating for three seconds would cover nine times the distance (32 = 9).
Would that it were so simple. This example only works when initial velocity is zero. Displacement is proportional to the square of time when acceleration is constant and initial velocity is zero. A true general statement would have to take into account any initial velocity and how the velocity was changing. This results in a terribly messy proportionality statement. Displacement is directly proportional to time and proportional to the square of time when acceleration is constant. A function that is both linear and square is said to be quadratic, which allows us to compact the previous statement considerably. Displacement is a quadratic function of time when acceleration is constant
Proportionality statements are useful, but not as general as equations. We still don't know what the constants of proportionality are for this problem. One way to figure them out is to use algebra.
Start with the definition of average velocity.
| v = | ∆s |
| ∆t |
Expand ∆s to s − s0 and condense ∆t to t.
| v = | s − s0 |
| t |
Solve for position.
s = s0 + vt[a]
To continue, we need to resort to a little trick known as the mean speed theorem or the Merton rule. I prefer the latter since the rule can be applied to any quantity that changes at a uniform rate — not just speed. The Merton rule was first published in 1335 at Merton College, Oxford by the English philosopher, mathematician, logician, and calculator William Heytesbury (1313–1372). When the rate of change of a quantity is constant, its average value is halfway between its final and initial values.
v = ½(v + v0)[4]
Substitute the first equation of motion [1] into this equation [4] and simplify with the intent of eliminating v.
| v = ½[(v0 + at) + v0] v = ½(2v0 + at) v = v0 + ½at[b] |
Now substitute [b] into [a] to eliminate v [vee bar].
s = s0 + (v0 + ½at)t
And finally, solve for s as a function of t.
s = s0 + v0t + ½at2[2]
This is the second equation of motion. It's written like a polynomial — a constant term (s0), followed by a first order term (v0t), followed by a second order term (½at2). Since the highest order is 2, it's more correct to call it a quadratic.
The symbol s0 [ess nought] is often thought of as the initial position. The symbol s is the position some time t later. You could call it the final position if you wished. The change in position (∆s) is called the displacement or distance (depending on circumstances) and some people prefer writing the second equation of motion like this.
∆s = v0t + ½at2[2]
velocity-position
The first two equations of motion each describe one kinematic variable as a function of time. In essence…
- Velocity is directly proportional to time when acceleration is constant (v ∝ t).
- Displacement is proportional to time squared when acceleration is constant (∆s ∝ t2).
Combining these two statements gives rise to a third — one that is independent of time. By substitution, it should be apparent that…
- Displacement is proportional to velocity squared when acceleration is constant (∆s ∝ v2).
This statement is particularly relevant to driving safety. When you double the speed of a car, it takes four times more distance to stop it. Triple the speed and you'll need nine times more distance. This is a good rule of thumb to remember.
The conceptual introduction is done. Time to derive the formal equation.
method 1
Combine the first two equations together in a manner that will eliminate time as a variable. The easiest way to do this is to start with the first equation of motion…
v = v0 + at[1]
solve it for time…
| t = | v − v0 |
| a |
and then substitute it into the second equation of motion…
s = s0 + v0t + ½at2[2]
like this…
| s = | s0 + v0 | ⎛ ⎜ ⎝ | v − v0 | ⎞ ⎟ ⎠ | + ½a | ⎛ ⎜ ⎝ | v − v0 | ⎞2 ⎟ ⎠ |
| a | a |
| s − s0 = | vv0 − v02 | + | v2 − 2vv0 + v02 |
| a | 2a |
| 2a(s − s0) = 2(vv0 − v02) + (v2 − 2vv0 + v02) |
| 2a(s − s0) = v2 − v02 |
Make velocity squared the subject and we're done.
v2 = v02 + 2a(s − s0)[3]
This is the third equation of motion. Once again, the symbol s0 [ess nought] is the initial position and s is the position some time t later. If you prefer, you may write the equation using ∆s — the change in position, displacement, or distance as the situation merits.
v2 = v02 + 2a∆s[3]
method 2
The harder way to derive this equation is to start with the second equation of motion in this form…
∆s = v0t + ½at2[2]
and solve it for time. This is not an easy job since the equation is quadratic. Root - the clockwork expansion crack. Rearrange terms like this…
½at2 + v0t − ∆s = 0
and compare it to the general form for a quadratic.
ax2 + bx + c = 0
The solutions to this are given by the famous equation…
| x = | −b ± √(b2 − 4ac) |
| 2a |
Replace the symbols in the general equation with the equivalent symbols from our rearranged second equation of motion…
| t = | −v0 ± √[v02 − 4(½a)(∆s)] |
| 2(½a) |
clean it up a bit…
| t = | −v0 ± √(v02 − 2a∆s) |
| a |
and then substitute it back into the first equation of motion.
v = v0 + at[1]
| v = v0 + a | ⎛ ⎜ ⎝ | −v0 ± √(v02 − 2a∆s) | ⎞ ⎟ ⎠ |
| a |
Stuff cancels and we get this…
v = ±√(v02 + 2a∆s)
Square both sides and we're done.
v2 = v02 + 2a∆s[3]
That wasn't so bad now, was it?
calculus derivations
Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. By definition, acceleration is the first derivative of velocity with respect to time. Take the operation in that definition and reverse it. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. This gives us the velocity-time equation. If we assume acceleration is constant, we get the so-called first equation of motion [1].
| a | = |
| |||||||||
| dv | = | a dt | |||||||||
| = |
| |||||||||
| v − v0 | = | at | |||||||||
| v | = | v0 + at [1] | |||||||||
Again by definition, velocity is the first derivative of position with respect to time. Reverse this operation. Instead of differentiating position to find velocity, integrate velocity to find position. This gives us the position-time equation for constant acceleration, also known as the second equation of motion [2].
| v | = |
| ||||||||||
| ds | = | vdt | ||||||||||
| ds | = | (v0 + at) dt | ||||||||||
| = |
| ||||||||||
| s − s0 | = | v0t + ½at2 | ||||||||||
| s | = | s0 + v0t + ½at2 [2] | ||||||||||
Unlike the first and second equations of motion, there is no obvious way to derive the third equation of motion (the one that relates velocity to position) using calculus. We can't just reverse engineer it from a definition. We need to play a rather sophisticated trick.
The first equation of motion relates velocity to time. We essentially derived it from this derivative…
| dv | = a |
| dt |
The second equation of motion relates position to time. It came from this derivative…
| ds | = v |
| dt |
The third equation of motion relates velocity to position. By logical extension, it should come from a derivative that looks like this…
| dv | = ? |
| ds |
But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1 (dtdt) and a special version of algebra (algebra with infinitesimals). Look what happens when we do this. We get one derivative equal to acceleration (dvdt) and another derivative equal to the inverse of velocity (dtds).
| dv | = | dv | 1 |
| ds | ds | ||
| dv | = | dv | dt |
| ds | ds | dt | |
| dv | = | dv | dt |
| ds | dt | ds | |
| dv | = | a | 1 |
| ds | v |
Next step, separation of variables. Get things that are similar together and integrate them. Here's what we get when acceleration is constant…
| = |
| ||||||||||
| v dv | = | a ds | ||||||||||
| = |
| ||||||||||
| ½(v2 − v02) | = | a(s − s0) | ||||||||||
| v2 | = | v02 + 2a(s − s0) [3] | ||||||||||
Certainly a clever solution, and it wasn't all that more difficult than the first two derivations. However, it really only worked because acceleration was constant — constant in time and constant in space. If acceleration varied in any way, this method would be uncomfortably difficult. We'd be back to using algebra just to save our sanity. Not that there's anything wrong with that. Algebra works and sanity is worth saving.
| v = | v0 + at | [1] |
| + | ||
| s = | s0 + v0t + ½at2 | [2] |
| = | ||
| v2 = | v02 + 2a(s − s0) | [3] |