Simple pendulum It consist of a small object suspended from the end of a light weight cord. The motion of a simple pendulum swinging back and forth with negligible friction If the restoring force F is proportional to the displacement x ,the motion will be simple harmonic motion. The restoring force is the net force on the end of the bob (mass at the end of the pendulum) and equal to the component of the weight mg tangent to the arc.
S.H.M F=-mgsin ϴ For small angles sin equal angle F=-kx and F=- mgϴ , ϴ=X/L K=mg/L or k/m=g/L
S.H.M The time constant T derived as follows:- F=-mgx/L = -mω2r ω2=g/L ω =2πf 4π2f2=g/L f2 =(1/4) π2 (g/L) f=(1/2π) The time period T = 1/f T=2π
المحاضرة 5Gravity Kepller’slaws: The law of orbits:- All planets move in elliptical orbits with the sun at one focus of the ellipse The law of areas:- A line that connect a planet to the sun sweeps out equal areas at equal times The law of periods:-The square of the period of any planet is proportional to the cube of the semi-major axis of its orbit.
Gravity T2 = Kr3 where k depend on mass of the sun and k= 2.97x10-19 s2 /m3 r : semi major axis of orbit T : time period
Gravity Newton’s Law for gravity :- Between any two points in space, there is forces of attraction . This force depend on the masses and the distance between them so that F=Gxm1xm2/r2 where G is a constant which is called gravitational constant and equal to:- 6.67x 10-11 Nm2/kg2
Gravity When a mass m rotates about a heavy mass M the force of attraction (gravity) is F=GmM/r2 It has an equivalent inertial force mv2/r=mω2r v star m r Sun M
gravity From these forces newton obtained kepller ‘s third law F=GXMXm/r^2 =mω^2 xr T2=(4л2/GM)r3=kr3 where k=(4л2/GM) • Newton ‘s laws is applied to bodies if the distances between them is greater than their dimensions r dr R m F=0
gravity The gravitational forces on a particle of mass m inside a shell of mass M is always zero regardless of the position of the mass m inside. The net gravitational forces on mass m outside another spherical mass M is determined as if M is a point mass at the center F=0 if r <R RF=GmM/r2 if r ≥
GRAVITY EX1 Two trucks of 20 tons total weight are at a distance 5km apart. What is the force of attraction between them? What is the force of gravity with earth? Why we do not feel that they attract to one another Sol The force of attraction between the two trucks is F=Gm1m2/r2
gravity =6.67x10-1120x1000x20x1000/(5x1000)2 =106.72x10-11 The force of gravity between any truck and the earth is Fearth =m1xMearth/(r2) =6.67x10-11x20x1000x5.98x1024/(6.58x106)= =14.8x104 N Its clear that the earth gravitational force is much higher than the attractive force between the two trucks
gravity EX2 What is the radius of rotation of a synchronous satellite so that it rotates with about the earth axis with the same angular velocity ? Sol Since satellite must have the same periodic time of earth T=24x60x60= s T2= ( 4л2/GMearth)r3 Get r =4.23x107
gravity Ex3 Determine the mass of the sun given the distance from the sun is 1.5x1011 m, T given as 3.16x107 sec Sol T2 =(GMsun/4л2) r3 GET Msun T=3.16x107 Msun =2.0x1030 kg
gravity Gravitational Field Ғ=mğ From newton ‘s law the force between two masses is given as: GmM/r2 =mğ g = GM/r2 At the surface of earth r = R of the earth gat surface of the earth = 9.81 m/s^2
GRAVITY Gravitatational potential energy U It Is the work needed by the forces of gravity to locate a mass m in the field of another mass M ∆U=Ub-U∞ = U =-G xMxm/rb Is the work needed by the forces of gravity to locate one kg in the field of another mass M or It ‘s the gravitational potential energy per unit mass u=-GM/rb =U/m J/Kg
GRAVITY u=-GM/rb = U/m J/Kg Gravitational potential energy near the earth’S surface If a mass m is located at height h from the surface of earth then the potential energy is measured from the center of earth and equal to U0 = -GmMe /Re Uh-U0 =(GmMe)/(Re)2.h Uh-U0 =mg0h
gravity EX Two masses 600 and 800 kg apart at a distance of 25m What is the field of gravity at a point 20 m from mass 800 kg and 15 m from 600 kg? (Me =6x 1024 kg ,G = 6.62X10-11 Nm2/kg, Re =6380km)
gravity The field at point C is given as the vector sum of two fields g=g1+g2 abs gt= g1=Gx600 /152 =2.67G g2 =Gx800 / 202 = 2.0G gt=G =2.22X10-10 The potential energy is obtained if a unit mass is located at the point C
gravity U600 = -Gx1x600 /15 = - 40G U800 = - Gx1x800 / 20 = -40 G Ut =- G (40+40) = - 80 G = -5.34X109 J Conservation law of energy The summation of types of energies in a system is always constant For mechanical energy the sum of potential energy and kinetic energy of a body is always constant = Constant
gravity Satellites They are electronic devices designed to work in the space near to earth. There are two types namely: 1- Synchronous Satellites • They are mainly used in TV and internet communications and rotate around the earth with the same angular velocity of earth • They make one revolution around the earth axis in 23hr,56min , 4 sec.
gravity • They all rotate in one orbit over the equator of the earth • Their locations are always fixed to certain area on the earth surface. • The height of these satellites is 22223 mile. • The initial velocity of lunching from earth is determined from the conservation law of energy , so that it reaches its orbit with orbital velocity.
gravity (P.E +K.E)initial =(P.E+KE)final (-GxmxMe /Re )+0.5mV0^2 =(-GmMe/Re+h)+0.5mV^2 V2 =(V0)2 -(2GMe / Re )+ (2GMe /(Re +h)
satellites Asynchronous or polar satellites: • They are not fixed with respect to earth so they may have any orbit at any speed. • They are used for Arial photography , military and weather and research purpose • The radius of their orbits depends on the purpose and the function of the satellite and varies from 970 to 12000 Km.
satellites Escape Velocity Is the velocity which needed a mass m to escape from the gravitational field and reach infinity with zero velocity. Vescape = V0 = Where Me : mass of earth Re : radius of earth