Understanding Uniform Circular Motion and Gravitation Concepts
School
Concordia University**We aren't endorsed by this school
Course
CHEM 181
Subject
Physics
Date
Dec 12, 2024
Pages
38
Uploaded by DrOxide15789
10/1/221Reminders•Labs start this week!!•Tutorials Tu/W evenings•Office hours W 12:30-1:30•Peer Collab Th 3-5 •ANS #3 due Thursday evening3What was the least clear / most difficult part of the material on uniform circular motion and gravitation? 4
10/1/222Overview: Uniform circular motionDescribing rotation: rotation angle and angular velocityCentripetal acceleration and centripetal forcerole of friction, banked curvesFictitious forcesCentrifugal force, Coriolis force We will NOT use these!!Uniform circular motion in the universe: Newton’s law of gravitationKepler’s lawsPlanetary orbits, satellitesThat means going around in a circle at constant speed 5Describing circular motion࠵?࠵?= path length࠵?= radius࠵? =࠵?࠵?in units of radiansRotation angle: ࠵? =!"࠵?= rate of change of ࠵?= !"#$ %& ’(")*$ %& +,= &’A full revolution: ࠵?#$%%= &’ ""= 2࠵?radiansThis should allow you to convert between radians and degrees if you ever need to!2࠵?rad = 360°࠵? =instantaneous linear velocity (always tangential to the circle!)Angular velocity: or ࠵? = ࠵? ࠵?Period of one revolution: ࠵? =&’ "(=2࠵?࠵?Frequency:࠵? =)*=+&’or ࠵? = 2࠵? ࠵?NOTE units! ࠵? =,-./,࠵?= HzOmega, not w!6
10/1/223A.They are the same since they are on the same (solid) turntableB.The spider’s angular velocity is twice the ladybug’s C.The ladybug’s angular velocity is twice the spider’s D.I’m not sureA ladybug and a spider are sitting on a spinning turntable. The ladybug is halfway to the edge and the spider is at the edge. What can you say about their angular velocities?7A.They are the same since they are on the same (solid) turntableB.The spider’s angular velocity is twice the ladybug’s C.The ladybug’s angular velocity is twice the spider’s D.I’m not sureA ladybug and a spider are sitting on a spinning turntable. The ladybug is halfway to the edge and the spider is at the edge. What can you say about their angular velocities?A solid rotating object has the same angular velocity everywhere!࠵? = 2࠵? ࠵?where ࠵?is the frequency of rotationSome time Δ࠵?later, this is where the bugs would be…Note that their angular displacement Δ࠵?is the same (even though the path lengths are different)࠵? = Δ࠵?/Δ࠵?must be the same for both!8
10/1/224A.They are moving at the same speed since they are on the same (solid) record B.The spider’s speed is twice the ladybug’s speedC.The ladybug’s speed is twice the spider’s speedD.I’m not sureA ladybug and a spider are sitting on a spinning turntable. The ladybug is halfway to the edge and the spider is at the edge. What can you say about their linear velocity relative to the ground?9A.They are moving at the same speed since they are on the same (solid) record B.The spider’s speed is twice the ladybug’s speedC.The ladybug’s speed is twice the spider’s speedD.I’m not sureA ladybug and a spider are sitting on a spinning turntable. The ladybug is halfway to the edge and the spider is at the edge. What can you say about their linear velocity relative to the ground?࠵? = ࠵? ࠵?Since the radius of the spiders motion is much larger than that of the ladybug, and they’re moving at the same angular velocity ࠵?, the spider must have a larger linear velocity ࠵?10
10/1/225Suppose you whirl a yoyo in a circle at constant speed. Is the yoyo (i.e. the plastic part) undergoing any acceleration?A. NoB.YesC.I’m not sure11Suppose you whirl a yoyo in a circle at constant speed. Is the yoyo (i.e. the plastic part) undergoing any acceleration?The velocity hasn’t changed magnitude… but it has changed direction!A. NoB.YesC.I’m not sure12
10/1/226In roughly what directionis the average acceleration during this time interval?࠵?࠵?࠵?࠵?࠵?࠵? = ࠵?࠵?− ࠵?࠵?∝ ࠵?−Taking the limit: as the time interval considered gets smaller and smaller, you’ll find that the instantaneous acceleration points directly at the center of the circleABThe velocity hasn’t changed magnitude… but it has changed direction!CDE. I’m not sure13Uniform circular motion: centripetal accelerationAcceleration always points inwardsCentripetal (or “radial”) acceleration:࠵?࠵?࠵?=࠵?࠵?࠵?࠵?࠵?Δ࠵?=࠵?࠵?Δ࠵?×࠵?࠵?࠵?࠵?࠵? =࠵?7࠵?14
10/1/227While whirling your yoyo at constant speed, you get distracted and accidentally let go of it. What path does the yoyo follow? (Assume all motion is horizontal so you can neglect gravity)You let go now!ABCDEF15While whirling your yoyo at constant speed, you get distracted and accidentally let go of it. What path does the yoyo follow? (Assume all motion is horizontal so you can neglect gravity)You let go now!ABCDEFTangential velocity = initial velocityNo external forces => no acceleration Correct answer: C16
10/1/228Example: acceleration of the moonWhat is the acceleration of the moon in its orbit around the earth? Assume the distance from the earth to the moon is 3.84 × 102m, and the moon’s orbit has a period of 27.3 days.Not to scale!!3.84 × 102m࠵? =࠵?7࠵?Radius of earth/moon negligible⇒ ࠵? = 3.84×102m࠵? =distancetime=orbit circumferenceperiod࠵? =2࠵?࠵?࠵?=2࠵? ×3.84×102m27.3 days × (3600×24sday)࠵? =࠵?7࠵?= 0.00273ms7=13600࠵?An instrumental observation for development of the theory of gravity17Uniform circular motion: centripetal acceleration and forceAcceleration always points inwardsCentripetal (or “radial”) acceleration: ࠵? =8!,Centripetal force: the force required to keep an object of mass ࠵?in uniform circular motion:࠵? = ࠵?࠵? = ࠵?8!,directed inwardThere must be a net force to induce this acceleration!What causes this force? It could be a tension (e.g. the yoyo), friction (e.g. the ladybug and the spider), normal forces, gravitation…the list goes on!NOT centrifugal!!NOTE: “Centripetal” indicates the ROLE played by the force – it’s not a new type of force18
10/1/229A.Near but not exactly at the center B.Right at the edge C.Halfway between the center and the edge D.All positions require the same forceE.I’m not sureA ladybug is crawling around on a spinning turntable. At what location is the largest force required to keep the ladybug spinning with the turntable? 19࠵?9= ࠵?࠵?7࠵?A ladybug is crawling around on a spinning turntable. At what location is the largest force required to keep the ladybug spinning with the turntable? ࠵?9= ࠵?8!,This makes it look like the largest force is as ࠵? → 0…but is the speed ࠵?the same everywhere on the record?࠵? = ࠵?࠵?⇒largest centripetal force required at the largest radius! (answer B)The angular velocity ࠵?is the same everywhere on the turntable…the linear velocity ࠵?is not!20
10/1/2210A.The ladybug will slide offB.The spider will slide offC.The ladybug and spider slide off at the same timeD.The spider eats the ladybugE.I’m not sureA small ladybug and a huge spider (10 x the mass of the ladybug) are again sitting on a spinning turntable, both at the edge (thrill-seekers!). The turntable slowly starts to spin faster and faster. What happens first?You can assume the bugs have the same coefficients of friction on the turntable21Key idea: an object will first slip when the centripetal force required to keep it in circular motion exceeds the maximum static friction force࠵?9= ࠵?࠵?7࠵?࠵?+≤ ࠵?+࠵?࠵?Note: ࠵?!= ࠵?࠵?since no vertical acceleration.Why static friction and not kinetic friction?Slipping occurs when ࠵?7≥ ࠵?!89:࠵?࠵?7࠵? ≥ ࠵?+࠵?࠵?࠵?7≥ ࠵?+࠵?/࠵?Independent of massSo, if two objects sitting on a rotating table have the same coefficient of friction and rotate at the same radius, they will start to slip at the same angular velocity, regardless of their mass ࠵?>࠵?࠵?࠵??Center of turntableDoes this add up to give the centripetal force needed to keep the bug spinning?Free-body diagram viewed from the side:22
10/1/2211A student stands on their bathroom scale as they ride a Ferris wheel. On firm ground, they know the scale reads ‘X’ when they are on it. Which of the following is true as the Ferris wheel rotates uniformly?A.The scale will always indicate ‘X’ –their weight.B.The scale will indicate a larger weight than X at the top, and a smaller weight than X at the bottom. C.The scale will indicate a smaller weight than X at the top, and a larger weight than X at the bottom.D.I’m not sureKey idea: a scale measures normal force (how much you are pushing down on it)23A student stands on their bathroom scale as they ride a Ferris wheel. On firm ground, they know the scale reads ‘X’ when they are on it. Which of the following is true as the Ferris wheel rotates uniformly?A.The scale will always indicate ‘X’ –their weight.B.The scale will indicate a larger weight than X at the top, and a smaller weight than X at the bottom. C.The scale will indicate a smaller weight than X at the top, and a larger weight than X at the bottom.D.I’m not sureKey idea: a scale measures normal force (how much you are pushing down on it)࠵?࠵?࠵?>At the top of the ride there must be a net centripetal force downwards࠵?࠵?࠵?>At the top of the ride there must be a net centripetal force upwardsScale reads ࠵?!< ࠵?࠵?Scale reads ࠵?!> ࠵?࠵?24
10/1/2212One day you are standing in a spinning centrifuge holding a piece of graph paper, and the thought occurs to you that you could potentially describe your motion in a rotating coordinate frame.That will be easy! In the rotating coordinate frame I’m at rest, so there must be no net force on me!Hmm, in this rotating coordinate frame I can feel the normal force of the walls of the centrifuge pushing me inwards. That seems like a net horizontal force. Why am I not accelerating?Wait a second, this rotating coordinate frame is…accelerating…not inertial…Newton’s 2ndlaw doesn’t apply!Maybe I’ll stick to inertial reference frames25Centrifugal forceA ”fictitious force” that arises if you try to write Newton’s 2ndlaw in a rotating reference frameRemember, ࠵? = ࠵?࠵?only applies in inertial (non-accelerating) reference frames! Stick to inertial frames!In this non-inertial reference frame, you’ll find࠵?࠵? = ࠵? +other stuffAcceleration measured in the rotating reference frameSum of all real forces (gravity, contact)Fictictiousforces: centrifugal, coriolisYour carYour doughnutRemember the doughnut on the dashboard that looks like it’s moving as you go around a turn!26
10/1/221327What will actually crush every bone in Mr. Bond’s body?A.Centrifugal forceB.The contact force from the wall of the centrifugeC.The force of gravityD.I’m not sure28
10/1/2214So far in this course, we’ve seen how many real, distinct forces (i.e. causes for forces between objects)?A.Gravity, contactB.Gravity, contact, centripetalC.Gravity, contact, frictionD.Gravity, contact, friction, centripetalE.Gravity, contact, friction, centripetal, centrifugalF.I’m not sure29What are the real, distinct types of forces? TensionFrictionNormal forceGravityCentripetal forceCentrifugal forceCoriolis forceContact forcesThese are just specific types of contact force –they’re all caused by atomic-scale interactions between materials in contact with each otherNo, never! These are not real forces! They are invoked to try to explain apparent motion in accelerating reference framesThis isn’t a force you would draw on a free-body diagram: it’s what we call the net force that induces circular motion. It indicates ROLE, not NATURE of the force. It is composed of gravity and/or contact forces.Every force you draw on a free-body diagram is a gravitational or contact forceBetter to just talk about centripetal acceleration!30
10/1/2215Cornering, friction, and skiddingA car cornering (moving in a circular path) must have an acceleration of ࠵?&/࠵?towards the center of the circle, with a net force (provided by friction) in the same direction.࠵?࠵?࠵?>࠵?@,A9BACDIf ࠵?#"=7>=?@is too small, then ࠵?(your radius of curvature) will increase (you’ll start to skid and leave the road!)If you start to skid, then ࠵?!→ ࠵?A(smaller!) and it’s very difficult to recoverWouldn’t it be nice to have a centripetal force that doesn’t rely on friction? 34Banked curves࠵?࠵?࠵?>࠵?࠵?>࠵?࠵?By banking (angling) the road around a curve, the normal force gains a horizontal component!Can provide a centripetal force even in the absence of friction!࠵?No friction35
10/1/2216Banked curves࠵?࠵?࠵?>࠵?࠵?>࠵?࠵?By banking (angling) the road around a curve, the normal force gains a horizontal component!Can provide a centripetal force even in the absence of friction!࠵?If the car is not accelerating vertically, what is the horizontal component of the normal force?A.࠵?࠵? sin ࠵?B.࠵?࠵?/ cos ࠵?C.࠵?࠵? tan ࠵?D.࠵?࠵? cos ࠵?E.I’m not sureNo friction36Banked curves࠵?࠵?࠵?>࠵?࠵?>࠵?࠵?By banking (angling) the road around a curve, the normal force gains a horizontal component!Can provide a centripetal force even in the absence of friction!࠵?If the car is not accelerating vertically, what is the horizontal component of the normal force?A.࠵?࠵? sin ࠵?B.࠵?࠵?/ cos ࠵?C.࠵?࠵? tan ࠵?D.࠵?࠵? cos ࠵?E.I’m not sureThe horizontal component of the normal force is ⇒ ࠵?B= ࠵?࠵?/ cos ࠵?Idea:The vertical forces cancel࠵?࠵? = ࠵?Bcos ࠵?= ࠵?࠵?sin ࠵?cos ࠵?= ࠵?࠵? tan ࠵?࠵?Bsin ࠵?No friction࠵?>cos ࠵?࠵?>sin ࠵?37
10/1/2217Reminders•Labs start this week!!•Read the lab manual and general lab information documents ahead of time!•Peer Collab Th 3-5 •ANS #3 due Thursday evening•Monday Oct 3 is a holiday•No Peer Collab•M Lab sections moved to other slots next week•Perusall #5 due Tues Oct 4 1pm38Last timeAcceleration always points inwards࠵? = ࠵? ࠵?= ࠵?&࠵?࠵? = ࠵?&/࠵?Newton’s second law: ࠵?࠵?࠵?࠵?= ࠵? ࠵?We call the net force that induces circular motion a “centripetal” forceIt is composed of the sum of the real forces acting on the body (gravity, contact forces)࠵? = 2࠵? ࠵? =2࠵?࠵?Typical problem solving steps:1.Draw a free body diagram (do NOT include centripetal force on your FBD!)2.Apply Newton’s second law, setting total force equal to mass*centripetal acceleration39
10/1/2218Banked curves࠵?࠵?࠵?>࠵?࠵?࠵?By banking (angling) the road around a curve, the normal force gains a horizontal component!Can provide a centripetal force even in the absence of friction!࠵?>࠵?The horizontal component of the normal force is ⇒ ࠵?!= ࠵?࠵?/ cos ࠵?The vertical forces cancel࠵?࠵? = ࠵?!cos ࠵?= ࠵?࠵?sin ࠵?cos ࠵?= ࠵?࠵? tan ࠵?࠵?!sin ࠵?No friction࠵?>cos ࠵?࠵?>sin ࠵?NOTE: use vertical/horizontal coordinate system (not at angle ࠵?) since we know the net (centripetal) force is horizontal.If we adjust the angle perfectly, the car can take the curve without needing friction!࠵?࠵? tan ࠵? = ࠵?࠵?&࠵?tan ࠵? =࠵?&࠵?࠵?Independent of car mass (thankfully!)“Design speed” of the curve40A perfectly banked curve࠵?࠵?࠵?>࠵?tan ࠵? =࠵?7࠵?࠵?What is the banking angle ࠵?for a highway curve of radius of curvature of 200 m, with a design speed of 100 km/h? =(100kmh×1h3600s×1000mkm)7200 m (9.8ms7)= 0.394⇒ ࠵? = arctan 0.394 = 0.375 ࠵?࠵?࠵? = 21.5°(this would be scarily steep, which would cause people to slow down, which would cause problems (why?), so in reality you won’t see perfectly banked curves on a highway!)43
10/1/2219Challenge problem:Assume a curve of radius ࠵?is perfectly banked – that is, the banking angle ࠵?is given by tan ࠵? = ࠵?&/࠵?࠵?where ࠵?is design speed of the curve. Derive the expression for the minimal required coefficient of static friction, ࠵?!, for a speed ࠵?’ > ࠵?. Show that for ࠵?’ = ࠵?this correctly gives ࠵?!= 0. 44Challenge problem:Assume a curve of radius ࠵?is perfectly banked – that is, the banking angle ࠵?is given by tan ࠵? = ࠵?&/࠵?࠵?where ࠵?is design speed of the curve. Derive the expression for the minimal required coefficient of static friction, ࠵?!, for a speed ࠵?’ > ࠵?. Show that for ࠵?’ = ࠵?this correctly gives ࠵?!= 0. Which free-body diagram will be most helpful in solving this problem?࠵?࠵?࠵?>ABC࠵?O࠵?B:normal force࠵?F:centripetal force࠵?G: static friction force࠵?࠵?࠵?>࠵??࠵?࠵?࠵?>࠵??D࠵?࠵?࠵?>࠵??࠵??E. I’m not sure45
10/1/2220Challenge problem:࠵?࠵?࠵?>ABC࠵?O࠵?B:normal force࠵?F:centripetal force࠵?G: static friction force࠵?࠵?࠵?>࠵??࠵?࠵?࠵?>࠵??D࠵?࠵?࠵?>࠵??࠵??E. I’m not sureKey idea: the horizontal component of the normal force won’t be enough to provide the needed centripetal force at this higher velocity, so you’ll need an inward friction force as wellAnswer: CAssume a curve of radius ࠵?is perfectly banked – that is, the banking angle ࠵?is given by tan ࠵? = ࠵?&/࠵?࠵?where ࠵?is design speed of the curve. Derive the expression for the minimal required coefficient of static friction, ࠵?!, for a speed ࠵?’ > ࠵?. Show that for ࠵?’ = ࠵?this correctly gives ࠵?!= 0. Which free-body diagram will be most helpful in solving this problem?46Challenge problem:What equations will allow you to solve this problem?࠵?࠵?࠵?>࠵??A. Total perpendicular force = 0Total parallel force = ࠵? ࠵?&/࠵?B. Total perpendicular force = 0Total horizontal force = ࠵?࠵?H &/࠵?C. Total vertical force = 0Total parallel force = ࠵? ࠵?&/࠵?D. Total vertical force = 0Total horizontal force = ࠵?࠵?H &/࠵?E. I’m not sureverticalhorizontalperpendicularparallelAssume a curve of radius ࠵?is perfectly banked – that is, the banking angle ࠵?is given by tan ࠵? = ࠵?&/࠵?࠵?where ࠵?is design speed of the curve. Derive the expression for the minimal required coefficient of static friction, ࠵?!, for a speed ࠵?’ > ࠵?. Show that for ࠵?’ = ࠵?this correctly gives ࠵?!= 0. 47
10/1/2221Challenge problem: the physics࠵?࠵?Remember, the centripetal acceleration is horizontal, so you want to work with a vertical/horizontal coordinate frame.࠵? sin ࠵?࠵? cos ࠵?࠵?Gsin ࠵?࠵?Gcos ࠵?Vertical direction: no acceleration࠵?࠵? + ࠵?"࠵? sin ࠵? = ࠵? cos ࠵?Horizontal direction: centripetal acceleration࠵? sin ࠵? + ࠵?"࠵? cos ࠵? = ࠵?࠵?# $࠵?General math approach:solve for ࠵?and substitute it in the other equation, solve for ࠵?",and substitute tan ࠵? =%!&’.Assume a curve of radius ࠵?is perfectly banked – that is, the banking angle ࠵?is given by tan ࠵? = ࠵?&/࠵?࠵?where ࠵?is design speed of the curve. Derive the expression for the minimal required coefficient of static friction, ࠵?!, for a speed ࠵?’ > ࠵?. Show that for ࠵?’ = ࠵?this correctly gives ࠵?!= 0. ࠵?࠵?G࠵?G= ࠵?!࠵?48Challenge problem: the mathVertical direction: no acceleration࠵?࠵? + ࠵?"࠵? sin ࠵? = ࠵? cos ࠵?⇒ ࠵? =࠵?࠵?cos ࠵? − ࠵?"sin ࠵?Horizontal direction: centripetal acceleration࠵? sin ࠵? + ࠵?"࠵? cos ࠵? = ࠵?࠵?# $࠵?⇒࠵?࠵? sin ࠵?cos ࠵? − ࠵?"sin ࠵?+࠵?࠵?࠵?"cos ࠵?cos ࠵? − ࠵?"sin ࠵?= ࠵?࠵?# $࠵?⇒ ࠵? sin ࠵? + ࠵?࠵?"cos ࠵? =࠵?# $࠵?(cos ࠵? − ࠵?"sin ࠵? )⇒࠵?# $࠵?࠵?"sin ࠵? + ࠵?࠵?"cos ࠵? =࠵?# $࠵?cos ࠵? − ࠵? sin ࠵?Now solve for ࠵?"Now substitute ࠵?in the horizontal equation⇒ ࠵?"=࠵?# $࠵?cos ࠵? − ࠵? sin ࠵?࠵?# $࠵?sin ࠵? + ࠵? cos ࠵?=࠵?# $− ࠵?࠵? tan ࠵?࠵?# $tan ࠵? + ࠵?࠵?Now substitute tan ࠵? =%!&’࠵?"=࠵?# $− ࠵?࠵?࠵?$࠵?࠵?࠵?# $࠵?$࠵?࠵?+ ࠵?࠵?=࠵?࠵? (࠵?# $− ࠵?$)࠵?# $࠵?$+࠵?࠵?$Indeed, this goes to zero when ࠵?#= ࠵?49
10/1/2222A car moves at constant speed over a hilly road. Where does the car exert the greatest force on the road?A. Bottom of the hillB. Top of the hillC. On a flat part of the roadD. The force exerted by the car on the road is always the sameE. I’m not sure50A car moves at constant speed over a hilly road. Where does the car exert the greatest force on the road?At the top of the hill: ࠵?>࠵?࠵?Key idea:Newton’s third lawThe magnitude of the force exerted by the car on the road is the same as the magnitude of the force exerted by the road on the car – so let’s examine the force exerted by the road on the carNet centripetal force downwardsAt the bottom of the hill: ࠵?>࠵?࠵?Net centripetal force upwardsOn a flat section:࠵?>࠵?࠵?࠵?!< ࠵?࠵?࠵?!> ࠵?࠵?࠵?!= ࠵?࠵?࠵?!= ࠵?࠵? − ࠵?࠵?$/࠵?Note that the normal force can go to zero if you’re going fast enough! What happens?Correct answer: A 51
10/1/2223Which pulls harder gravitationally? The earth on the moon or the moon on the earth? Which body accelerates more?Not to scale.A.The Earth pulls harder on the Moon, so the Moon accelerates more B.The Moon pulls harder on the Earth, so the Earth accelerates more C.The two pull with the same force on each other, so the accelerations are the same D.None of the aboveE.I’m not sure52Which pulls harder gravitationally? The earth on the moon or the moon on the earth? Which body accelerates more?Not to scale.A.The Earth pulls harder on the Moon, so the Moon accelerates more B.The Moon pulls harder on the Earth, so the Earth accelerates more C.The two pull with the same force on each other, so the accelerations are the same D.None of the aboveE.I’m not sureNewton’s 3rdlaw: they exert equal and opposite gravitational forces on each other…but the moon accelerates a lot more because it has a smaller massAnswer: D53
10/1/2224Newton’s universal law of gravitationA leap of imagination:The force of gravity that we experience here on earth is the same force that keeps the moon in orbitAcceleration of the moon in its orbit (remember we calculated it!) ≈()*++࠵?Distance of moon = 3.84 × 10,m ≈(*+Earth’s radiusIf gravity ࠵?here on earth is the same force that causes the moon’s acceleration, that force must be falling off with distance like ࠵?]"9(∝)"-Other observations:On earth, all masses fall with the same acceleration due to gravity => the force due to gravity must scale with the object’s mass so that acceleration is independent of mass!࠵?]"9(∝8./0123"-࠵?࠵? = ࠵?’&4%= ࠵? ×other stuffThird law: an equal and opposite force is exerted by the object on the earth. By symmetry, the force must also depend on earth’s mass!࠵?]"9(∝8./0123815637"-∝means “proportional to” – i.e. equal to some constant times this54Newton’s universal law of gravitation࠵?]"9(= ࠵?B888-"-The force of gravity exerted by mass ࠵?)on a mass ࠵?&a distance ࠵?away…which is equal to the force exerted by mass ࠵?&on mass ࠵?), of course࠵?B= NewtonHs gravitational constant = 6.67 ×10_))N m&/kg&On earth’s surface, the dominant force of gravity on any object is coming from earth, ࠵?]"9(!$"#97‘= ࠵?B8 815637"15637-= ࠵? ࠵?࠵? = ࠵?B࠵?‘9">a࠵?‘9">a&The relationship between big G and little g:55
10/1/2225Three balls of mass 5M, 2M, and M are shown below. What is the direction of the net gravitational force on the ball of mass 2M due to the two other balls?0 m1 m2 m3 m4 m5 m6 m7 m5M2 M1 MA. B. C. The net gravitational force is zeroD. None of the aboveE. I’m not sure56Three balls of mass 5M, 2M, and M are shown below. What is the direction of the net gravitational force on the ball of mass 2M due to the two other balls?0 m1 m2 m3 m4 m5 m6 m7 m5M2 M1 MA. B. C. The net gravitational force is zeroD. None of the aboveE. I’m not sure࠵?’&4%= ࠵?(2࠵?)1࠵?2࠵?$= ࠵?2࠵?࠵?࠵?$×14࠵?’&4%= ࠵?2࠵?5࠵?4࠵?$= ࠵?(2࠵?)࠵?࠵?$×516j)k>)lso the net force is to the left (B)57
10/1/2226Measuring gravitation~ 10 feetCavendish (1798)Measured gravitational forces between suspended lead balls by observing the twisting motion induced by gravitational attraction~ 5 cmAnd many more…!58Consider a satellite orbiting earth as shown. What does its free-body diagram look like at this point in its orbit?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?9‘DB,A@abcdA. B. C. E. No forces -- it’s weightless! F. I’m not sure࠵?b,c8D. ࠵?,C9e‘B࠵?,C9e‘B59
10/1/2227Consider a satellite orbiting earth as shown. What does its free-body diagram look like at this point in its orbit?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?9‘DB,A@abcdA. B. C. E. No forces -- it’s weightless! F. I’m not sure࠵?b,c8D. ࠵?,C9e‘B࠵?,C9e‘BCorrect answer: DA satellite executing uniform circular motion around the earth has only one force acting on it (to a good approximation): earth’s gravityThat gravitational force must be equal to the satellite’s mass times its centripetal accelerationThat’s all that is needed to keep the satellite “up” in the sky!࠵?b,c8Note that ࠵?’&4%≠ ࠵?࠵?since the satellite could be far from earth’s surface60A note on weightlessness in orbitzAstronauts in orbit feel “weightless” because they are in freefall – there are no contact forces pushing on them (your experience of weight)Objects appear to float because they are in the same orbit as the astronaut (and the spaceship)The astronaut, spaceship, and apple stay in the same position relative to each other as they all “fall” in orbit around the earth61
10/1/2228The International Space Station (ISS) orbits earth at an altitude of approximately ℎ =400 km above sea level. What can you say about the magnitude of its acceleration ࠵?in orbit?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NOT TO SCALE!!ISSA.࠵? = 0B.࠵?is a bit smaller than ࠵?C.࠵?is a bit larger than ࠵?D.࠵?is a much smaller than ࠵?E.࠵?is a much larger than ࠵?F.I’m not sure64The International Space Station (ISS) orbits earth at an altitude of approximately ℎ =400 km above sea level. What can you say about the magnitude of its acceleration ࠵?in orbit?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NOT TO SCALE!!ISSA.࠵? = 0B.࠵?is a bit smaller than ࠵?C.࠵?is a bit larger than ࠵?D.࠵?is a much smaller than ࠵?E.࠵?is a much larger than ࠵?F.I’m not sure࠵?!࠵? ࠵?:࠵?$= ࠵?࠵?࠵? = ࠵?!࠵?:(࠵?:+ ℎ)$Common mistake with satellites: ࠵? = ࠵?:+ ℎ,not ℎ!࠵? = ࠵?!࠵?:࠵?:$Since ℎ ≪ ࠵?:, ࠵?is just a bit smaller than ࠵?࠵?b,c865
10/1/2229࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NOT TO SCALE!!NASA/STScIHSTISSThe International Space Station (ISS) orbits earth at an altitude of approximately ℎ =400 km above sea level. The Hubble Space Telescope (HST) orbits at approximately ℎ =550 km above sea level. What can you say about their relative orbital velocities ࠵?qGG, ࠵?rG*?A.࠵?qGG< ࠵?rG*B.࠵?qGG> ࠵?rG*C.࠵?qGG= ࠵?rG*D. Can’t tell without knowing their massesE.I’m not sureKey idea: For satellites, gravity is the only force acting on them, and it must be equal to their mass times their centripetal acceleration.68The International Space Station (ISS) orbits earth at an altitude of approximately ℎ =400 km above sea level. The Hubble Space Telescope (HST) orbits at approximately ℎ =550 km above sea level. What can you say about their relative orbital velocities ࠵?qGG, ࠵?rG*?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NOT TO SCALE!!NASA/STScIHSTISSA.࠵?qGG< ࠵?rG*B.࠵?qGG> ࠵?rG*C.࠵?qGG= ࠵?rG*D. Can’t tell without knowing their massesE.I’m not sure࠵?࠵?$࠵?= ࠵?!࠵? ࠵?:࠵?$࠵? =࠵?!࠵?:࠵?=࠵?!࠵?:࠵?:+ ℎThe higher up, the lower the velocityCommon mistake with satellites: ࠵? = ࠵?:+ ℎ,not ℎ!69
10/1/2230The International Space Station (ISS) orbits earth at an altitude of approximately ℎ =400 km above sea level. The Hubble Space Telescope (HST) orbits at approximately ℎ =550 km above sea level. What can you say about their relative orbital periods ࠵?qGG, ࠵?rG*?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NOT TO SCALE!!NASA/STScIHSTISSA.࠵?qGG< ࠵?rG*B.࠵?qGG> ࠵?rG*C.࠵?qGG= ࠵?rG*D. Can’t tell without knowing their massesE.I’m not sure࠵?࠵?$࠵?= ࠵?!࠵? ࠵?:࠵?$࠵? =࠵?!࠵?:࠵?=࠵?!࠵?:࠵?:+ ℎThe higher up, the lower the velocityCommon mistake with satellites: ࠵? = ࠵?:+ ℎ,not ℎ!70The International Space Station (ISS) orbits earth at an altitude of approximately ℎ =400 km above sea level. The Hubble Space Telescope (HST) orbits at approximately ℎ =550 km above sea level. What can you say about their relative orbital periods ࠵?qGG, ࠵?rG*?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NOT TO SCALE!!NASA/STScIHSTISSA.࠵?qGG< ࠵?rG*B.࠵?qGG> ࠵?rG*C.࠵?qGG= ࠵?rG*D. Can’t tell without knowing their massesE.I’m not sure࠵?࠵?$࠵?= ࠵?!࠵? ࠵?:࠵?$࠵? =࠵?!࠵?:࠵?=࠵?!࠵?:࠵?:+ ℎThe higher up, the lower the velocity࠵? =2࠵? ࠵?࠵?= 2࠵? ࠵?&;"<#The higher up, the longer the periodCommon mistake with satellites: ࠵? = ࠵?:+ ℎ,not ℎ!71
10/1/2231The International Space Station (ISS) orbits earth at an altitude of approximately ℎ =400 km above sea level. The Hubble Space Telescope (HST) orbits at approximately ℎ =550 km above sea level. What can you say about their relative orbital periods ࠵?qGG, ࠵?rG*?࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NOT TO SCALE!!NASA/STScIHSTISS࠵?࠵?$࠵?= ࠵?!࠵? ࠵?:࠵?$࠵? =࠵?!࠵?:࠵?=࠵?!࠵?:࠵?:+ ℎThe higher up, the lower the velocity࠵? =2࠵? ࠵?࠵?= 2࠵? ࠵?&;"<#The higher up, the longer the periodCommon mistake with satellites: ࠵? = ࠵?:+ ℎ,not ℎ!NOTE: There is a maximum velocity and minimum period for satellites as ℎ → 0!࠵?<4==࠵?!࠵?:࠵?:≈ 7900 m/s࠵?<>?= 2࠵? ࠵?:&#;"<#≈ 85minutes72Challenge questionExplain how geosynchronous satellites work (satellites that appear to float above the same spot on earth at all times).They appear motionless in our sky – are they, actually?What are their altitudes and speeds?Can they be at any latitude?73
10/1/2232Challenge questionExplain how geosynchronous satellites work (satellites that appear to float above the same spot on earth at all times).They appear motionless in our sky – are they, actually?No! They’re orbiting with a 24-hour period that matches the earth’s rotation!Can they be at any latitude?No! They’ll only stay over the same spot on earth if they are at the equator (otherwise they will drift north/south) 74Challenge questionExplain how geosynchronous satellites work (satellites that appear to float above the same spot on earth at all times).They appear motionless in our sky – are they, actually?No! They’re orbiting with a 24-hour period that matches the earth’s rotation!Can they be at any latitude?No! They’ll only stay over the same spot on earth if they are at the equator (otherwise they will drift north/south) What are their altitudes and speeds?࠵?࠵?$࠵?= ࠵?!࠵? ࠵?:࠵?$⇒ ࠵? =࠵?!࠵?:࠵?࠵? =2࠵? ࠵?࠵?= 2࠵? ࠵?&;"<#=24 hours࠵? = ࠵?:+ ℎ =࠵?࠵?:࠵?!2࠵?$/)≈ 4.22 × 10Am⇒ ℎ ≈ 35900km⇒ ࠵? =࠵?!<#&≈ 3070m/s75
10/1/2233Planetary motion, Kepler’s laws, and NewtonKey idea: Kepler’s laws were empirical, based on observation – they came more than 50 years before Newton’s theory of gravitation! 1. The orbit of a planet about a central body is an ellipse, with the central body (e.g. sun) at one focus࠵?B࠵?C࠵?C+ ࠵?B= ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?2. The orbit sweeps out equal areas in equal timesOpenstax3. Any two planets orbiting the same body will have periods (࠵?(, ࠵?$) and average distances (࠵?(, ࠵?$) related by࠵?($࠵?$$=࠵?()࠵?$)Newton’s theory of gravitation explained WHY the orbits behaved this way…and much more!Only true for tiny masses ࠵?orbiting a much larger mass ࠵?, such that ࠵?is essentially stationaryOnly works for isolated system – it fails there are any other masses around exerting gravitational forcesThis was the first major occurrence of a theoretical (mathematical) analysis that gave a deeper understanding of an empirically-observed relationship76Relating Kepler’s and Newton’s Laws: an exampleKepler: Any two objects orbiting the same massive body will have periods (࠵?(, ࠵?$) and average distances (࠵?(, ࠵?$) related by࠵?($࠵?$$=࠵?()࠵?$)࠵?࠵?= ࠵?࠵?࠵?࠵? ࠵?࠵?NASA/STScIHSTISSRecall: using Newton’s laws we derived the period of a satellite at radius ࠵?from the earth࠵? = 2࠵? ࠵?&;!<"࠵?DEE$=2࠵? ࠵?DEE$࠵?DEE࠵?!࠵?:࠵?FEG$=2࠵? ࠵?FEG$࠵?FEG࠵?!࠵?:⇒࠵?FEG$࠵?DEE$=࠵?DEE)࠵?FEG$Newton’s law of gravitation can why Kepler’s laws are true!Everything else cancels since they’re orbiting the same body (earth)77
10/1/2234Will an object weigh more (i.e. on a scale) at the equator or at the poles? (Assume a perfectly spherical Earth)A.At the equatorB.At the polesC.It will weigh the same everywhereD.I’m not sure78Will an object weigh more (i.e. on a scale) at the equator or at the poles? (Assume a perfectly spherical Earth)A.At the equatorB.At the polesC.It will weigh the same everywhereD.I’m not sure࠵?Centripetal acceleration ࠵? = ࠵?$࠵?:Net force ࠵?H= ࠵?࠵? = ࠵? ࠵?$࠵?:࠵?b,c8࠵?>At the equator: net force required to keep you on earth!At the poles: no net force required to keep you on earthFor a spherical planet, ࠵?’&4%is the same everywhere on the surface (symmetry!)Earth is rotating…࠵?b,c8࠵?>Remember, a scale measures normal force!Correct answer: BSo your normal force (what the scale reads) is lower at the equator79
10/1/2235Challenge questionHow would a spinning space station (with astronauts inside) ”simulate” gravity on earth? (for simplicity, assume the space station is far from any massive bodies) What would the astronaut feel? Could they tell the difference by e.g. dropping an apple?Blue lines = walls of the space station80Challenge questionHow would a spinning space station (with astronauts inside) ”simulate” gravity on earth? (for simplicity, assume the space station is far from any massive bodies) What would the astronaut feel? Could they tell the difference by e.g. dropping an apple?Feels the normal force from the floor that induces centripetal acceleration. If ࠵?࠵?$/࠵?is equal to their usual weight ࠵?࠵?, it will feel like they are standing on earth.Apple will continue to move in a straight line at the tangential velocity it has when the astronaut let go of it. It will eventually hit the spaceship floor.By that point, the astronaut will have rotated…so that the apple lands at their feet.81
10/1/2236A past exam question: A skier is skiing as shown in the figure. At the instant she is on top of the bump, her speed is ࠵?and she remains in contact with the ground. Neglect friction and air resistance.(a)Draw a free-body diagram indicating the forces on her at that moment. Label all forces in the FBD.(b) Write down the correct equation (including signs) describing her situation, combining the terms ࠵?࠵?, ࠵?࠵?&/࠵?, and ࠵?B.Radius ࠵?࠵?82What do your answers look like?Radius ࠵?࠵?࠵?࠵?࠵?>࠵?B− ࠵?࠵? =࠵?࠵?&࠵?࠵?࠵?7࠵?࠵?>࠵?B−࠵?࠵?&࠵?= ࠵?࠵?࠵?>࠵?࠵? − ࠵?B=࠵?࠵?&࠵?࠵?࠵?ABCD. Something else83
10/1/2237Key idea: The net force on the skier must match the centripetal force needed to keep her going (at least instantaneously) in uniform circular motionRadius ࠵?࠵?࠵?࠵?࠵?>There are only two real forces on the skier: gravity and normal forceThey need to add up to give you ࠵?࠵?&/࠵?in the downward directionIf “down” is positive: ࠵?࠵? − ࠵?B= ࠵?࠵?&/࠵?If “up” is positive: −࠵?࠵? + ࠵?B= −࠵?࠵?&/࠵?84Suppose there was a planet with exactly the same density as earth, but only half the radius. The mass of a uniform sphere is given by ࠵? =l’}࠵? ࠵?}, where ࠵?is the density, so this planet would have 1/8ththe mass of earth.What would be the ratio of gravitational acceleration ࠵?~on the surface of this planet compared to gravity on earth (࠵?°)? A.b"b#=opB.b"b#=oqC.b"b#=o7D.b"b#= 1E. I’m not sure85
10/1/2238Suppose there was a planet with exactly the same density as earth, but only half the radius. The mass of a uniform sphere is given by ࠵? =l’}࠵? ࠵?}, where ࠵?is the density, so this planet would have 1/8ththe mass of earth.What would be the ratio of gravitational acceleration ࠵?~on the surface of this planet compared to gravity on earth (࠵?°)? A.b"b#=opB.b"b#=oqC.b"b#=o7D.b"b#= 1E. I’m not sure=12࠵?°࠵?~= ࠵?B࠵?~࠵?~&= ࠵?B1/8 ࠵?°࠵?°/2&=12࠵?B࠵?°࠵?°&࠵?]"9(= ࠵?B࠵? ࠵?~࠵?~&= ࠵? ࠵?~86