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Sect. 5-6: Newton’s Universal Law of Gravitation

Sect. 5-6: Newton’s Universal Law of Gravitation. This cartoon mixes 2 legends: 1. The legend of Newton , the apple & gravity which led to Newton’s Universal Law of Gravitation . 2 2. The legend of William Tell & the apple.

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Sect. 5-6: Newton’s Universal Law of Gravitation

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  1. Sect. 5-6: Newton’s Universal Law of Gravitation This cartoon mixes 2 legends: 1. The legend of Newton, the apple & gravity which led to Newton’s Universal Law of Gravitation. 2 2. The legend of William Tell & the apple.

  2. It was very SIGNIFICANT & PROFOUNDin the 1600's when Sir Isaac Newtonfirst wrote Newton's Universal Law of Gravitation! • This was done at the age of about 30. It was this, more than any of his other achievements, which caused him to be well-known in the world scientific community of the late 1600's. • He used this law, along with Newton's 2nd Law(his 2nd Law!) plus Calculus, which he also (co-) invented, to PROVE that The orbits of the planets around the sun mustbe ellipses. • For simplicity, we assume in Ch. 5 that these orbits are circular. • Ch. 5 fits THE COURSE THEME OFNEWTON'S LAWS OF MOTIONbecause he used his Gravitation Law& his 2nd Lawin his analysis of planetary motion. His prediction that planetary orbits are elliptical is in excellent agreement with Kepler's analysis of observational data & with Kepler's empirical laws of planetary motion.

  3. When Newton first wrote the Universal Law of Gravitation, it was the first time, anyone had EVERwritten a theoretical expression (physics in math form) & used it toPREDICT something that is in agreement with observations! For this reason, Newton's formulation of his Universal Gravitation Lawis considered the BEGINNING OF THEORETICAL PHYSICS. • It also gave Newton his major “claim to fame”. After this, he was considered to be a “major leader” in science & math among his peers. • In modern times, this, plus the many other things he did, have led to the consensus that Sir Isaac Newton was the GREATEST SCIENTIST WHO EVER LIVED

  4. Newton’s Law of Universal Gravitation • This is an EXPERIMENTAL LAWdescribing the gravitational force of attractionbetween 2 objects. • Newton’s reasoning: the Gravitational force of attraction between 2 large objects (Earth - Moon, etc.) is the SAME force as the attraction of objects to the Earth. • Apple story: This is likely not a true historical account, but the reasoning discussed there is correct. This story is probably legend rather than fact.

  5. Newton’s Question:If the force of gravity is being exerted on objects on Earth, what is the origin of that force? Newton’s realization was that this force must come from the Earth. He further realized that this force must be what keeps the Moon in its orbit.

  6. The gravitational force on you is half of a Newton’s 3rd Law pair: Earth exerts a downward force on you, & you exert an upward force on Earth. When there is such a large difference in the 2 masses, the reaction force (force you exert on the Earth) is undetectable, but for 2 objects with masses closer in size to each other, it can be significant.  Must be true from Newton’s 3rd Law! The Force of Attractionbetween 2 small masses is the same as the force between Earth & Moon, Earth & Sun, etc.

  7. By observing planetary orbits, Newton concluded that the gravitational force decreases as the inverse of the square of the distance r between the masses. Newton’s Universal Law of Gravitation: “Every particle in the Universe attracts every other particle in the Universe with a force that is proportional to the product of their massesand inversely proportional to the square of the distance between them: F12 = -F21 [(m1m2)/r2] The direction of this force is along the line joining the 2 masses.  Must be true from Newton’s 3rd Law

  8. The FORCE between 2 small masses has the same origin (Gravity) as the FORCE between the Earth & the Moon, the Earth & the Sun, etc.  From Newton’s 3rd Law!

  9. This force is written as: • G a constant, • G the Universal Gravitational Constant • G is measured & is the same forALL objects. Gmust be small! Newton’s Universal Gravitation Law • Measurement of G in the lab is tedious & sensitive because it is so small. • First done by Cavendish in 1789. • A modern version of the Cavendish experiment: Two small masses are fixed at ends of a light horizontal rod. Two larger masses were placed near the smaller ones. • The angle of rotation is measured. • Use Newton’s 2nd Law to get the vector force between the masses. Relate to the angle of rotation & then extract G.  Cavendish Measurement Apparatus

  10. G =the Universal Gravitational Constant • Measurements find, in SI Units: • The force given above is strictly valid only for: • Very small masses m1& m2 (point masses) • Uniform spheres • For other objects: Need integral calculus!

  11. The Universal Law of Gravitation is an example of an inverse square law • The magnitude of the force varies as the inverse square of the separation of the particles • The law can also be expressed in vector form The negative sign means it’s an attractive force • Aren’t we glad it’s not repulsive?

  12. Comments Force exerted by particle 1 on particle 2 Force exerted by particle 2 on particle 1 21 = - This tells us that the forces form a Newton’s 3rd Lawaction-reaction pair, as expected. 21 The negative sign in the above vector equation tells us that particle 2 is attracted toward particle 1

  13. Gravity is a “field force” that always exists between 2 masses, regardless of the medium between them. The gravitational force decreases rapidly as the distance between the 2 masses increases This is an obvious consequence of the inverse square law More Comments

  14. Example 5-10: The Gravitational Force Between 2 People A 50-kg person & a 70-kg person are sitting on a bench close to each other. Estimate the magnitude of the gravitational force each exerts on the other.

  15. Example 5-11: Spacecraft at 2rE • A spacecraft at an altitude of twice the Earth radius. • Earth Radius:rE = 6320 km • Earth Mass:ME = 5.98  1024 kg • FG = G(mME/r2) • At the surface (r = rE) • FG = weight • = mg = G[mME/(rE)2] • At r = 2rE • FG = G[mME/(2rE)2] • = (¼)mg = 4900 N m Earth, mass ME

  16. Example 5-12: Force on the Moon Find the net force on the Moon due to the gravitational attraction of both the Earth & the Sun, assuming they are at right angles to each other. ME= 5.99  1024kg MM=7.35 1022kg MS = 1.99  1030 kg rME = 3.85  108 m rMS = 1.5  1011 m F = FME + FMS (vector sum!)

  17. F = FME + FMS (vector sum!) FME = G [(MMME)/(rME)2] = 1.99  1020 N FMS = G [(MMMS)/(rMS)2] = 4.34  1020 N F = [ (FME)2 + (FMS)2](½) = 4.77  1020 N tan(θ) = 1.99/4.34  θ = 24.6º

  18. Gravitational Force Due to a Mass Distribution • It can be shown with integral calculus that: The gravitational force exerted by a spherically symmetric mass distribution of uniform density on a particle outside the distribution isthe same as if the entire mass of the distribution were concentrated at the center. • So, assuming that the Earth is such a sphere, the gravitational force exerted by the Earth on a particle of mass m on or near the Earth’s surface is FG = G[(mME)/r2]; MEEarth Mass, rEEarth Radius • Similarly, to treat the gravitational force due to large spherical shaped objects, can show with calculus, that: 1) If a (point) particle is outside a thin spherical shell, the gravitational force on the particle is the same as if all the mass of the sphere were at center of the shell. • 2) If a (point) particle is inside a thin spherical shell, the gravitational force on the particle iszero. So, we can model a sphere as a series of thin shells. For a mass outside any large spherically symmetric mass, the gravitational force acts as though all the mass of the sphere is at the sphere’s center.

  19. Vector Form of the Universal Gravitation Law In vector form, The figure gives the directions of the displacement & force vectors. If there are many particles, the total force is the vector sum of the individual forces:

  20. Example: Billiards (Pool) • 3 billiard (pool) balls, masses m1 = m2 = m3 = 0.3 kg on a table as in the figure. Triangle sides: a = 0.4 m, b = 0.3 m, c = 0.5 m. Calculate the magnitude & direction of the totalgravitational force F on m1due to m2& m3. Note: Gravitational force is a vector, so we have to add the vectorsF21 & F31to get the vector F(using the vector addition methods of earlier). F = F21 + F31 Using components: Fx = F21x + F31x = 0 + 6.67  10-11 Fy = F21y + F31y = 3.75  10-11 N + 0 So,F = [(Fx)2 + (Fy)2]½ = 3.75  10-11 N tanθ = 0.562, θ = 29.3º

  21. Sect. 5-7: Gravity Near the Earth’s Surface The Gravitational Acceleration g g and d The Gravitational Constant G

  22. G vs. g • Obviously, it’s very important to distinguish between G and g! • They are obviously very different physical quantities! • G  The Universal Gravitational Constant • It is the same everywhere in the Universe G = 6.673  10-11 N∙m2/kg2 ALWAYS at every location anywhere • gThe Acceleration due to Gravity g = 9.80 m/s2(approximately!) on the Earth’s surface. g varies with location

  23. g in terms of G m • Consider an object on Earth’s surface: mE = mass of the Earth (say, known) rE = radius of the Earth (known) m = mass of the object (known) • Assumethat the Earth is a uniform,perfectsphere. The weight of m is FG = mg • The Gravitational force on m is FG = G[(mmE)/(rE)2] Setting these equal gives: All quantities on the right are measured! mE g = 9.8 m/s2

  24. Using the same process, we can “Weigh” the Earth! (Determine it’s mass). On Earth’s surface, equate the usual weight of mass m to the Newton’s Gravitation Law form for the gravitational force: m mE Knowing g = 9.8 m/s2 & the radius of the Earth rE, the mass of the Earth can be calculated: All quantities on the right are measured!

  25. Effective Acceleration Due to Gravity • The acceleration due to gravity at a distance r from Earth’s center. • Write gravitational force as: FG = G[(mME)/r2]  mg (effective weight) g effective acceleration due to gravity. SO :g = G (ME)/r2 ME

  26. Altitude Dependence of g • If an object is some distance h above the Earth’s surface, r becomes RE + h. Again, set the gravitational force equal to mg: G[(mME)/r2]  mg This gives: ME • This shows that gdecreases with increasing altitude • As r ® , the weight of the object approaches zero • Example 5-13,g on Mt. Everest

  27. Altitude Dependence of g Lubbock, TX: Altitude: h  3300 ft  1100 m  g  9.798 m/s2 Mt. Everest: Altitude: h  8.8 km  g  9.77 m/s2

  28. Example: Effect of Earth’s Rotation on g Assuming Earth is perfect sphere, determine how Earth’s rotation affects the value of g at equator compared to its value at poles. Newton’s 2nd Law: ∑F = ma = W – mg W = mg – ma At the pole, no acceleration, a = 0, W = mg At the equator, centripetal acceleration: aR = [(v2)/(rE)] = W = mg - m[(v2)/(rE)] = mg g = g - [(v2)/(rE)] = 0.037 m/s2 (v = (2πrE)/T = 4.64  102 m/s, T = 1 day = 8.64  104 s)

  29. “Weighing” the Sun! • We’ve “weighed” the Earth,now lets “Weigh” the Sun!!(Determine it’s mass).Assume: Earth & Sun are perfectuniform spheres & Earth orbit is a perfect circle (actually its an ellipse). • Note: For Earth, Mass ME= 5.99  1024kg The orbit period is T = 1 yr  3 107 s The orbit radius r = 1.5  1011 m So, the orbit velocity is v = (2πr/T),v  3 104 m/s • Gravitational Force between Earth & Sun: FG = G[(MSME)/r2] Circular orbit is circular  Centripetal acceleration Newton’s 2nd Law gives: ∑F = FG = MEa = MEac = ME(v2)/r OR:G[(MSME)/r2] = ME(v2)/r. If the Sun mass is unknown, solve for it: MS = (v2r)/G  2  1030 kg  3.3  105ME

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