Newtons Law Cartoon Newtons Law Cartoon Funny

Newton's Universal Law of Gravitation This drawing mixes 2 legends: i. The legend of Newton, the apple & gravity which led to The Universal Law of Gravitation. 2. The legend of William Tell & the apple.

• It was very Pregnant & PROFOUND in the 1600'due south when Sir Isaac Newton first wrote Newton'southward Universal Law of Gravitation! • This was washed at the age of well-nigh 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'southward.

Newton'southward Universal Law of Gravitation • Newton used this law, forth with Newton's ii nd Constabulary (his 2 nd Law!) plus Calculus, which he likewise (co-) invented, to Bear witness that The orbits of the planets around the sun MUST exist ellipses. • For simplicity, we assume in the following that these orbits are round.

• The topic of Gravitation fits THE COURSE THEME OF NEWTON'S LAWS OF Move • Newton used his Gravitation Law & his 2 nd Constabulary in his analysis of planetary motion. His prediction that planetary orbits are elliptical is in excellent agreement with Kepler'due south assay of observational data & with Kepler'south empirical laws of planetary move.

• When Newton get-go wrote the Universal Police force of Gravitation, it was the beginning time, anyone had Ever written a theoretical expression (physics in math form) & used information technology to PREDICT something in understanding with observations! • For this reason, Newton'due south formulation of his Universal Gravitation Law is considered Commencement OF THEORETICAL PHYSICS. to be the

The Universal Law of Gravitation • Also gave Newton his major "claim to fame". Afterwards this, he was considered to be a "major leader" in scientific discipline & math amidst his peers. In mod times, this, plus the many other things he did, have led to the consensus that Sir Isaac Newton was the GREATEST SCIENTIST WHO E'er LIVED

Newton's Universal Law of Gravitation • This is an EXPERIMENTAL LAW describing the gravitational force of attraction between 2 objects. • Newton'due south reasoning: The Gravitational Force of Allure betwixt ii large objects (Globe - Moon, etc. ) is the SAME force as the attraction of objects to the Earth.

Newton's Universal Law of Gravitation The Gravitational Force of Allure between 2 large objects (Globe - Moon, etc. ) is the Aforementioned force as the allure of objects to the Earth. Apple Story: • This is probable non a true historical business relationship, only the reasoning discussed there is right. This story is probably fable rather than fact.

Newton's Question: • If the force of gravity is being exerted on objects on Earth, what is the origin of that force? • Newton realized that this strength must come from the Globe. • He further realized that this force is what keeps the Moon in its orbit.

• The gravitational strength on you is half of a Newton's 3 rd Constabulary Pair • Earth exerts a down force on you, & you exert an upward force on World. When in that location is such a big difference in the two masses, the reaction strength (force you exert on the World) is undetectable, but for 2 objects with masses closer in size to each other, it can be pregnant. This must be true from Newton's 3 rd Law!

Always a Newton's 3 rd Law Pair! Newton's Reasoning • The Force of Allure betwixt 2 small masses is the same equally the forcefulness between Earth & Moon, Earth & Sun, etc.

• Past observing planetary orbits & analyzing other's observations, Newton ended that the gravitational force decreases as the changed foursquare of the distance r between the 2 masses. Newton'due south Universal Police of Gravitation: • "Every particle in the Universe attracts every other particle in the Universe with a force that is proportional to the product of the masses & inversely proportional to the square of the altitude betwixt them"

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 & inversely proportional to the square of the distance betwixt them: F 12 = -F 21 [(m ane m 2)/r 2] F 12 must = – F 21 from Newton's 3 rd Law! • The management of this force is along the line joining the two masses.

• The Strength between 2 small masses has the same origin (Gravity) as the Strength between the Earth & the Moon, the Globe & the Sun, etc. From Newton's 3 rd Law!

Newton'southward Universal Gravitation Law • This force is written as: G a abiding G Universal Gravitational Constant • Thou is measured & is the aforementioned for ALL objects. G must exist small-scale!

G Universal Gravitational Constant • Thou is measured & is the same for ALL objects. Grand must be small! • Measurement of G in the lab is irksome & sensitive because it is then small. – First done by Cavendish in 1789. Cavendish Measurement Apparatus

G Universal Gravitational Constant • Modern version of the Cavendish experiment: Two small masses are stock-still at ends of a light horizontal rod. • Two larger masses are placed most the Cavendish smaller ones. The angle of rotation is Measurement nd measured. Utilize Newton's 2 Law to Apparatus get the vector forcefulness between the masses. Relate to the angle of rotation & and then excerpt Yard.

G = Universal Gravitational Constant • Measurements discover, in SI Units: • The strength given above is strictly valid only for: – Very modest (point) masses 1000 one & one thousand 2 – Compatible spheres • For other objects: Demand integral calculus!

• The Universal Law of Gravitation is an instance of an inverse square law – The force magnitude varies every bit the inverse square of the separation of the particles • The police force can be expressed in vector grade. • The negative sign means it'southward an attractive force! Aren't nosotros glad it's not repulsive?

Comments F 12 Force exerted by particle 1 on particle 2 F 21 Forcefulness exerted past particle 2 on particle 1 Its also true that F 21 = - F 12 • This tells us that the forces grade a Newton's 3 rd Police force activity-reaction pair, as expected. • The negative sign in the vector equation tells usa that particle ii is attracted toward particle 1

More Comments • Gravity is a "field force" that always exists between two masses, regardless of the medium between them. The gravitational strength decreases speedily as the altitude between the 2 masses increases. (This is an obvious outcome of the changed foursquare police)

Solar Organisation Data (needed for calculations) Encounter likewise Table 5. 1, page 147 or our volume

Solar Arrangement Information

Example Gravitational Force Between 2 People • A m 1 = fifty -kg person & grand 2 = seventy -kg person are sitting on a bench close to each other. • Guess the magnitude of the gravitational force each exerts on the other.

Case Gravitational Force Between 2 People • A g 1 = 50 -kg person & thousand 2 = seventy -kg person are sitting on a bench close to each other. • Estimate the magnitude of the gravitational force each exerts on the other. • Solution: Suppose r = 1. 0 thou. • The strength is then: F = (Gm 1 m ii)/(r two) = (half dozen. 67 10 -xi)(50)(seventy)/(one)two F ii. 33 10 -7 N • Too pocket-sized to be noticed!

Example: Spacecraft at 2 r. East • A spacecraft of mass 1000 = 2000 kg at an distance of twice the Globe radius. Numbers from tables: Earth, Mass ME m Earth Radius: r. Due east = 6320 km Mass: ME = 5. 98 1024 kg FG = G(1000. ME/r 2) At r = two r. E FG = G[m. ME/(2 r. Due east two )] = 4900 N

Example: Strength on the Moon • Find the cyberspace 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 1024 kg, MM= 7. 35 1022 kg MS = 1. 99 1030 kg r. ME = 3. 85 108 chiliad, r. MS = 1. five 1011 grand F = FME + FMS (A vector sum!)

F = FME + FMS (A vector sum!) FME = K [(MMME)/(r. ME)2] = 1. 99 1020 North FMS = Thousand [(MMMS)/(r. MS)2] = four. 34 1020 N F = [ (FME)2 + (FMS)ii](½) = 4. 77 1020 Due north tan(θ) = one. 99/iv. 34 θ = 24. 6º

Gravitational Force Due to a Mass Distribution • Information technology can be shown with integral calculus that: The gravitational force exerted past a spherically symmetric mass distribution of uniform density on a particle outside the distribution is the same as if the unabridged mass of the distribution were full-bodied at the middle. • Then, assuming that the Globe is such a sphere, the gravitational force exerted by the Globe on a particle of mass one thousand on or almost the Globe's surface is FG = G[(m. ME)/r two] ME Earth Mass, r. East World Radius

FG = G[(m. ME)/r 2] ME Earth Mass, r. E Earth Radius • Similarly, to treat the gravitational force due to large spherical shaped objects, we tin can testify with calculus, that: ane. If a (point) particle is outside a sparse spherical trounce, the gravitational force on the particle is the same every bit if all the mass of the sphere were at middle of the shell. two. If a (bespeak) particle is inside a thin spherical shell, the gravitational force on the particle is cypher. So, we tin model a sphere equally a series of thin shells. For a mass exterior whatsoever large spherically symmetric mass, the gravitational force acts equally though all the mass of the sphere is at the sphere'due south center.

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

Example: Billiards (Pool) • three billiard (pool) balls, masses g 1 = m 2 = g 3 = 0. iii kg on a table as in the effigy. Triangle sides: a = 0. 4 thou, b = 0. 3 thou, c = 0. 5 m. • Calculate the magnitude & management of the total gravitational force F on chiliad 1 due to 1000 2 & m iii. • Note: Gravitational strength 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) • 3 billiard (pool) balls, masses grand 1 = thou 2 = 1000 3 = 0. 3 kg on a table as in the figure. Triangle sides: a = 0. four m, b = 0. 3 m, c = 0. five m. • Calculate the magnitude & direction of the total gravitational forcefulness F on m 1 due to chiliad 2 & 1000 three. • Annotation: Gravitational force is a vector, so nosotros have to add the vectors F 21 & F 31 to get the vector F (using the vector improver methods of before). F = F 21 + F 31 • Using components: Fx = F 21 ten + F 31 x = 0 + half dozen. 67 10 -11 N Fy = F 21 y + F 31 y = three. 75 10 -eleven N • So, F = [(Fx)two + (Fy)ii]½ = 3. 75 ten -xi N tanθ = 0. 562, θ = 29. 3º

Gravity Nearly the Earth's Surface The Gravitational Acceleration g and The Gravitational Constant One thousand

Thousand vs. yard • Plainly, it's very important to distinguish between G and k! – They are clearly very unlike physical quantities! G Universal Gravitational Constant • It'southward the aforementioned everywhere in the Universe G = 6. 673 -11 10 two 2 North∙m /kg ALWAYS at every location anywhere g The Acceleration Due to Gravity • g = ix. 80 thousand/s 2 (approximately!) on the Earth'south surface. g varies with location

thousand in terms of 1000 • Consider a mass m on Earth'south surface: m. E = mass of Earth (say, known) r. E = radius of Earth (known) g = mass of the object (known) • Assume that the Earth is a uniform, perfect sphere. • The weight of thou is FG = mg • The Gravitational strength on m is FG = G[(mm. E)/(r. Eastward)2] • Setting these equal gives: • All quantities on the correct are measured! m m. E yard = 9. viii m/s two

• Using the same process, we can yard "Weigh" the Earth! (Decide it'due south mass). • On Earth'due south surface, equate the usual weight of mass chiliad to the m. East Newton's Gravitation Police class for the gravitational forcefulness: All quantities on the right are measured! • Knowing one thousand = 9. viii m/southward 2 & the radius of the World r. Eastward, the mass of the Earth g. E tin be calculated: chiliad. E vi 1024 kg

Effective Acceleration Due to Gravity • Consider the acceleration due to gravity at a distance r from Earth's center. • Write the gravitational strength every bit: FG = G[(k. ME)/r 2] mg ( "Effective Weight" of k) g Effective acceleration due to gravity. SO : thou = Grand(ME 2 )/r ME

Distance Dependence of g • If an object is some distance h higher up the Earth's surface, r becomes RE + h. Once again, set the gravitational forcefulness equal to mg : Thousand[(grand. ME • This gives: )/r ii] mg ME • This shows that g decreases with increasing altitude • Equally r ® , the weight of the object approaches zero Example, g on Mt. Everest

Altitude Dependence of g Lubbock, TX: Altitude: h 3300 ft 1100 m grand 9. 798 m/s ii Mt. Everest: Altitude: h 8. 8 km g 9. 77 m/s 2

Example: Effect of World's Rotation on g • Presume that World is a perfect sphere. • Determine how its rotation affects the value of thousand at the equator compared to its value at the poles. • Let towards the middle of the Earth exist positive y. nd 2 Newton's Police force is: ∑F = ma. R = mg – West and then West = mg – ma. R = mg – m(v 2/r)

Instance: Effect of Earth'due south Rotation on k Newton's 2 nd Law: ∑F = ma. R = mg – Westward so W = mg – ma. R W = mg – m(v ii/r) • At the pole, v = 0, so W – mg = 0 & W = mg • At the equator, the centripetal acceleration is a. R = [(v 2)/(r. Due east)] • So, the constructive weight of a mass g there is: W = mg – ma. R = mg - g[(v 2)/(r. E)] = mg

Example: Effect of Earth's Rotation on g • Effective weight of mass m at equator: W = mg – ma. R = mg - m[(v 2)/(r. E)] = mg Menstruation = T = ane twenty-four hours = viii. 64 104 s • And then the speed v there is five = (2πr. E)/T = iv. 64 102 g/s • The centripetal dispatch at the equator is a. C = [(v 2)/(r. Eastward)] = 0. 037 m/south 2 • Effective gravitational acceleration at equator: thousand = k - [(v 2)/(r. E)] = 9. 8 - 0. 037 = 9. 763 m/s 2

"Weighing" the Sun! • We've "weighed" the Earth, now lets "Weigh" the Sun!! (Make up one's mind it's mass). • Presume: Earth & Sun are perfect compatible spheres & Earth orbit is a perfect circumvolve (actually its an ellipse). • Note: For Earth: Mass ME = 5. 99 1024 kg The orbit period is T = 1 year 3 107 due south The orbit radius r = 1. 5 1011 k The orbit velocity v = (2πr/T), = five 3 104 thousand/s

"Weighing" the Lord's day! • Annotation: For Earth: Mass ME = v. 99 1024 kg The orbit period is T = i yr 3 107 s The orbit radius r = 1. 5 1011 m The orbit velocity v = (2πr/T), = 5 3 104 m/south • The Gravitational Force between the Earth & the Sun: FG = G[(MSME)/r 2] • Circular orbit Centripetal acceleration • Newton's 2 nd Law for the World gives: ∑F = FG = MEac = ME(v two)/r OR: Thou[(MSME)/r 2] = ME(v 2)/r. • If the Dominicus mass is unknown, solve for it: MS = (5 2 r)/G ii 1030 kg three. three 105 ME

Gravitation and the Moon'due south Orbit • The Moon follows an approximately circular orbit around the Earth • There is a force required for this motion • Gravity supplies the Centripetal Force

The Moon'southward Motion • We've assumed that the Moon orbits a "fixed" Earth – This is a good approximation – Information technology ignores World's motion effectually the Dominicus • The Earth and Moon actually both orbit their heart of mass – Nosotros can recollect of the Earth as orbiting the Moon – The circle of the World'south motility is very small compared to the Moon's orbit

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