# Derivations of Formula

## Unit 1 : Rotational Motion and Astrophysics

### Integration of Acceleration with respect to time :- ________________________________________________________________________

### Integration of Velocity with respect to time :- ________________________________________________________________________

### Square both sides of Equation of Motion 1, then substitute in Equation of Motion 2 :- ________________________________________________________________________

### Equation 4 - Radial Acceleration ### The time (Δt) can be found using the following formula :- ### The Change in Velocity (Δv) can be found by the following approximation :-  ### The Average Acceleration of the Particle can then be found by the following :- ### The above formula gives the Acceleration over a large time period. If the value of θ is made to tend towards zero, the value of sinθ ≈ θ and the Instantaneous Acceleration at a point can be found :-  ________________________________________________________________________

### Equation 5 - Centripetal Force ________________________________________________________________________

### The diagram below shows an object of mass m rotating around a central point P at a distance of r. ### Gravitational Force :- ### Centripetal Force :- ### By combining the above formulae together and by then substituting Velocity, the Period can be found as follows :- _______________________________________________________________________

### However, by convention, we state that Gravitational Potential Energy is equal to 0 J at infinity, and as such , the formula is given as :- _______________________________________________________________________

### The Kinetic Energy of a satellite can be found be deriving a form of Ek = 1/2mv2 , based upon a Centripetal Force calculation :- _______________________________________________________________________

### The combined Energy of a satellite can be found by the addition of the Kinetic Energy and the Gravitational Potential Energy :- _______________________________________________________________________

### The Potential Energy of an Object in a Gravitational field is given by :- ### As the Gravitational Potential Energy is defined as equal to 0 J at Infinity, therefore to escape the Gravitational Potential Well, the object must be given Energy equal to :- ### Ek + Ep = 0 _______________________________________________________________________

### Equation 12 - Schwartzchild Radius (Event Horizon of a Black Hole) _______________________________________________________________________

## Unit 2 : Quanta and Waves

### The displacement on the vertical axis is equal to :- ### By differentiation, an equation of the velocity can be obtained :- _______________________________________________________________________

### By differentiation of SHM Velocity, the acceleration can also be found :- _______________________________________________________________________

### By combining the two equations for SHM Velocity and Acceleration and by using the trig identity below, the following can be derived :-   _______________________________________________________________________

### The velocity of a particle undergoing SHM was found by :- ### If this formula is combined with the Kinetic Energy formula, a method to find the Kinetic Energy of the particle at any given point in its motion can be derived :- _______________________________________________________________________

### If an Ideal system (no Energy loss) ### Therefore at the equilibrium point of the oscillation, Etot = Ek + 0 it can be shown that :- ### By substituting into the general equation above, the Potential Energy can be shown by :- _______________________________________________________________________

### y = a sin 2 π ( ft - x/λ)

_______________________________________________________________________

### If we apply Snell's Law a formula for the Brewster's Angle can be derived :- _______________________________________________________________________

## Unit 3 : Electromagnetism

### The Electric Field Strength (E) at a given point is defined as the force experienced by a unit positive charge at that point. ### By substituting the value for force gained using Coulomb's Law, the following formula for Electric field strength can be derived :- ### Therefore by substituting into the above Electric field strength formula :- _______________________________________________________________________

### Also previously, the Electric field strength could also be found using :- ### By combining the above formulae an expression for Electrostatic Potential (V) can be derived :- ### Integrating the above gives :-  _______________________________________________________________________

### Equation 24 - Closest Approach of Charged Particles _______________________________________________________________________

### The Magnetic field generated by wire 1 is given by :- ### By substitution of the Force calculation above, the Force applied to wire two can be derived :- ### As these fields are not based upon a point source, an extended description as a base unit is required - the Force per unit length :- _______________________________________________________________________

### By using the Centripetal Force calculations from Unit 1, the radius of curvature of the Charge can be found :-  ### Also, by applying calculations for Angular Velocity from the Circular Motion section of Unit 1, the Period and Frequency of the rotation can be derived :- _______________________________________________________________________