Physics 106 Lecture 9 Newton’s Law of Gravitation SJ 7th Ed.: Chap 13.1 to 2, 13.4 to 5

• Historical overview • N’Newton’s inverse-square law of gravi iitation ƒ Force ƒ Gravitational acceleration “g” • Superposition • Gravitation near the Earth’s surface • Gravitation inside the Earth (concentric shells) • Gravitational potential energy ƒ Related to the force by integration ƒ A conservative force means it is path independent ƒ Escape velocity

Gravitation – Introduction Why do things fall? Why doesn’t everything fall to the center of the Earth? What holds the Earth (and the rest of the Universe) together? Why are there stars, planets and galaxies, not just dilute gas?

ƒ Aristotle – Earthly physics is different from celestial physics ƒ Kepler – 3 laws of planetary motion, Sun at the center. Numerical fit/no theory ƒ Newton – English, 1665 (age 23) • Physical Laws are the same everywhere in the universe (same laws for legendary falling apple and planets in solar orbit, etc). • Invented differential and integral calculus (so did Liebnitz) • Proposed the law of “universal gravitation” • Deduced Kepler’s laws of planetary motion • Revolutionized “Enlightenment” thought for 250 years ƒ Reason ÅÆ prediction and control, versus faith and speculation ƒ Revolutionary view of clockwork, deterministic universe (now dated) ƒ Einstein - Newton + 250 years (1915, age 35) ƒ General Relativity – mass is a form of concentrated energy (E=mc2), gravitation is a distortion of space-time that bends light and permits black holes (gravitational collapse). ƒ Planck, Bohr, Heisenberg, et al – Quantum mechanics (1900–27) ƒ Energy & angular momentum come in fixed bundles (quanta): atomic orbits, spin, photons, etc. ƒ Particle-wave duality: determinism breaks down. ƒ There should be a “graviton” (quantum gravity particle). No progress yet. ƒ Current Issues ƒ Dark Matter – Luminous mass of galaxies is too small to explain stars’ orbits. ƒ Dark Energy and inflation – Possible anti-gravity at long range fuels accelerating expansion of the universe, and also early Big Bang.

1 Gravitation – Basic Concepts 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 G m1m2 For a pair of masses: | F12 | = r 2 • Inertial mass: 12 ƒ Measures resistance to acceleration, e.g.: F = ma. ƒ Measures response to gravitational acceleration - a field. • Gravitational mass: ƒ Mass is the source of the gravitational acceleration field - always ƒ Gravitational mass measures strength of a gravitational field produced.

• Duality/Equivalence: ƒ Every bit of mass acts as both inertial and gravitational mass with the same value of m in each role. • Gravitational Force ƒ No contact needed: “action at a distance”. ƒ Cannot be screened out, unlike electrical forces. ƒ Always attractive unlike electrical forces (except for “dark energy”, maybe). ƒ Very weak compared to electrical forces. Too small to notice between most human- scale objects and smaller (e.g., p+ and e- ). Gravitation is long range, has cosmological effects over long times. But it is a weak force on the human scale.

Newton’s Law of Universal Gravitation G 3rd law pair of forces Force on Gravitational Constant F -11 2 2 m1 21 m due to m G=6.67x10 Nm /kg G 2 1 G F12 r12 G G m1m2 Displacement F12 = − rˆ12 m2 2 from m1 to m2 r12 Unit vector • Force is between pairs of point masses • Symmetric in m & m so F = - F G along r12 1 2 12 21 r • Not screened or affected by other bodies ˆ 12 ˆ • Easy to miss between masses near a r12 ≡ G = −r21 third large mass (e.g. on Earth surface) r12

IF m2 IS REMOVED, IS ANYTHING AT POINT 2 DIFFERENT BECAUSE m1 IS STILL AT POINT 1? FIELD transmits the force (no contact, action at a distance) G G G Gm F m g g 1 rˆ Acceleration field 12 ≡ 2 12 ⇒ 12 = − 2 12 r12

Inverse square law: A=4πr2 as sphere grows field field at location Earth (or force) r of m2 due to m1 x area is constant

2 Finding the Value of G

• Henry Cavendish first measured G directly (1798) • Two masses m are fixed at the ends of a light horizontal rod (torsion pendulum) • Two large masses M were placed near the small ones • The angle of rotation was measured • Results were fitted into Newton’s Law

G=6.67x10-11 N.m2/kg2 G versus g: • G is the universal gravitational constant, the same everywhere

• g = ag is the acceleration due to gravity. It varies by location. • g = 9.80 m/s2 at the surface of the Earth

Why was the Law of Gravitation not obvious (except to Newton). How big are gravitational forces between ordinary objects?

G m m 1 Newton is about the force needed to | F |= 1 2 12 2 support 100 grams of mass on the Earth r12

m1 m2 r12 F12 1 kg 1 kg sandwich 1 meter 6.67x10-11 N. a liter of soda 100 kg 100 kg 1 meter 6.67x10-7 N. a person another person

106 kg 106 kg 100 meters 0.67 N.

a ship another ship still hard to detect CliConclusion: • G is very small, so…need huge masses to get perceptible forces

Does gravitation play a role in atomic physics & chemistry?

9.1x10-31 kg 1.7x10-27 kg 5x10-11 meter 4x10-47 N. electron proton orbit radius

3 Superposition: The net force on a point mass when there are many others nearby is the vector sum of the forces taken one pair at a time m G G G G G G 2 r Fon 1 = ∑Fi,1 = F2,1 + F3,1 + F4,1 12 i≠1 G F21 G r All gravitational effects are between 13 m m1 G 3 pairs of masses. No known effects F31 depend directly on 3 or more masses. G G r14 F41 m4 m m = m 4 Example: G 2 3 r' m4 = m5 G G K G m2 r m1 r m3 Fon m1 = m1 gat m1 = 0 by symmetry Gm G g ≡ a = - i r' i,1 gi,1 2 r1,i

m5

For continuous mass G G distributions, integrate Fon 1 = ∫ dF1 mass dist

Numerical Example: y

Find net force on m1 due to m2 and m3 m2 • Use superposition a • BasicG forces atG right angles m3 2a F12 Fnet,j = ∑Fi,j x F m1 i≠ j 13

4 Superposition for a triangle

‰ 9.1. In the sketch, equal masses are placed at the vertices of an equilateral triangle, each of whose sides equals “s”. In which direction would the top-most chunk of mass try to accelerate (ignore the Earth’s gravity) with the bottom two held in place?

m

s s A) B) C) D) E) a = 0 +

s m m

‰ 9.2 . Another chunk of mass is placed at the exact center of the triangle in the sketch. In which direction does it tend to accelerate?

G G Fnet,j = ∑Fi,j i≠ j

Shell Theorem: superposition for masses with spherical symmetry

1. For a test mass OUTSIDE of a uniform spherical shell of mass, the shell’s gravitational force (field) is the same as that of a point mass concentrated at the shell’s mass center m r m r x x

Same for a solid sphe re (e.g., Earth , Sun) via ne sted shell s m r r r x x ++x

2. For a test mass INSIDE of a uniform spherical shell of mass, the shell’s gravitational force (field) is zero

m • Obvious by symmetry for center x x • Elsewhere, need to integrate over sphere

3. For a solid sphere, the force on a test mass INSIDE includes only the mass closer to the CM than the test mass.

m • Example: On surface, measure acceleration g a x distance r from center V = 4 πr 3 • Example: Halfway to center, ag = g/2 sphere 3

5 Gravitation near the surface of the Earth: What do “g” and “weight = mg” mean? m • Earth’s mass acts as like a point mass me at the center (by the Shell Theorem) h • Radius of Earth = re • Object with mass m is at altitude h…

re …above the surface, so r = re+h • Weight W = mag with acceleration ggyiven by Newtons Law of Gravitation (any altitude) me G G me a = − rˆ at any altitude g 2 (re + h)

When m is “on or h << re or, in other words re + h ≈ re near the surface: G G me where 2 g ≡ ag ≅ 2 g ≅ 9.8 m/s re

Example: Use the above to find the mass of the Earth, given: 2 2 3 • g = 9.8 m/s (measure in lab) g re 9.8 x 6370 x 10 -11 3 2 ⇒ me = = • G = 6.67x10 m /kg.s (lab) G 6.67 x 10-11 • re = 6370 km (average - measure) 24 me = 5.98 x 10 kg

Altitude dependence of g

ƒ Weight decreases with altitude h ƒ The work needed to increase Δh declines, since weight decreases

6 Free fall acceleration

9.3 What is the magnitude of the free-fall acceleration at a point that is

a distance 2re above the surface of the Earth, where re is the radius of the Earth?

a) 4.8 m/s2 b) 1.1 m/s2 c) 3.3 m/s2 d) 2.5 m/s2 e) 6.5 m/s2

G me a = at any altitude g 2 (re + h)

g = 9.8 m/s2

Gravitational “field” transmits the force

ƒ A piece of mass m1 placed somewhere creates a “gravitational field” that has values described by some function g1(r) everywhere in space. ƒ Another piece of mass m2 feels a force proportional to g1(r) and in the same direction, also proportional to m . 2 G G G m m F = m g (r) = − 1 2 rˆ Conceppgts for g-fields: 12 2 1 2 12 r12 ƒ No contact needed: “action at a distance”. ƒ Acceleration Field created by gravitational mass transmits the force as a distortion of space that another (inertial) mass responds to. ƒ Gravitational field is “Conservative” (i.e. can have a potential energy function). ƒ g cannot be screened out, unlike electrical fields. ƒ g is always attractive (except cosmologically, maybe), unlike electrical fields. G ƒ The field g (r) is the gravitational force per unit mass G F GM 1 g ˆ gr==−2 created by mass m 1, present at all points whether or not mr there is a test mass m2 located there ƒ The gravitational field vectors point in the direction of the acceleration a particle would experience if placed in the field at each point. Field lines help to visualize strength and direction. ƒ close together Æ strong field, ƒ direction Æ force on test mass

7 Gravitational Potential Energy ΔU

ΔU = mgΔh fails unless ag is constant – force depends on r Work done by G G Definition: potential gravity on a test dW ≡ F • dr = − dU g g energy mass m moved change through dr force varies displacement along path Choose: gravitational potential = zero at R = ∞ where the force = zero i.e.: U(R) = Æ 0 as R Æ ∞. Mass M creates the g-field. Integrate along a radial path from R to infinity

∞ ∞G G ∞GmM ΔU ≡ − dW = − F • dr = − dr ∫R ∫R g ∫R 2 GmM r U = − ∞dr GmM ∞ g = − GmM = | R ∫R r2 r R ƒ The gravitational potential energy between any two particles Note: varies as 1/R. The force varies as 1/R2 NOT R 2 ƒ The potential energy is negative because the force is attractive and we chose the potential energy to be zero at infinite separation. ƒ Another form of energy (external work or kinetic energy) is converted when the potential energy and separation between masses increase.

Gravitational Potential Energy Mutual potential energy of a system of many particles

shared, sum over Utotal = Uij(rij) ∑ all possible pairings all pairs The total gravitational potential energy of the system is the sum over all pairs of particles. Gravitational potential energy obeys the superposition principle

Example m2 3 possible pairs

r12 r23 ⎧ ⎫ ⎪Gm1m2 Gm1m3 Gm2m3 ⎪ Utotal = U12 + U13 + U23 = −⎨ + + ⎬ r13 ⎪ r r r ⎪ m1 m3 ⎩ 12 13 23 ⎭

The slope of theG potentialG energy curve is related to the force. Recall: dW ≡ Fg • dr = − dUg

G Slope of potential force due to dUg energy function gravitation Fg = − (derivative, dr gradient) minus

8 Gravitational Potential Energy, cont

• As a particle moves from A to B, its gravitational potential energy changes by ΔU

• But the mechanical energy remains constant, idindepend ent of path , so l ong as no oth er force is acting

Emech = K + U(r)

• Graph of the gravitational potential energy U versus r for

an object above the Earth’s Emech2 surface • The potential energy goes to zero as r approaches infinity • The mechanical energy may be Emech1 positive, negative, or zero

Conservation of mechanical energy with gravitation

• Emech determines whether motion is bound, free, or at escape threshold

Emech is constant FREE

U =0 r r r Emech = K + Ug(r) g 1 2

Gmem U (r) = − always negative Emech g 1 r Turning KE point r U(r1) 2 For Emech < 0, particle is bound and KE2 = 0 cannot escape. It cannot move beyond

a turning point (e.g., r2) BOUND

For Emech > 0, particle is free. It can

E Gm U = − e reach r = infinity and still have some KE g r left G dUg Emech = 0 is t he escape condi ti on. A F = − particle at any location r would need at g dr least KE = -Ug(r) to move off to the right The tangent to the potential and never return. energy graph measures the How much energy does it cost per kilogram to gravitational force escape completely from the surface of the Earth? Gmex1 Ug = - = 17.4 KWH/kg re

9 Escape speed formula – derivation and example

Escape condition for object of mass m from the surface: 1 2 Gmem Emech ≡ K + Ug = 0 = mvesc - 2 re The mass m cancels:

1 2 Gme 2Gme 0 = vesc - vesc = = 2gre 2 re re

Example: Find the escape speed from the Earth’s surface 2 g = 9.8 m/s re = 6370 km 6 vesc = 2x9.8x6.370x10 = 11,100 m/ s ≈ 7 mi/s Example: Jupiter has 300 times the Earth’s mass and 10 times the Earth’s diameter. How does the escape velocity for Jupiter compare to that for the Earth?

2.G.mjup 2.G.me.300 v jup = = = 30 ve ≈ 5.5 ve rjup re.10

How much does g vary across objects near the earth’s surface?

dr ~ size of object OtiOptiona l Topic

depends on ratio of object size to Earths radius

conclusion: can treat g as constant

10 When is it valid to approximate Ug by mgΔh?

Optional Topic

Answer: when Δh << re

11