Editing Newton's laws of motion (section)

===Examples===
The study of the behavior of massive bodies using Newton's laws is known as Newtonian mechanics. Some example problems in Newtonian mechanics are particularly noteworthy for conceptual or historical reasons.

====Uniformly accelerated motion====
{{Main|Free fall|Projectile motion}}
[[Image:Bouncing ball strobe edit.jpg|thumb|upright=1.3|A [[bouncing ball]] photographed at 25 frames per second using a [[stroboscope|stroboscopic flash]]. In between bounces, the ball's height as a function of time is close to being a [[parabola]], deviating from a parabolic arc because of air resistance, spin, and deformation into a non-spherical shape upon impact.]]
If a body falls from rest near the surface of the Earth, then in the absence of air resistance, it will accelerate at a constant rate. This is known as [[free fall]]. The speed attained during free fall is proportional to the elapsed time, and the distance traveled is proportional to the square of the elapsed time.<ref>{{Cite journal |last=Nicodemi |first=Olympia |author-link=Olympia Nicodemi |date=2010-02-01 |title=Galileo and Oresme: Who Is Modern? Who Is Medieval? |url=https://doi.org/10.4169/002557010X479965 |journal=[[Mathematics Magazine]] |volume=83 |issue=1 |pages=24–32 |doi=10.4169/002557010X479965 |s2cid=122113958 |issn=0025-570X}}</ref> Importantly, the acceleration is the same for all bodies, independently of their mass. This follows from combining Newton's second law of motion with his [[Newton's law of universal gravitation|law of universal gravitation]]. The latter states that the magnitude of the gravitational force from the Earth upon the body is
<math display="block">F = \frac{GMm}{r^2} ,</math>
where <math>m</math> is the mass of the falling body, <math>M</math> is the mass of the Earth, <math>G</math> is Newton's constant, and <math>r</math> is the distance from the center of the Earth to the body's location, which is very nearly the radius of the Earth. Setting this equal to <math>ma</math>, the body's mass <math>m</math> cancels from both sides of the equation, leaving an acceleration that depends upon <math>G</math>, <math>M</math>, and <math>r</math>, and <math>r</math> can be taken to be constant. This particular value of acceleration is typically denoted <math>g</math>:
<math display="block">g = \frac{GM}{r^2} \approx \mathrm{9.8 ~m/s^2}.</math>

If the body is not released from rest but instead launched upwards and/or horizontally with nonzero velocity, then free fall becomes [[projectile motion]].<ref>{{cite web|url=https://webhome.phy.duke.edu/~schol/phy361/faqs/faq3/ |first=Kate |last=Scholberg |author-link=Kate Scholberg |access-date=2022-01-16 |title=Frequently Asked Questions: Projectile Motion |website=Physics 361 |year=2020}}</ref> When air resistance can be neglected, projectiles follow [[parabola]]-shaped trajectories, because gravity affects the body's vertical motion and not its horizontal. At the peak of the projectile's trajectory, its vertical velocity is zero, but its acceleration is <math>g</math> downwards, as it is at all times. Setting the wrong vector equal to zero is a common confusion among physics students.<ref>{{Cite journal |last1=Carli |first1=Marta |last2=Lippiello |first2=Stefania |last3=Pantano |first3=Ornella |last4=Perona |first4=Mario |last5=Tormen |first5=Giuseppe |date=2020-03-19 |title=Testing students ability to use derivatives, integrals, and vectors in a purely mathematical context and in a physical context |journal=[[Physical Review Physics Education Research]] |language=en |volume=16 |issue=1 |pages=010111 |doi=10.1103/PhysRevPhysEducRes.16.010111 |bibcode=2020PRPER..16a0111C |s2cid=215832738 |issn=2469-9896|doi-access=free |hdl=11577/3340932 |hdl-access=free }}</ref>

====Uniform circular motion====
{{Main|Circular motion}}
[[File:Binary system orbit q=3 e=0.gif|thumb|Two objects in uniform circular motion, orbiting around the [[barycenter]] (center of mass of both objects)]]
When a body is in uniform circular motion, the force on it changes the direction of its motion but not its speed. For a body moving in a circle of radius <math>r</math> at a constant speed <math>v</math>, its acceleration has a magnitude<math display="block">a = \frac{v^2}{r}</math>and is directed toward the center of the circle.{{refn|group=note|Among the many textbook explanations of this are Frautschi et al.<ref name=":0" />{{Rp|page=104}} and Boas.<ref name="Boas">{{Cite book |last=Boas |first=Mary L. |title=Mathematical Methods in the Physical Sciences |title-link=Mathematical Methods in the Physical Sciences |date=2006 |publisher=Wiley |isbn=978-0-471-19826-0 |edition=3rd |location=Hoboken, NJ |oclc=61332593 |author-link=Mary L. Boas}}</ref>{{Rp|page=287}}}} The force required to sustain this acceleration, called the [[centripetal force]], is therefore also directed toward the center of the circle and has magnitude <math>mv^2/r</math>. Many [[orbit]]s, such as that of the Moon around the Earth, can be approximated by uniform circular motion. In such cases, the centripetal force is gravity, and by Newton's law of universal gravitation has magnitude <math>GMm/r^2</math>, where <math>M</math> is the mass of the larger body being orbited. Therefore, the mass of a body can be calculated from observations of another body orbiting around it.<ref>{{Cite book |last=Brown |first=Mike |title-link=How I Killed Pluto and Why It Had It Coming |title=How I Killed Pluto and Why It Had It Coming |date=2010 |publisher=Spiegel & Grau |isbn=978-0-385-53108-5 |edition=1st |location=New York |oclc=495271396 |author-link=Mike Brown (astronomer)}}</ref>{{Rp|page=130}}

[[Newton's cannonball]] is a [[thought experiment]] that interpolates between projectile motion and uniform circular motion. A cannonball that is lobbed weakly off the edge of a tall cliff will hit the ground in the same amount of time as if it were dropped from rest, because the force of gravity only affects the cannonball's momentum in the downward direction, and its effect is not diminished by horizontal movement. If the cannonball is launched with a greater initial horizontal velocity, then it will travel farther before it hits the ground, but it will still hit the ground in the same amount of time. However, if the cannonball is launched with an even larger initial velocity, then the curvature of the Earth becomes significant: the ground itself will curve away from the falling cannonball. A very fast cannonball will fall away from the inertial straight-line trajectory at the same rate that the Earth curves away beneath it; in other words, it will be in orbit (imagining that it is not slowed by air resistance or obstacles).<ref>{{Cite journal |last1=Topper |first1=D. |last2=Vincent |first2=D. E. |date=1999-01-01 |title=An analysis of Newton's projectile diagram |url=https://iopscience.iop.org/article/10.1088/0143-0807/20/1/018 |journal=[[European Journal of Physics]] |volume=20 |issue=1 |pages=59–66 |doi=10.1088/0143-0807/20/1/018 |bibcode=1999EJPh...20...59T |s2cid=250883796 |issn=0143-0807}}</ref>

====Harmonic motion====
{{Main|Harmonic oscillator}}
[[Image:Animated-mass-spring.gif|right|frame|An undamped [[spring–mass system]] undergoes simple harmonic motion.]]
Consider a body of mass <math>m</math> able to move along the <math>x</math> axis, and suppose an equilibrium point exists at the position <math>x = 0</math>. That is, at <math>x = 0</math>, the net force upon the body is the zero vector, and by Newton's second law, the body will not accelerate. If the force upon the body is proportional to the displacement from the equilibrium point, and directed to the equilibrium point, then the body will perform [[simple harmonic motion]]. Writing the force as <math>F = -kx</math>, Newton's second law becomes
<math display="block">m\frac{d^2 x}{dt^2} = -kx \, .</math>
This differential equation has the solution
<math display="block">x(t) = A \cos \omega t + B \sin \omega t \, </math>
where the frequency <math>\omega</math> is equal to <math>\sqrt{k/m}</math>, and the constants <math>A</math> and <math>B</math> can be calculated knowing, for example, the position and velocity the body has at a given time, like <math>t = 0</math>.

One reason that the harmonic oscillator is a conceptually important example is that it is good approximation for many systems near a stable mechanical equilibrium.{{refn|group=note|Among the many textbook treatments of this point are Hand and Finch<ref name="hand-finch">{{Cite book|last1=Hand|first1=Louis N.|url=https://www.worldcat.org/oclc/37903527|title=Analytical Mechanics|last2=Finch|first2=Janet D.|date=1998|publisher=Cambridge University Press|isbn=0-521-57327-0|location=Cambridge|oclc=37903527}}</ref>{{Rp|page=81}} and also Kleppner and Kolenkow.<ref name="Kleppner">{{Cite book|last1=Kleppner|first1=Daniel|url=https://books.google.com/books?id=Hmqvhu7s4foC|title=An introduction to mechanics|last2=Kolenkow|first2=Robert J.|date=2014|publisher=Cambridge University Press|isbn=978-0-521-19811-0|edition=2nd|location=Cambridge|oclc=854617117}}</ref>{{Rp|page=103}}}} For example, a [[pendulum]] has a stable equilibrium in the vertical position: if motionless there, it will remain there, and if pushed slightly, it will swing back and forth. Neglecting air resistance and friction in the pivot, the force upon the pendulum is gravity, and Newton's second law becomes <math display="block">\frac{d^2\theta}{dt^2} = -\frac{g}{L} \sin\theta,</math>where <math>L</math> is the length of the pendulum and <math>\theta</math> is its angle from the vertical. When the angle <math>\theta</math> is small, the [[Sine and cosine|sine]] of <math>\theta</math> is nearly equal to <math>\theta</math> (see [[small-angle approximation]]), and so this expression simplifies to the equation for a simple harmonic oscillator with frequency <math>\omega = \sqrt{g/L}</math>.

A harmonic oscillator can be ''damped,'' often by friction or viscous drag, in which case energy bleeds out of the oscillator and the amplitude of the oscillations decreases over time. Also, a harmonic oscillator can be ''driven'' by an applied force, which can lead to the phenomenon of [[resonance]].<ref>{{Cite journal|last1=Billah|first1=K. Yusuf|last2=Scanlan|first2=Robert H.|date=1991-02-01|title=Resonance, Tacoma Narrows bridge failure, and undergraduate physics textbooks|url=http://www.ketchum.org/billah/Billah-Scanlan.pdf|journal=[[American Journal of Physics]] |volume=59|issue=2|pages=118–124|doi=10.1119/1.16590|issn=0002-9505|bibcode=1991AmJPh..59..118B}}</ref>

====Objects with variable mass====
{{main|Variable-mass system}}
[[File:Space Shuttle Atlantis launches from KSC on STS-132 side view.jpg|thumb|Rockets, like the [[Space Shuttle Atlantis|Space Shuttle ''Atlantis'']], expel mass during operation. This means that the mass being pushed, the rocket and its remaining onboard fuel supply, is constantly changing.]]
Newtonian physics treats matter as being neither created nor destroyed, though it may be rearranged. It can be the case that an object of interest gains or loses mass because matter is added to or removed from it. In such a situation, Newton's laws can be applied to the individual pieces of matter, keeping track of which pieces belong to the object of interest over time. For instance, if a rocket of mass <math>M(t)</math>, moving at velocity <math>\mathbf{v}(t)</math>, ejects matter at a velocity <math>\mathbf{u}</math> relative to the rocket, then<ref name=Plastino-1992/>
<math display="block">\mathbf{F} = M \frac{d\mathbf{v}}{dt} - \mathbf{u} \frac{dM}{dt} \, </math>
where <math>\mathbf{F}</math> is the net external force (e.g., a planet's gravitational pull).<ref name="Kleppner" />{{rp|139}}

====Fan and sail====
[[File:Newtonssailboat.jpg|thumb|A boat equipped with a fan and a sail]]
The fan and sail example is a situation studied in discussions of Newton's third law.<ref name="j940">{{cite journal | last=Wilson | first=Jerry D. | title=LETTERS: Newton's Sailboat | journal=[[The Physics Teacher]] | volume=10 | issue=6 | date=1972-09-01 | issn=0031-921X | doi=10.1119/1.2352231 | pages=300| bibcode=1972PhTea..10..300W }}</ref> In the situation, a [[fan (machine)|fan]] is attached to a cart or a [[sailboat]] and blows on its sail. From the third law, one would reason that the force of the air pushing in one direction would cancel out the force done by the fan on the sail, leaving the entire apparatus stationary. However, because the system is not entirely enclosed, there are conditions in which the vessel will move; for example, if the sail is built in a manner that redirects the majority of the airflow back towards the fan, the net force will result in the vessel moving forward.<ref name="Stocklmayer"/><ref name="l303">{{cite journal | last=Clark | first=Robert Beck | title=The answer is obvious, Isn't it? | journal=The Physics Teacher | volume=24 | issue=1 | date=1986-01-01 | issn=0031-921X | doi=10.1119/1.2341931 | pages=38–39| bibcode=1986PhTea..24...38C }}</ref>