Gravity

Gravity is very simple under Newtonian mechanics. Every object with mass attracts every other object with mass. The size of this force depends on the masses of the objects and the square of the distance between them. The equation looks like this:

F = G(mM/r^2)

F = Force
G = Big G
M = mass of 1st object
m = mass of 2nd object
r = distance between objects

G (called 'big gee') is Newton's gravitational constant, with a value of approximately:

 G=6.67*10^-11 m^3 kg^-1 s^2

This is an extremely small number, so gravity is very weak on most scales. Pairs of objects have to be very massive or very close together (the scale of micrometres) before the gravitational attraction is appreciable.

So what's the force between the Earth and the Sun? The relevant data are:

Distance between the Earth and the Sun: 1.4960×10^11 m
Mass of the Earth: 5.9736×10^24 kg

Mass of the Sun: 1.9891×10^30 kg

Plug these numbers into the equation and we get the force as being about 3.53*10^22 N.

This is a pretty large number, but what do we get for things on a more comprehensible scale? The same process for two people of mass 70kg standing 1 metre apart yields a force of about 10^-7 N. Hardly anything at all, which is why we almost never notice gravitational attraction between objects on the earth.

If you divide the both sides of this equation by m then you get an expression for 'Force per unit mass'. I.e. how much force a 1kg object would experience at this distance from the other object. This quantity is called g (aka 'little gee'), the 'gravitational field strength'.

g = F/m = G(M/r^2)

At the surface of the Earth we get a value of about 9.81 N/kg from this equation.

Fields

Newton's law of gravitation predicts that every objects pulls on every other object. The gravitational field of an object is a way of describing the way it pulls on other objects. The field can be represented by field lines, where the direction of the line indicates the direction in which an object would feel a force and the relative density of lines indicates how strong the force is.

NB: A field line diagram gives only a qualitative analysis of the gravitational field. It can only tell you that 'the field is stronger here than there'; it can't tell you 'the field is precisely this strong here'.

Planetary bodies which are approximately spherical can be thought of as being 'point-like' with all their mass concentrated at their centre. This is because for any part of the planet which isn't at the centre, there's another part on the other side of the planet. The resultant gravity of these two objects can be more simply thought of as coming from a single object of double the mass halfway between them. Do this for every part of the planet and you get it all averaging out to a single object at the centre of the planet with the same mass as the entire planet.

For a point like object the field around it is 'radial' with field lines coming in from all directions and getting further apart the further they are from the object. This is what you'd expect: the further away an object is, the smaller the gravitational field strength gets because we are dividing by r squared.

Now imagine zooming in on that picture to an area about the size of your house. From this perspective, the field lines look like they're equally spaced and all going up because we are looking at a scale where the depth and curvature of the Earth are essentially negligible. Field lines being equally spaced and unidirectional is called a 'uniform field' because the value of g is the same everywhere.

Whether you consider the Earth's gravitational field to be uniform or radial is ultimately a question of perspective and scale: if you're considering objects bouncing around close to the surface of the Earth, then it's fine to approximate the field as uniform. If you're considering things on an astronomical scale like orbiting planets and satellites then it will be better to think of it as radial.

Potential

You may remember gravitational potential energy: the energy put into something when you work against gravity to move it up away from the Earth and the energy you get back when it falls down again. Potential is defined as the gravitational potential energy per unit mass, given the symbol V. This is similar to gravitational force: F is the size of the force on a particular object and g is the general force per unit mass of any object. GPE is the potential energy of a particular object and V is the potential energy per unit mass of any object.

The main use of the idea of potential is this: as objects move between points of different potential, they gain or lose energy. This means that if you know the potential at two points in space you can work out how much energy it will take to move a certain mass between them. If it comes out negative then your object will gain energy, probably speeding up as a result. Think about an object falling towards the Earth. As it descends, the force of gravity from the Earth acts on it, speeding it up. Another way to think about this is that it has moved from a point of one potential to a point of lower potential, so it must have gained energy, in this case the kinetic energy of it plummeting downwards.

It is worth remembering that there is no absolute standard of potential or potential energy. This is to say that where the point of 'zero' potential is is ultimately arbitrary as it is only the change in potential from point to point that is of any physical significance. This means that it is fine to have negative potentials, as long as you know where your zero-point (often called the datum) is. Two commonly used conventions are:

• That the zero point is at an infinite distance away from the objects concerned such that all potentials are negative. This one is usually used when considering radial fields.
• That the zero point is at the surface of the Earth such that points above it have positive potential. This is most useful when considering the roughly uniform field near the surface of the Earth.

Recognize that under both conventions potential goes up as one gets further from Earth. This makes sense as to get further away from an object you need to do work against its gravitational field, so your potential energy, and hence the potential at that point, must increase.

Potential for a uniform field

In a uniform field the value of g is the same everywhere, so to raise an object up all we need to do is oppose gravity with a force of the same size. For an object of mass m, this force will have to be of magnitude mg. The work done by this force raising the object through a distance of h will be mgh. So the potential energy of an object of mass m is mgh. To find the potential energy per unit mass we just divide this by m. So the potential of a point at a distance h from the zero-point in a uniform gravitational field is given by:

V = gh

V = potential

g = gravitational field strength

h = distance from datum

Potential for a radial field

With this important result:

V = G(M/r)

V = potential

G = big G

M = mass of object to which gravitational field is due

r = distance from that object

If you're using the convention of zero potential at infinity then you get

This is a bit of a fudge; strictly speaking one should consider the integral of all the little changes in potenial from the centre of the planet to the distance r. However the result is identical so let's just get on with it.

You now know how to calculate the potential due to the gravitational field of point like objects. What if you want to calculate the potential due to two objects? You don't have to for A level. Which is good.

Because it's one complicated motherfucker.

Simple Harmonic Motion

Simple harmonic motion can be used to describe the behavior of simple oscillating systems. Classic examples include pendulums and springs (basically anything that goes up and down or left and right periodically).

The defining condition for SHM is the following:

Acceleration is proportional to displacement from a point of equilibrium and always directed towards that point

Displayed mathematically:

a = -kx

a = force
x = displacement
k = some positive constant

A mass on a spring

Disclaimer: I'm using the convention for springs where tension is expressed in terms of the natural length and a modulus of elasticity. The same logic applies if you use the simpler relation T= kd. Just substitute k for λ/l.

Take for example, a mass on a light spring:

m = mass
g = gravitational field strength
l = natural length of the spring
λ = modulus of elasticity of the spring
e = extension of the spring
In this position, the mass is in equilibrium. The tension in the spring, λe/l, equals the weight of the mass, mg. Now what happens if we stretch the mass downwards? The extension of the spring will increase slightly, so the force of tension on the mass will also increase. Now the mass is no longer in equilibrium: there is a resultant force on it acting upwards. So it will accelerate upwards.

The mass will accelerate upwards until it has returned to the equilibrium position. It is now back where it started, with one important difference: it has some upwards velocity. At this point the mass will move upwards and we get a new picture:

Now the has traveled mass above the point of equilibrium. This means that the extension of the spring has gone down, so the force of tension on the mass has also gone down. There is now a resultant force downwards as the weight of the mass is greater than the tension from the spring. The mass now accelerates downwards until it returns to the equilibrium position with some downward velocity. The same thing happens again. And again. And again. In theory. In practice, energy almost always leaks away from the system.

This is simple harmonic motion: the mass moves down, is accelerated upwards, moves up, is accelerated downwards, moves down, and so on. The maximum value of the displacement is called the amplitude of the motion and is usually denoted by A.

Disclaimer: the rest of this section is devoted to a derivation of the defining SHM equation. There's nothing beyond GCSE maths and I think that it's useful to be able to get your head around it, but you don't need to in order to get full marks on the AQA Physics A2. Feel free to skip to the next section, entitled 'Important Equations', if you don't care about the maths.

It is possible to derive the defining SHM equation by considering a point in the motion of the mass-spring system. If we take downwards to be the positive direction then for a point in this picture:

Defining x as the extension of the spring, we get the following:

With the important result:

Where x* is an adjusted form of the extension which gives displacement from the equilibrium position:

x*=x-(glm/λ)

This is the defining equation for SHM, with x* as the displacement.

It is worth noting that when x*=0:

Which is the result you'd expect: when our displacement is zero, we are in the equilibrium position and there is no resultant force.

Another way of thinking about this is to say that the factor by which we adjusted x, glm/λ, is exactly the value which x would need to take for the system to be in equilibrium. I.e. when the spring has extended by glm/λ, the tension in the spring exactly equals the weight of the mass.

Important Equations

The defining SHM equation is this:

a = -kx

The value of k will depend on the exact nature of the SHM system concerned. We found earlier that for a mass on a spring, k=λ/lm. You should always be able to find a value of k by considering the forces at work. A general result for a mass on a pendulum is k=g/l where l is the length of the pendulum.

The maximum value of the acceleration occurs when displacement is at it's maximum value. I.e. when x=A

a(max) = -kA

Another important equation which gives the frequency of a SHM system is this:

2πf=√k

f=(√k)/2π

f = frequency in Hz or s^-1

Which means that you can always replace k with (2πf)^2 in any equation.

To find the displacement of the system after a certain length of time t use the following equation:

x = Acos(2πft)

t = time in s

Similarly to find the velocity for a given displacement:

v = 2πf√(A^2-x^2)

The maximum velocity is achieved when the system is at it's equilibrium point. This is fairly intuitive: think at what point on a swing are you travelling fastest? Plugging x=0 into the above equation yields:

v(max) = 2πfA

Finally, to find the period of the system (how long it takes to move through a single complete oscillation) simply recall that:

T = 1/f

And so,

T = 2π/(√k)

All of these equations can be derived from the defining SHM equation, but it's quite difficult and I won't be doing it here (FP3 differential equations methods required).

Sine and Cosine Waves

Cosine

The formula for displacement as a function of time is:

x = Acos(2πft) = Acos(2πt/T)

The cosine function varies between 1 and -1, so the maximum and minimum values of x as time varies are A and -A respectively. 

The maximum values of the cos function are when it's domain (2πt/T), = any integer multiple of 2π (0, 2π, 4π etc). This will occur when:

t = 0, corresponding to the system being released at maximum displacement
t = T, corresponding to the system having completed a single full oscillation and having returned to it's starting position
t = nT where n is a positive integer, corresponding to the system having completed n cycles

The same occurs for the minimum values of the cos function, just shifted by T/2. This should make sense as T/2 is the time it takes for the system to move from being at amplitude on one side to at amplitude on the other side (i.e. half one complete cycle).

So this is how the displacement equation works: the cos function dictates where in the oscillation the system is and then you multiply by A to get the actual displacement in space.

Sine

You can also use a sine function instead.

x = Asin(2πft) = Asin(2πt/T)

In this case the graph will look more like this:

This function behaves in exactly the same way, except it is shifted so that it passes through the origin. This means that when t = 0, x also = 0, so the system is at the equilibrium position at the start of the motion.

Use the cosine version when the system is moved to amplitude and then released, like us extending the spring and then letting go. Use the sine version when the system is initially at equilibrium and given some impulse at t = 0.

Angular Frequency

SHM is closely related to circular motion. If you look at the diagram below you will see that the object on the right is undergoing circular motion with constant angular speed. If you look only at it's vertical displacement (the object on the left) then you get something undergoing SHM.

If you look at the SHM this way, then one complete oscillation is analogous to a complete turn around the circle, i.e. the object has turned through an angle of 2π. It is for this reason that we talk about SHM systems as having an 'angular frequency'. It's your angular speed around the hypothetical circle in 'phase space'.

The angular frequency is this:

ω = 2πf = 2π/T

The expression 2πf crops up a lot in the SHM equations, so it is often replaced with ω.

Finally, memorize this equation. It will probably save your life in the exam.

k = ω^2 = (2πf)^2

Acceleration and Velocity

We saw earlier how to calculate the acceleration and velocity of the system as functions of displacement. I.e. we can calculate how fast the system is moving and how much it is accelerating given how far it is from the equilibrium position. But what if we want to look at acceleration and velocity as functions of time?

Recall the equation for displacement as a function of time (using the angular frequency substitution explained just above):

x = Acos(ωt)

If we differentiate this with respect to time we get an equation for velocity as a function of time. This requires C3 maths (differentiating trig and the chain rule).

v = -Aωsin(ωt)

Differentiate again and we get acceleration as a function of time.

a = -A(ω^2)cos(ωt)

If you draw out these graphs you will see that whenever x is positive, a is negative. This is exactly what you'd expect given the earlier equation a=-kx.

Similarly you will see that v = 0 whenever x is at a maximum or minimum. This corresponds to the fact that when the system is at amplitude it will be instantaneously at rest. Also, when x = 0, v will be at a maximum or minimum. This also corroborates with the reasoning we used earlier to find the formula for maximum speed: on a swing you are at your fastest at the bottom of the arc, i.e. when your displacement is zero!

Resonance

For any SHM system we found that there is a frequency it will tend to oscillate at. Given the defining SHM equation:

a = -kx

We saw that this frequency, f, is given by:

f=(√k)/2π

This also called the 'natural frequency' of the SHM system. Left to it's own devices the system will always try to oscillate at this frequency. Now think about a child on a swing. If you push the swing out then it will tend to oscillate at it's natural frequency. But how can you get it to swing even further out?

Pushing in one direction won't help; you'll just end up pinning the swing on one side with it not moving at all. So you have to push periodically, sometimes pushing hard and sometimes not pushing at all (or if you're really committed, sometimes pulling).

Consider what would happen if you applied your force at any old frequency. Sometimes you'll be pushing the swing in the same direction as it is moving and you'll speed it up. But sometimes you'll be pushing against it, slowing it down.

So what's the right frequency to push at? The natural frequency of the swing! If you push at this frequency then you'll always be pushing with the swing, never against it. The fancy way of saying this is that you'll be 'in phase' with the swing.

This is the effect called resonance: a force applied to a SHM system at it's natural frequency which causes the amplitude of the motion to constantly increase. This can be a very bad thing: if we did it to the mass on the spring from earlier we may end up stretching the spring so much that it breaks. If you apply a periodic force in the form of a sound compression wave at the the right frequency then you can cause glass to oscillate so violently that it breaks.

Momentum and Force

What is Momentum?

Momentum is a measure of 'oomph'.

 An object with lots of mass will have more momentum than a lighter one travelling at the same speed. Similarly for two objects with the same mass, the faster one has more momentum. The instantaneous momentum of an object is given by the simple equation:

p = mv

p = momentum

m = mass

v = velocity

[NB: an underlined quantity indicates that it's a vector, i.e. it has both magnitude and direction]

Momentum in what direction?

Momentum is a vector, which means that it has a direction. So if you're talking about an object's momentum, it could well be important to specify which direction that momentum is in. For example, this object has some velocity v at some angle θ to the horizontal.

You could say that this object has momentum mv at the same angle to the horizontal. Alternatively you could split it up into component parts like this:

Now we would say that the object has mvsinθ up-down momentum and mvcosθ left-right momentum. This is called expressing the vector in component form, because there is an up-down component and a left-right component.

It is perfectly possible to have negative values for momentum; all it means is that your object is travelling backwards. So if I defined 'North' as positive and an object has -x momentum, that means that it has x momentum South.

Momentum Changes

You will have heard that momentum is always conserved. I'll show you the very simple derivation of that law in a moment. For now I want to go through Newton's 1st three laws of motion and clarify what they all mean.

Newton's 1st

"If there is no net force on an object, then its velocity is constant"

This one is fairly intuitive. If you don't push an object in any way then it will not change the way it moves. It is worth noting that this is where Newton deviated from the tradition. It had been accepted for thousands of years, since Aristotle, that a force is necessary to keep an object moving and that objects will tend towards being at rest if left to their own devices. The reason they thought this is that they could not see lots of hidden forces which do tend to slow objects down like air resistance and friction. If you took those away you would see the truth of Newton's 1st: no force means no change in velocity.

Newton's 2nd

"Force is proportional to the rate of change of momentum"

Velocity is just how much your position in space (your displacement) changes in a given interval of time. Similarly, force is how much your momentum changes in a given interval of time. The faster your momentum changes, the more force you will feel. The less force you experience, the less your momentum will change per unit time. So force can be expressed as follows:

F = dp/dt = d(mv)/dt

F = force

p = momentum

m = mass

t = time

If mass is constant, which it usually is, then it can be taken outside of the derivative and we get:

F = m dv/dt 

F = ma

a = acceleration

So the more massive an object is, the less it will accelerate under a given force.

Once again, force is a vector, which means we have to specify a direction. Remember our particle from earlier? If we apply a force in the -ve left-right direction then his momentum will change. But because our force is only in the left-right direction, its up-down momentum will remain unchanged, no matter how much force we use or for how long.

If this force is applied for some length of time then the velocity of the particle, expressed in component form looks like this:

You can see that our left-right velocity has changed to A, but our up-down velocity has not. So how do we calculate A? Remember back to Newton's 2nd law: F= dp/dt. If I told you that a car was travelling at 10 m/s for 30s then how would you calculate it's displacement? Simple, you just multiply the car's velocity (it's rate of change of displacement) by the amount of time it was travelling for. Similarly, to calculate how much was the momentum change of our particle just multiply the force (it's rate of change of momentum) by how much time the force acted for:

ΔP = Ft

So our change in momentum is equal to Ft. A specific change in momentum is also called an impulse. Now, back to calculating A. A is our final velocity, so our final momentum must be equal to Am. We also know that this quantity Am is just our initial momentum combined with the change in momentum. Since our change was negative (in the opposite direction to where we were initially traveling)  we can calculate Am by subtracting the change in momentum from the initial momentum:

Am = mvcosθ - Ft

Plug in values and you can calculate A! Having done this you can go back from component form to a single resultant vector.

It is also possible to solve this kind of problem geometrically using something called a momentum triangle. Write down an arrow for your initial momentum, tack on another arrow for your change in momentum (or impulse) and wherever you end up will be your resultant final momentum.

Newton's 3rd

"To every action there is an equal and opposite reaction"

Simply put: if you push me, I push you back. For any force (and that is ANY force) there is a reaction force to it acting on the other object in the other direction with exactly the same magnitude. Let's look at an example:

I'm currently sitting on a chair, so the chair is pushing up on me. What's the 3rd law pair? It's the downward force that the chair feels from me. Similarly the chair is standing on the ground with the ground pushing up on it. What's the 3rd law pair? The downward force which the Earth feels from the chair. 

A question you may well be asking is 'If every chair on the earth is pushing it down then why doesn't the Earth accelerate off in some direction?'. The answer is still Newton's 3rd law. Consider my weight. It's a force which acts downwards on me. So what's the 3rd law pair to this force? In this case it's the gravitational pull of me on the Earth upwards. Take note: the force acts on the Earth, not on me. 

If the chair and I stay still then the forces on us must be perfectly balanced (remember Newton's 1st). This means that the force on me upwards from the earth must be exactly equal in size to the force of my weight on me downwards. But similarly, the 3rd law pair forces on the Earth will be balanced: the downward force from me and the chair will exactly equal the upward pull of gravitation on the Earth. So the Earth doesn't accelerate off in any direction because those pushing forces from all those people sitting on chairs will be exactly equaled by gravitational pulling forces from those same people.

If the forces on me aren't balanced, say I've jumped and accelerated up, then the Earth will accelerate a bit. But remember back to Newton's 2nd. F=ma. The Earth is incredibly massive, so that little bit of extra momentum that I give it will not accelerate it by very much.

Conservation of Momentum

Momentum is always conserved. Whenever the momentum of one object changes, the momentum of some other object must change to accommodate this. If it ever looks like momentum isn't conserved then it's because you're looking at a non-closed system.

I'm sitting in my car and I start moving. I've gained some momentum, so something else must have lost some momentum. But it looks like nothing else has changed in momentum, so how has momentum been conserved? The answer is that the Earth will have sped up in the opposite direction. If we just look at the car then momentum is not conserved because it is not a closed system as there are forces external to the system acting on it. If we consider the Earth and the car then momentum is conserved as the two taken together form a closed system.

But why is momentum conserved? It comes down to Newton's laws. The 2nd law shows us that, in a given stretch of time, a force is equivalent to a change in momentum. The 3rd law shows that a force in one direction will be exactly balanced by another force in the opposite direction. But if a force is equivalent to a change of momentum, this means that any change of momentum in one direction will be exactly balanced by a change of momentum in the opposite direction. This means that the overall momentum of a closed system never changes: momentum is conserved. Done mathematically:

If body A exerts a force F on body B for t seconds then body B will gain Ft momentum.

By Newton's 3rd, body B also exerts an opposite force on body A for the same length of time. So body A will gain -Ft momentum

The total change of momentum for both bodies is given by the sum of their individual changes:

Δp = Ft - Ft = 0
So the overall change in momentum is always zero for any interaction between bodies.