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 & PROFOUND in the 1600's when Sir Isaac Newton first 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 2 nd Law (his 2 nd Law!) plus Calculus, which he also (co-) invented, to PROVE that The orbits of the planets around the sun MUST be ellipses. –For simplicity, we assume in Ch. 5 that these orbits are circular. Ch. 5 fits THE COURSE THEME OF NEWTON'S LAWS OF MOTION because he used his Gravitation Law & his 2 nd Law in 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 EVER written a theoretical expression (physics in math form) & used it to PREDICT something that is in agreement with observations! For this reason, Newton's formulation of his Universal Gravitation Law is 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. This is an EXPERIMENTAL LAW describing the gravitational force of attraction between 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. Newton’s Law of Universal Gravitation

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 Force of Attraction between 2 small masses is the same as the force between Earth & Moon, Earth & Sun, etc.  Must be true from Newton’s 3 rd Law! The gravitational force on you is half of a Newton’s 3 rd 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.

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 masses and inversely proportional to the square of the distance between them: F 12 = -F 21  [(m 1 m 2 )/r 2 ] The direction of this force is along the line joining the 2 masses.  Must be true from Newton’s 3 rd 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 3 rd Law!

9. 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 2 nd Law to get the vector force between the masses. Relate to the angle of rotation & then extract G. This force is written as: G  a constant, G  the Universal Gravitational Constant G is measured & is the same for ALL objects. G must be small!  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 m 1 & m 2 (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 This tells us that the forces form a Newton’s 3 rd Law action-reaction pair, as expected. The negative sign in the above vector equation tells us that particle 2 is attracted toward particle 1 21 = - 21

13. More Comments 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

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 2r E m Earth, mass M E Earth Radius: r E = 6320 km Earth Mass: M E = 5.98  10 24 kg F G = G(mM E /r 2 ) At the surface (r = r E ) F G = weight = mg = G[mM E /(r E ) 2 ] At r = 2r E F G = G[mM E /(2r E ) 2 ] = (¼)mg = 4900 N A spacecraft at an altitude of twice the Earth radius.

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. M E = 5.99  10 24 kg M M =7.35  10 22 kg M S = 1.99  10 30 kg r ME = 3.85  10 8 m r MS = 1.5  10 11 m F = F ME + F MS (vector sum!)

17. F = F ME + F MS (vector sum!) F ME = G [(M M M E )/(r ME ) 2 ] = 1.99  10 20 N F MS = G [(M M M S )/(r MS ) 2 ] = 4.34  10 20 N F = [ (F ME ) 2 + (F MS ) 2 ] (½) = 4.77  10 20 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 is the 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 F G = G[(mM E )/r 2 ]; M E  Earth Mass, r E  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 is zero. 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. 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: Vector Form of the Universal Gravitation Law

20. 3 billiard (pool) balls, masses m 1 = m 2 = m 3 = 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 total gravitational force F on m 1 due to m 2 & m 3. Note: Gravitational force is a vector, so we have to add the vectors F 21 & F 31 to get the vector F (using the vector addition methods of earlier). F = F 21 + F 31 Example: Billiards (Pool) Using components: F x = F 21x + F 31x = 0 + 6.67  10 -11 F y = F 21y + F 31y = 3.75  10 -11 N + 0 So, F = [(F x ) 2 + (F y ) 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. 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∙m 2 /kg 2 ALWAYS at every location anywhere g  The Acceleration due to Gravity g = 9.80 m/s 2 (approximately!) on the Earth’s surface. g varies with location G vs. g

23. Consider an object on Earth’s surface: m E = mass of the Earth (say, known) r E = radius of the Earth (known) m = mass of the object (known) Assume that the Earth is a uniform, perfect sphere. The weight of m is F G = mg The Gravitational force on m is F G = G[(mm E )/(r E ) 2 ] Setting these equal gives: All quantities on the right are measured! g in terms of G m mEmE g = 9.8 m/s 2

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: mEmE m All quantities on the right are measured! Knowing g = 9.8 m/s 2 & the radius of the Earth r E, the mass of the Earth can be calculated:

25. Effective Acceleration Due to Gravity The acceleration due to gravity at a distance r from Earth’s center. Write gravitational force as: F G = G[(mM E )/r 2 ]  mg (effective weight) g  effective acceleration due to gravity. SO : g = G (M E )/r 2 MEME

26. If an object is some distance h above the Earth’s surface, r becomes R E + h. Again, set the gravitational force equal to mg: G[(mM E )/r 2 ]  mg This gives: Altitude Dependence of g MEME This shows that g decreases 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/s 2 Mt. Everest: Altitude: h  8.8 km  g  9.77 m/s 2

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 2 nd Law : ∑F = ma = W – mg W = mg – ma At the pole, no acceleration, a = 0, W = mg At the equator, centripetal acceleration: a R = [(v 2 )/(r E )] = W = mg - m[(v 2 )/(r E )] = mg g = g - [(v 2 )/(r E )] = 0.037 m/s 2 (v = (2πr E )/T = 4.64  10 2 m/s, T = 1 day = 8.64  10 4 s)

29. “Weighing” the Sun! We’ve “weighed” the Earth, now lets “Weigh” the Sun!! (Determine it’s mass). Assume: Earth & Sun are perfect uniform spheres & Earth orbit is a perfect circle (actually its an ellipse). Note: For Earth, Mass M E = 5.99  10 24 kg The orbit period is T = 1 yr  3  10 7 s The orbit radius r = 1.5  10 11 m So, the orbit velocity is v = (2πr/T), v  3  10 4 m/s Gravitational Force between Earth & Sun: F G = G[(M S M E )/r 2 ] Circular orbit is circular  Centripetal acceleration Newton’s 2 nd Law gives: ∑F = F G = M E a = M E a c = M E (v 2 )/r OR: G[(M S M E )/r 2 ] = M E (v 2 )/r. If the Sun mass is unknown, solve for it: M S = (v 2 r)/G  2  10 30 kg  3.3  10 5 M E