Transcript المحاضرة 5Gravity

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
Gravity 5 ‫المحاضرة‬
Kepller’s laws:
1. The law of orbits:- All planets move in
elliptical orbits with the sun at one
focus of the ellipse
2. The law of areas:- A line that connect a
planet to the sun sweeps out equal areas at
equal times
3. 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
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
star
r
Sun M
v
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
dr
m
r
F=0
R
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
F=GmM/r2
if
r ≥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
g at 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