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 ϴFor small angles sin equal angleF=-kx and F=- mgϴ , ϴ=X/LK=mg/L or k/m=g/L

S.H.M

The time constant T derived as follows-:

F=-mgx/L = -mω2r

ω2=g/Lω =2πf 4π2f2=g/L

f2 =(1/4) π2 (g/L) f=(1/2π)

The time period T = 1/f

T=2π

5Gravityالمحاضرة Kepller’s laws:

1. The law of orbits:- All planets move in elliptical orbits with the sun at one focus of the ellipse2.The law of areas:- A line that connect a planet to the sun sweeps out equal areas at equal times3.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

GravityNewton’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

GravityWhen a mass m rotates about a heavy mass M the force of attraction (gravity) is

F=GmM/r2

It has an equivalent inertial forcemv2/r=mω2r

Sun M

r

starv

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

Rm

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

R F=GmM/r2 if r ≥

GRAVITYEX1

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 anotherSolThe 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 isFearth =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

gravityEX2

What is the radius of rotation of a synchronous satellite so that it rotates with about the earth

axis with the same angular velocity?

SolSince satellite must have the same periodic time of earth

T=24x60x60= sT2= ( 4л2/GMearth)r3

Get r =4.23x107

gravityEx3

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

gravityGravitational 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

g at surface of the earth = 9.81 m/s^2 

GRAVITYGravitatational potential energy UIt Is the work needed by the forces of gravity to locatea 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

gravityEX 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)

gravityThe field at point C is given as the vector sum of two fields

g=g1+g2 

abs gt=

g1=Gx600 /152 =2.67Gg2 =Gx800 / 202 = 2.0G

gt=G =2.22X10-10

The potential energy is obtained if a unit mass is located at the point C

 

gravityU600 = -Gx1x600 /15 = - 40G

U800 = - Gx1x800 / 20 = -40 G

Ut =- G (40+40) = - 80 G = -5.34X109 J

Conservation law of energyThe summation of types of energies in a system is always constantFor mechanical energy the sum of potential energy

and kinetic energy of a body is always constant = Constant

gravitySatellitesThey 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)

satellitesAsynchronous 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.

satellitesEscape 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 earthRe : radius of earth