3-Newton`s law of gravity قانون نيوتن للثقالة

Download Report

Transcript 3-Newton`s law of gravity قانون نيوتن للثقالة

3- Newton's law of gravity
‫قانون نيوتن للثقالة‬
Galileo Galilei
Using a telescope he
made, Galileo
Moons of Jupiter.
Phases of Venus.
His findings supported
a Copernican model.
He spent the end of
his life under “house
arrest” for his beliefs.
Johannes Kepler
German astronomer (1571–1630)
Kepler has try to
deduce a
model for the
motion of the
Isaac Newton (1642-1727).
• " Every particle in the Universe
attracts every other particle with a
force that is directly proportional
to the product of their masses and
inversely proportional to the
square of the distance between
them ".
‫• أي جسيم في الكون يجذب أي جسيم‬
‫آخر بقوة تتناسب طرديا ً مع‬
‫مضروب كتلتيهما وتتناسب عكسيا‬
‫مع مربع المسافة فيما بينهما‪.‬‬
If the particles have
masses m1 and m2 and
are separated by a
distance r, the
magnitude of this
gravitational force is:
m2 ‫ و‬m1 ‫إذا كانت كتلة الجسيمان‬
‫ فإن‬، r ‫وكان يفصالهما مسافة‬
:‫مقدار قوة التثاقل يكون‬
m1 m2
• where G is a universal constant called the
universal gravitational constant which
has been measured experimentally.
• The value of G depends on the system of
units used, its value in SI units is:
.ً ‫ ثابت عام يسمى ثابت التثاقل العام وهو مقاس معمليا‬G‫حيث‬
‫ على نظام الوحدات المستخدمة وقيمته في‬G‫تعتمد قيمة‬
:‫النظام الدولي‬
G = 6.672 x 10-11 N. m2 / kg2
• The force law is an:
inverse-square law
‫قانون التربيع العكسي‬
because the magnitude of the
force varies as the inverse square
of the separation of the particles.
• We can express this force in
vector form ‫ شكل اتجاهي‬by
defining a unit vector
r12 ‫متجه الوحدة‬
• Because this unit vector is in the
direction of the displacement vector
r12 directed from m1 to m2, the force
exerted on m2, by m1 is :
F21 = - (G ( m1 m2 ) / r122 ) r12
• Likewise, by Newton's third law
the force exerted on m1 by m2,
designated F12, is equal in
magnitude to F21 and in the
opposite direction.
• That is these forces form an actionreaction pair ‫زوج من قوى الفعل ورد الفعل‬
F12 = F21
• the gravitational force exerted by a
finite-size, spherically symmetric
mass distribution on a particle
outside the sphere is the same as
if the entire mass of the sphere
were concentrated at its center
‫• تعمل القوى كما لو أن كتلة الكرة‬
. ‫مركزة في مركزها‬
• For example, the force exerted by the
Earth on a particle of mass m at the
Earth's surface has the magnitude
Fg = G ( mE m ) / RE2
mE is the Earth's mass ‫ كتلة األرض‬and
RE is the Earth's radius ‫نصف قطر األرض‬
• This force is directed toward the
center of the Earth
‫موجهة نحو مركز األرض‬
• At points inside the earth:
We would find that the force
decreases as we approach the
Exactly at the center the
gravitational force on a body
would be zero.
4-Measurement of the gravitational constant
‫قياس ثابت التثاقل العام‬
The universal gravitational constant, G, was
measured by Henry Cavendish in 1798
5- Weight and gravitational force
‫الوزن وقوة التثاقل‬
If g is the magnitude of the free-fall
acceleration, and since the force on a
freely falling body of mass m near the
surface of the Earth is given by
F = m g, we can equate
m g = G ( mE m / RE2 )
G mE
Using the facts that g = 9.80 m/s2 at
the Earth's surface and RE = 6.38 x
10)6( m, we find that
mE = 5.98 x 10)24( kg.
• From this result, the average density
of the Earth is calculated to be :
• ρE = mE / VE = mE / ( 4/3 π RE3 )
= 5.98 x 10 24
( 4/3 π 6.38 x 106 m
= 5500 kg/m3 = 5.5 g/cm3
• Since this value is higher than
the density of most rocks at the
Earth's surface (density of
granite =
3 g/cm3),
• we conclude that the inner core
of the Earth has a density much
higher than the average value.
‫• بما أن تلك الكثافة قيمتها أعلى من كثافة‬
‫معظم المواد الصخرية على األرض فإننا‬
‫نستنتج من ذلك أن القلب الداخلي لألرض له‬
‫كثافة أعلى من القيمة المتوسطة لكثافة‬
• The magnitude of the gravitational
force acting on this mass is:
• Fg = G ( ME m / r2 )
= G ( ME m / ( RE + h )2 )
• If the body is in free-fall, then
Fg = mg' and we see that g', the freefall acceleration experienced by an
object at the altitude h, is
g' = G mE / r2
= G mE / ( RE + h ) 2
• Thus, it follows that g' decreases with
increasing altitude ‫تقل عجلة الجاذبية كلما‬
‫ ارتفعنا عن سطح األرض‬.
• Since the true weight of an object is
mg , we see that as r →∞, the true
weight approaches zero.
6- The Gravitational Field ‫مجال‬
• When a particle of mass m is
placed at a point where the
field is the vector g, the
particle experiences a force
Fg = m g.
• the gravitational field is defined
g = Fg / m
• consider an object of mass m
near the Earth's surface.
The gravitational force on the
object is directed toward the
center of the Earth and has a
magnitude (m g).
• Since the gravitational force on
the object has a magnitude :
(G mE m) / r2
field g is
G mE
where r is a unit vector pointing
radially outward from the Earth, and
the minus sign indicates that the field
points toward the center of the Earth
and is always opposite to r
• We have used the same symbol
‫ نفس الرمز‬g for gravitational field
magnitude that we used earlier for
the acceleration of free fall.
The units of the two quantities are
the same ‫ الكميتان لهما نفس الوحدات‬.
• A ring-shaped body with radius a
has total mass M. Find the
gravitational field at point p, at a
distance x from the center of the
ring, along the line through the
center and perpendicular to the
plane of the ring.
• We imagine the ring as being
divided into small segments Δs,
each with mass ΔM. At point P
each segment produces a
gravitational field Δg with
• Δg = )G ΔM( / r2 = )G ΔM( / )x2 + a2)
• The component of this field along the xaxis is given by : Δgx = - Δg cosφ =
• - G ΔM .
• x2 + a2
(x2 + a2) ½
= - G ΔM x
(x2 + a2)3/2
• we simply sum all the ΔM 's.
This sum is equal to the total
mass M.
GM x
(x  a )
7-Gravitatiuonal Potential
energy‫طاقة الوضع للجاذبية‬
• we know that the earth's
gravitational force on a body of
mass m, at any point outside
the earth, is given by
• w = fg = (G m mE ) / r2
We compute the work Wgrav done by the
gravitational force when r changes
from r1 to r2
W grav
Fr dr
Thus Wgrav is given by:
W grav
 G m mE
G m mE
G m mE
• Wgrav = U1 - U2 where U1 and U2
are the potential energies of
positions 1 and 2 . So Comparing
this with the eq. of Wgrav gives:
G m mE
8- Kepler's laws ‫قوانين كبلر‬
•The complete analysis is
summarized in three statements,
known as Kepler's laws:
• l. All planets move in elliptical orbits with
the Sun at one of the focal
‫كل الكواكب تتحرك في مدارات‬
.‫على شكل قطع ناقص تقع الشمس في إحدى بؤرتيه‬
• 2. The radius vector drawn from the Sun
to a planet sweeps out equal areas in
equal time intervals.
‫• المتجه القطري المخطوط ما بين الشمس و كوكب ما يمسح‬
. ‫مساحات متساوية في أزمنة متساوية‬
• 3. The square of the orbital period of any
planet is proportional the cube of the semi
major axis of the elliptical orbit.
‫ربع الزمن الدوري ألي كوكب يتناسب مع مكعب‬
. ‫المحور األفقي للمدار الذي على شكل قطع ناقص‬