Mechanics 2 Flashcards
Kepler’ First Law
All the planets move in elliptical orbits with the Sun at one focus
Kepler’s Second Law
A line joining any planet to the sun sweeps out equal areas in equal times
Kepler’s Third Law
The square of the period of any planet is proportional to the cube of the planets mean distance from the sun
T² = Cr ³
C has almost the same value for all planets
C = 4π²/GMs * r ³
Newton’s Law of Gravity
Every point particle exerts an attractive force on every other point that is proportional to the masses of the two particles and inversely proportional to the square of the distance separating them
F12 = - Gm1m2/ r12²
Gravitational Potential Energy
Gravitational energy for a system consisting of a particle mass m outside a spherical object of mass M and at a distance r from its centre
U(r) = - GMm/r
the potential energy approaches zero as r approaches infinity
Mechanical Energy
The mechanical energy E for a system consisting of a particle of mass m outside a spherically symmetric object of mass M and at a distance r from its centre
E = 1/2 mv² - GMm/r
Escape Speed
For a given value of r, the speed of the particle for which E=0 is called the escape speed, ve
i.e. if v=ve then E=0
Classification of Orbits
if E<0 , then the system is bound and the orbit is an ellipse (or circle which is a type of ellipse)
If E≥0 , then the system is unbound and the orbit is hyperbola (or a parabola if E=0)
Gravitational Field
Definition
g = Fg/m
Gravitational Field Due to Earth
g = - GMe / r² , for r≥Re
Gravitational Field Due to a Thin Spherical Shell
Outside the shell, the field is the same as if all the mass were concentrated at the centre. The field inside the shell is zero.
g = - GM/r² for r>R g = 0 for r
Conditions for the Equilibrium of a Rigid Object
1) the net external force acting on the object is zero.
2) the net external torque about any point must be zero.
Stability of the Equilibrium of a Rigid Object
The equilibrium of an object can be classified as stable, unstable or neutral. An object resting on some surface will be in equilibrium if its centre of gravity lies over its base of support.
Stability can be improved by lowering the centre of gravity or increasing the width of the base.
Gravitational Force and Centre of Gravity
The force of gravity exerted on the various parts of an object can be replaced by a single force, the total gravitational force, acting at the centre of gravity.
For an object in a uniform gravitational field, the centre of gravity coincides with the centre of mass.
Couples
A pair of equal and opposite forces constitutes a couple.
Torque Produced by a Couple
τ = FD
D = the distance between the lines of action of the two forces
Conditions for Static Equilibrium in an Accelerated Reference Frame
1) ΣF = m ,acm where acm is the acceleration of the centre of mass, which is also the acceleration of the reference frame
2) Στcm = 0 ,the sum of external torques about the centre of mass must be zero
Young’s Modulus
γ = Stress / Strain = (F/A) / (ΔL/L)
Shear Modulus
Ms = Shear Stress / Shear Strain = (Fs/A)/(ΔX/L) = (Fs/A) / tanθ
Density
ρ = dm/dV
Density and Temperature Dependence
The densities of most solids and liquids are approximately independent of temperature and pressure whereas those of gases depend strongly on both temperature and pressure
Specific Gravity
The specific gravity of a substance is the ratio of its density to that of another substance, usually water
Pressure
P = F/A
Units of Pressure
1 atm = 760 mmHg = 760 torr = 101.325kPa
1 bar = 100kPa
