Syllabus
Law of motion; conservation of energy and momentum, application to rotating frames, centripetal and Coriolis acceleration; Motion under a central force; Conservation of angular momentum, Kepler's Law; Fields and potentials; Gravitational field and potential due to spherical bodies, Gauss and Poisson equations, gravitational self-energy; Two-body problem; Reduced mass; Rutherford scattering; Centre of mass and laboratory.
Laws of Motion
Law of inertia
The rate of change of momentum is the external force applied
Equal and opposite reaction for every action.
Momentum
In a system of $j$ number of particles, two forces could exist: internal and external. The internal forces cancel each other and only external force is left. The total force on the system will be the rate of change of momentum of the system.
$ \Sigma \frac{d \vec{P_j}}{dt} = \Sigma f_j^{external} \implies f_{external} = \frac{dP}{dt} $
Centre of Mass
$M$ is the mass of the system and $\vec{R}$ is the position vector of the centre of the system. Thus, the centre of mass of the system having $j$ particles with total mass $M$ is given by
$ \vec{R} = \frac{ \Sigma M_j \vec{r_j} }{M} $
position of the centre of mass to the centre of mass is zero. so if no external forces exist and the momentum does not change, thus we will have the centre of mass moving in uniform velocity.
Momentum, flow of mass and motion of rocket
If a rocket moves with a velocity $\vec{v}$ and dust enters it with velocity $\vec{u}$ at the rate of $\frac{dm}{dt}$, the necessary force $\vec{F}$ to keep the rocket in same velocity is given by,
$ \vec{F} = ( \vec{v} - \vec{u} ) \frac{dm}{dt} $
If the rocket of mass $M$ is moving with velocity $\vec{v}$ and it is ejecting mass $\Delta m$ with velocity $\vec{u}$, the rate of change of momentum is given by,
$ \frac{d \vec{P}}{dt} = M \frac{d \vec{v}}{dt} - \vec{u} \frac{dM}{dt} $
as, $ \frac{dm}{dt} = - \frac{dM}{dt} $
For a rocket in free space, the rate of change of momentum is zero, so the final velocity of the rocket is given by,
$ \vec{v_f} = - \vec{u} \ln \frac{M_i}{M_f}$
For a rocket in a gravitational field, the final velocity is given by,
$ \vec{v_f} = \vec{u} \ln \frac{M_i}{M_f} - g t_f $
where $t_f$ is the time to final velocity, so if the time to burn fuel is less, the greater will be the final velocity.
Energy
The work-energy theorem gives that the difference in kinetic energy between two states is work done to move between the system,
$ KE_b - KE_a = W_{ba} $
The escape velocity of any planet is independent of the mass, as given by,
$ V_{escape} = \sqrt{2gR_e} $
where $R_e$ is the radius of the planet.
In a conservative force field, the external forces are zero. The total mechanical energy of the system is given by the sum of potential and kinetic energy. This potential energy depends upon only the object's position in the force field and the kinetic energy depends on the object's velocity. so any loss of potential energy is a gain of kinetic energy and vice versa, thus their sum is conserved, so as the total mechanical energy of the system.
$T E = K E + P E$
Angular Momentum is given by
$ \vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v} $
Torque is given by
$ \vec{N} = \vec{r} \times \vec{F} = \frac{d \vec{L}}{dt} $
For a conservative force $\vec{F}$, its curl must be zero and there will be scalar function $V$ exists such that,
$ \vec{F} = - \nabla V$ and $ \nabla \times \vec{F} = 0$