3- Newton's law of gravity قانون نيوتن للثقالة Galileo Galilei (1564-1641)...

3- Newton's law of gravity

للثقالة نيوتن قانون

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.

Johannes Kepler German astronomer (1571–1630)

Kepler has try to deduce a mathematical model for the motion of the planets.

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 ".

يجذب • الكون في جسيم أيتتناسب بقوة آخر جسيم أيكتلتيهما مضروب مع & طرديا

مربع مع عكسيا وتتناسب. بينهما فيما المسافة

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

• 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

• 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.

• We can express this force in vector form اتجاهي by شكلdefining a unit vector

r12 الوحدة متجه

• 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 action-reaction pair ورد الفعل قوى من زوجالفعل

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

كتلة • أن لو كما القوى تعملمركزها الكرة في .مركزة

• 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

• This force is directed toward the center of the Earth

األرض مركز نحو موجهة

• 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.

4-Measurement of the gravitational constantالعام قياس التثاقل ثابت

The universal gravitational constant, G, was measured by Henry Cavendish in 1798

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 )

2

E

E

R

mGg

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 )3

= 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.

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.

• 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 free-fall 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 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.

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.

• the gravitational field is defined by:

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

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

• 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

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.

• 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.

• 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

• we simply sum all the ΔM 's. This sum is equal to the total mass M.

23

22x

)ax(

xMGg

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

We compute the work Wgrav done by the gravitational force when r changes

from r1 to r2

2

1

r

r

rgravdrFW

Thus Wgrav is given by:

1

E

2

E

r

r

2Egrav r

mmG

r

mmG

r

drmmGW

2

1

• 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:

r

mmGU E

8- Kepler's laws كبلر قوانين

•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 points. شكل على مدارات في تتحرك الكواكب كل

بؤرتيه إحدى في الشمس تقع ناقص .قطع

• 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

• 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