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3- Newton's law of gravity
للثقالة نيوتن قانون
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Galileo Galilei(1564-1641)
Using a telescope he Using a telescope he made, Galileo made, Galileo observed:observed:
Moons of Jupiter.Moons of Jupiter.Phases of Venus.Phases of Venus.
His findings supported His findings supported a Copernican model.a Copernican model.He spent the end of He spent the end of his life under “house his life under “house arrest” for his beliefs.arrest” for his beliefs.
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Johannes Kepler German astronomer (1571–1630)
Kepler has try to deduce a mathematical model for the motion of the planets.
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Isaac Newton (1642-1727).
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• " 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 ".
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يجذب • الكون في جسيم أيتتناسب بقوة آخر جسيم أيكتلتيهما مضروب مع & طرديا
مربع مع عكسيا وتتناسب. بينهما فيما المسافة
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If the particles have masses m1 and m2 and are separated by a distance r, the magnitude of this gravitational force is:
الجسيمان كتلة كانت و m1إذاm2 مسافة يفصالهما ، rوكان
مقدار التثاقل فإن : يكون قوة
2
21
g r
mm GF
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• 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:
يسمى Gحيث عام العام ثابت التثاقل ثابت . قيمة تعتمد & معمليا مقاس نظام Gوهو على
النظام في وقيمته المستخدمة الوحدات:الدولي
G = 6.672 x 10-11 N. m2 / kg2
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• The force law is an:
inverse-square law
العكسي التربيع قانون
because the magnitude of the force varies as the inverse square of the separation of the particles.
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• We can express this force in vector form اتجاهي by شكلdefining a unit vector
r12 الوحدة متجه
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• 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
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• 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.
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• That is these forces form an action-reaction pair ورد الفعل قوى من زوجالفعل
F12 = F21
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• 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
كتلة • أن لو كما القوى تعملمركزها الكرة في .مركزة
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• 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 األرض and كتلةRE is the Earth's radius قطر نصف األرض
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• This force is directed toward the center of the Earth
األرض مركز نحو موجهة
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• At points inside the earth:
We would find that the force decreases as we approach the center.
Exactly at the center the gravitational force on a body would be zero.
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4-Measurement of the gravitational constantالعام قياس التثاقل ثابت
The universal gravitational constant, G, was measured by Henry Cavendish in 1798
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5- Weight and gravitational forceالتثاقل وقوة الوزن
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 )
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2
E
E
R
mGg
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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.
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• 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 )3
= 5500 kg/m3 = 5.5 g/cm3
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• 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.
من • أعلى قيمتها الكثافة تلك أن بماعلى الصخرية المواد معظم كثافة
أن ذلك من نستنتج فإننا األرضكثافة له لألرض الداخلي القلب
لكثافة المتوسطة القيمة من أعلى األرض.
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• The magnitude of the gravitational force acting on this mass is:
• Fg = G ( ME m / r2 )
= G ( ME m / ( RE + h )2 )
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• If the body is in free-fall, then
Fg = mg' and we see that g', the free-fall acceleration experienced by an object at the altitude h, is
g' = G mE / r2
= G mE / ( RE + h ) 2
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• Thus, it follows that g' decreases with increasing altitude الجاذبية تقل عجلة
األرض سطح عن ارتفعنا . كلما
• Since the true weight of an object is mg , we see that as r →∞, the true weight approaches zero.
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6- The Gravitational Field مجال التثاقل
• 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.
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• the gravitational field is defined by:
g = Fg / m
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• 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).
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• Since the gravitational force on the object has a magnitude :
(G mE m) / r2
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field g is
rr
mG
m
Fg
2
Eg
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
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• We have used the same symbol الرمز g for gravitational field نفس
magnitude that we used earlier for the acceleration of free fall.
The units of the two quantities are the same نفس لهما الكميتان. الوحدات
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Example• 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.
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• 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 magnitude.
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• Δg = (G ΔM) / r2 = (G ΔM) / (x2 + a2)• The component of this field along the x-
axis is given by : Δgx = - Δg cosφ =• - G ΔM . x • x2 + a2 (x2 + a2) ½
• = - G ΔM x• (x2 + a2)3/2
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• we simply sum all the ΔM 's. This sum is equal to the total mass M.
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23
22x
)ax(
xMGg
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7-Gravitatiuonal Potential energy الوضع للجاذبية طاقة
• 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
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We compute the work Wgrav done by the gravitational force when r changes
from r1 to r2
2
1
r
r
rgravdrFW
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Thus Wgrav is given by:
1
E
2
E
r
r
2Egrav r
mmG
r
mmG
r
drmmGW
2
1
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• 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:
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r
mmGU E
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8- Kepler's laws كبلر قوانين
•The complete analysis is summarized in three statements, known as Kepler's laws:
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• l. All planets move in elliptical orbits with the Sun at one of the focal points. شكل على مدارات في تتحرك الكواكب كل
بؤرتيه إحدى في الشمس تقع ناقص .قطع
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• 2. The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals.
و • الشمس بين ما المخطوط القطري المتجهأزمنة في متساوية مساحات يمسح ما كوكب
.متساوية
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• 3. The square of the orbital period of any planet is proportional the cube of the semi major axis of the elliptical orbit.
مع • يتناسب كوكب ألي الدوري الزمن ربعشكل على الذي للمدار األفقي المحور مكعب
ناقص . قطع