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3- Newton's law of gravity قانون نيوتن للثقالة Galileo Galilei (1564-1641) Using a telescope he made, Galileo observed: Moons of Jupiter. Phases of Venus.

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Presentation on theme: "3- Newton's law of gravity قانون نيوتن للثقالة Galileo Galilei (1564-1641) Using a telescope he made, Galileo observed: Moons of Jupiter. Phases of Venus."— Presentation transcript:


2 3- Newton's law of gravity قانون نيوتن للثقالة

3 Galileo Galilei ( ) Using a telescope he made, Galileo observed: 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.

4 Johannes Kepler German astronomer (1571–1630) Kepler has try to deduce a mathematical model for the motion of the planets.

5 Isaac Newton ( ).

6 " 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 ".

7 أي جسيم في الكون يجذب أي جسيم آخر بقوة تتناسب طردياً مع مضروب كتلتيهما وتتناسب عكسيا مع مربع المسافة فيما بينهما.

8 If the particles have masses m 1 and m 2 and are separated by a distance r, the magnitude of this gravitational force is: إذا كانت كتلة الجسيمان m 1 و m 2 وكان يفصلاهما مسافة r ، فإن مقدار قوة التثاقل يكون:

9 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: حيثG ثابت عام يسمى ثابت التثاقل العام وهو مقاس معملياً. تعتمد قيمةG على نظام الوحدات المستخدمة وقيمته في النظام الدولي: G = x N. m2 / kg2

10 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.

11 We can express this force in vector form شكل اتجاهي by defining a unit vector r 12 متجه الوحدة

12 Because this unit vector is in the direction of the displacement vector r 12 directed from m1 to m 2, the force exerted on m 2, by m 1 is : F 21 = - (G ( m 1 m 2 ) / r 12 2 ) r 12

13 Likewise, by Newton's third law the force exerted on m 1 by m 2, designated F 12, is equal in magnitude to F 21 and in the opposite direction.

14 That is these forces form an action- reaction pair زوج من قوى الفعل ورد الفعل F 12 = F 21

15 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.

16 For example, the force exerted by the Earth on a particle of mass m at the Earth's surface has the magnitude Fg = G ( m E m ) / R E 2 m E is the Earth's mass كتلة الأرض and R E is the Earth's radius نصف قطر الأرض

17 This force is directed toward the center of the Earth موجهة نحو مركز الأرض

18 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.

19 4-Measurement of the gravitational constant قياس ثابت التثاقل العام The universal gravitational constant, G, was measured by Henry Cavendish in 1798

20 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 ( m E m / R E 2 )



23 Using the facts that g = 9.80 m/s2 at the Earth's surface and R E = 6.38 x 10)6( m, we find that m E = 5.98 x 10)24( kg.

24 From this result, the average density of the Earth is calculated to be : ρ E = m E / V E = m E / ( 4/3 π R E 3 ) = 5.98 x / ( 4/3 π 6.38 x 106 m )3 = 5500 kg/m3 = 5.5 g/cm3

25 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 الأرض فإننا نستنتج من ذلك أن القلب الداخلي للأرض له كثافة أعلى من القيمة المتوسطة لكثافة الأرض.

26 The magnitude of the gravitational force acting on this mass is: Fg = G ( M E m / r2 ) = G ( M E m / ( R E + h )2 )

27 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 m E / r2 = G m E / ( R E + h ) 2

28 Thus, it follows that g' decreases with increasing altitude تقل عجلة الجاذبية كلما ارتفعنا عن سطح الأرض. Since the true weight of an object is mg, we see that as r →∞, the true weight approaches zero.

29 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.

30 the gravitational field is defined by: g = Fg / m

31 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).

32 Since the gravitational force on the object has a magnitude : (G m E m) / r2

33 field g is 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

34 We have used the same symbol نفس الرمز g 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.


36 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.


38 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.

39 Δg = (G ΔM) / r2 = (G ΔM) / (x2 + a2) The component of this field along the x- axis is given by : Δg x = - Δg cosφ = - G ΔM. x x2 + a2 (x2 + a2) ½ = - G ΔM x (x2 + a2)3/2

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


42 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 m E ) / r2

43 We compute the work W grav done by the gravitational force when r changes from r 1 to r 2

44 Thus W grav is given by:

45 W grav = U 1 - U 2 where U 1 and U 2 are the potential energies of positions 1 and 2. So Comparing this with the eq. of Wgrav gives:


47 8- Kepler's laws قوانين كبلر The complete analysis is summarized in three statements, known as Kepler's laws:

48 l. All planets move in elliptical orbits with the Sun at one of the focal points. كل الكواكب تتحرك في مدارات على شكل قطع ناقص تقع الشمس في إحدى بؤرتيه.

49 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.

50 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 لأي كوكب يتناسب مع مكعب المحور الأفقي للمدار الذي على شكل قطع ناقص.

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