1,682 Matching Annotations
  1. Jan 2021
    1. providing the pistons are at the same vertical height

      so that there is no pressure differential due to \(\Delta z\) elevation difference, i.e., \(\rho g\Delta z\)

    1. the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure

      ...because the designers know how to use calculus, not rough averages

  2. Dec 2020
    1. PHET EXPLORATIONS: GAS PROPERTIES

      Excellent simulations. I highly recommend them. It is especially nifty to view the diffusion simulation, with data enabled, and watch both sides of the container converge to an equilibrium temperature.

    2. Note that the forces are larger underneath, due to greater depth, giving a net upward or buoyant force that is balanced by the weight of the swimmer.

      Important: buoyancy forces.

    3. In contrast, the same force applied to an area the size of the sharp end of a needle

      A friend of mine in the Air Force did a lot of loading of tanks into the big C5-A Galaxy cargo plane. He said that, to save weight, the floor of the cargo bay was not that stout, but that a tank, with its huge weight spread over the big wide treads, would not damage the floor, and a woman walking in high heel shoes would puncture the floor!

    4. weather

      E.g., in relation to hurricanes. On the Weather Channel etc., the intensity of a hurricane or tropical storm is usually described by a single number, its central pressure. Hurricane Dorian in 2019 had a central pressure of 910 millibars at its most intense on Sept. 1, 2019 over the Bahamas. Normal atmospheric pressure at sea level on a fair day is higher, 1013.25 millibars. Notice how small a pressure variation that is,

      $$\frac{\Delta p}{1013.25\, mb}=\frac{103.25\, mb}{1013.25\, mb}=0.102$$

      i.e., about 10%.

    1. Substituting and into the expression for mass gives

      This is a lot of mass. \(1.000\text{ metric ton }=1000\, kg\), so this reservoir mass is \(2\times 10^9\, \text{tons}\), two billion metric tone. This is a lot of mass.

    1. essentially in contact

      NOT. They are interacting and the interaction is stable against small spatial perturbations. I.e., it will oscillate around its equilibrium and rapidly radiate off the excess energy to settle back down to the equilibrium distance.

    1. when scientists began to discover that they are different manifestations of the same force

      Hans Christian Ørsted in Denmark, and Michael Faraday in England made this discovery.

    2. electromagnetic forces would completely overwhelm the gravitational force

      And, in fact, the electromagnetic forces produced by the muscle fibers in your quadriceps muscles DO overwhelm the gravitational force of the Earth, every time you go up a stairway!

    3. time actually slows down

      Time does NOT slow down. However, time measurements by two different observers disagree if the two observers are timing the same events from different positions relative to the "very massive" body, like a black hole.

    4. Our weight is the gravitational force due to the entire Earth acting on us.

      It is interesting to think about other large, nearby objects that pull gravitationally on us.

      $$\text{You }\approx50 - 100\,kg \\ \updownarrow \\ \text{Earth }\approx 6\times 10^24\, kg,\, \\ @ r=6.371\times 10^6 \,m$$

      vs.

      $$\text{You on the sidewalk @ 350 Fifth Ave., NYC } \\ \approx 50 - 100\,kg\\ \updownarrow \\ \text{Empire State Building }\approx 3.31\times 10^8\, kg,\\ \text{w/ center of mass, elevation } h\approx 200\,m$$

      How do those forces compare?

    1. at any single point

      Instead of an average velocity relative to two points:

      1. \(\left(x_1,\, y_1\right)\) @ time \(t_1\)
      2. \(\left(x_2,\, y_2\right)\) @ time \(t_2\)

      where the average speed is

      $$\vec{\bar{v}}=\left(\frac{x_2-x_1}{t_2-t_1}\right)\hat{i}+\left(\frac{y_2-y_1}{t_2-t_1}\right)\hat{j}$$

      When you take the instantaneous limit, \(\Delta t\longrightarrow 0\) , the other two Δs decrease, also, but the ratios \(\frac{\Delta x}{\Delta t}\) and \(\frac{\Delta y}{\Delta t}\) do not vanish; they converge to a finite value, the components \(v_x\) and \(v_y\)of the instantaneous velocity at time t, where \(t_1<t<t_2\),</p>

      $$\vec{v}\left(t\right)=v_x \left(t\right)\hat{i}+v_y \left(t\right)\hat{j}$$

      Compare to the \(v\left(t\right)\) formula with the instantaneous limit in the next paragraph.

    1. How fast an object is moving is given by the rate at which the position changes with time.

      Two different parameterizations of the path \(\Gamma\), one by increments of (approximately) equal distance, \(t\), and a second parameterization, \(\color{red}\theta\) , as it might be for an aggressive race car driver, with equal increments of time.

    2. other units of length

      Like a light-second

      $$1.00\, LS= 3.00\times 10^8\, m$$

      or the Earth-Sun distance, which we call the astronomical unit,

      $$1.00\, AU = 499\, LS$$

      etc.

    3. ΔxΔx\Delta x means change in position (final position less initial position)

      For any \(\Delta\) the order of subtraction is always

      $$\text{final}-\text{initial}$$

      $$\text{later}-\text{earlier}$$

      and if some kind of sequence exists, \(\left(u_1, u_2, u_3 \ldots\right) \) $$\Delta u=un - u{n-1}$$

    4. Since displacement indicates direction, it is a vector and can be either positive or negative, depending on the choice of positive direction.

      Tricky but important point.

    5. we use reference frames that are not stationary but are in motion relative to Earth

      ...in motion relative to Earth or to any other object. For instance, when docking your capsule to a spacecraft on orbit, you are moving and your target, the spacecraft, is a moving target.

      So your sense of steering includes a moving frame of reference. In fact, you are in a frame of reference that is in free fall!

    6. frame of reference

      In the usual sense, we can imagine three mutually perpendicular rigid meter sticks...

      "starr-160503-4526-Triticum_aestivum-Glenn_seedheads_and_meter_stick_for_height-Hawea_Pl_Olinda-Maui" by Starr Environmental is licensed with CC BY 2.0. To view a copy of this license, visit https://creativecommons.org/licenses/by/2.0/

      ...but it could be a set of rigid curves, like on this sundial.

      "Israel-05367 - Sundial" by archer10 (Dennis) is licensed with CC BY-SA 2.0. To view a copy of this license, visit https://creativecommons.org/licenses/by-sa/2.0/

    7. When you’re in motion, the basic questions to ask are: Where are you? Where are you going? How fast are you getting there?

      This is a set of basic questions that apply to a LOT of the scientific enterprise.

    1. How the wave is reflected at the boundary of the medium depends on the boundary conditions

      Boundary conditions, so important!! Just like initial conditions being important.

  3. Nov 2020
    1. That shape is determined by the total energy and angular momentum of the system, with the center of mass of the system located at the focus.

      SHAPES OF ORBITS determined by energy and angular momentum states:

      1. Center of mass, again.
      2. The effective potential energy which we know from applying basic planetary angular momentum concepts. $$V_{eff}\left(r\right)=\frac{L^2}{2mr^2}+-\frac{GMm}{r}$$

    1. Conservation of Angular Momentum of a Collision

      This example is not even a trick shot in a circus -- i.e., why would anybody want to do this. But it is still illustrative

    2. Dismount from a High Bar

      An interesting exercise. The gymnast slows down his spin just in time to land extended to his full height. Requires a lot of practice, probably.

    1. with r being the distance from the point particle to the axis of rotation.

      This is an important distinction to make for each and every moment of inertia calculation. The rotation axis

      1. might or might not go through the center of mass,
      2. it might be aligned along the length of a rod-shaped object or perpendicular to the long axis of the rod shaped axis,
      3. it might be along the edge or a rectangular box-shaped object,
      4. or it might be any weird axis you need to study,
      5. and you might have an irregular object without even any symmetry. Many symmetries in these objects

      Asteroid Eros, very irregular

  4. Oct 2020
    1. It is important to remember that potential energy is a property of the interactions between objects in a chosen system, and not just a property of each object.

      This will be abstracted in PHY2049, when studying the topic of electrostatic potential, which is defined as a property of a charged object or an array of charged objects. Electrostatic potential is not measured in Joules. It is measured in Volts.

    2. For example, the lowest height in a problem is usually defined as zero potential energy, or if an object is in space, the farthest point away from the system is often defined as zero potential energy

      Good to keep in mind, even in PHY2049

    1. Compare the work required to accelerate a car of mass 2000 kg from 30.0 to 40.0 km/h with that required for an acceleration from 50.0 to 60.0 km/h.

      Good workout

    2. By using the work-energy theorem, you did not have to solve a differential equation to determine the height.

      True here, but not always the case on a complicated problem.

    3. We can also get this result from the work-energy theorem.

      This is the primary reason why I refrain from teaching this kinematic equation earlier in a semester

    4. The importance of the work-energy theorem, and the further generalizations to which it leads, is that it makes some types of calculations much simpler to accomplish than they would be by trying to solve Newton’s second law.

      A much more interesting way to view the universe.

  5. pressbooks.online.ucf.edu pressbooks.online.ucf.edu
    1. Work Done by a Variable Force over a Curved Path

      This example is a little bit unphysical, in that keeping an object on a parabola requires some F = ma. There must be some kind of constraint, like a parabolic track that keeps an object on the track.The external \(\color{red}\vec{F}\) here would be in addition to the constraint forces. So analyzing the forces would be tricky. E.g., you'd also need to know a temporal parameter for the path, which would give you the magnitude of the tangent velocity vector at any time in the path. SO... we will ignore this example!

      It is a nice workout, however.

    2. paths

      This diagram is slightly blooped up. Path (b) is just the hypotenuse. Path (a), we may assume, is the drop of 1 m folowed by a traverse right of 3 m.

    1. For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the unit vector ^ii^\mathbf{\hat{i}}, and the scalar y-component of a vector is its dot product with the unit vector ^jj^\mathbf{\hat{j}}:

      One can use this idea to compute components of a vector in a new coordinate system (p, q). Just take the dot product of the vector with each new unit vector,

      $$\vec{A}\cdot \hat{p}=A_p$$

      etc.

    2. We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components.

      This is a helpful skill to use from time to time: getting products of two vectors in terms of each vector's components.

    1. NASA researchers test a model plane in a wind tunnel. (credit: NASA/Ames)

      Mary Jackson, the engineer in Hidden Figures, worked at a wind tunnel just like this one, at NASA/Langley in Virginia. Long may she reign!

    1. A higher coefficient would also allow the car to negotiate the curve at a higher speed

      Racing tires are designed to have much larger coefficients of friction, even \(\mu_s >1\).

    2. In Motion in Two and Three Dimensions, we examined the basic concepts of circular motion. An object undergoing circular motion, like one of the race cars shown at the beginning of this chapter, must be accelerating because it is changing the direction of its velocity. We proved that this centrally directed acceleration, called centripetal acceleration, is given by the formula ac=v2rac=v2r{a}_{\text{c}}=\frac{{v}^{2}}{r} where v is the velocity of the object, directed along a tangent line to the curve at any instant. If we know the angular velocity ωω\omega, then we can use ac=rω2.ac=rω2.{a}_{\text{c}}=r{\omega }^{2}. Angular velocity gives the rate at which the object is turning through the curve, in units of rad/s. This acceleration acts along the radius of the curved path and is thus also referred to as a radial acceleration.

      Very nice synopsis for uniform circular motion.

    3. particularly low pressures.

      Incorrect. The pressure gradient from the center of a hurricane (maybe 900 millibars) outward to a region of fair weather (1023 millibars), a distance of a few hundred miles, is about 10%, not particularly big, yet it has enormous effects over that distance

    1. Figure 5.26 The weight of a tightrope walker causes a wire to sag by 5.0∘5.0∘5.0^\circ.

      We worked on an asymmetric version of this in Session 11, 9/29/20

  6. Sep 2020
    1. forces. Zooming in deeper, your body’s atoms are held together by electrical forces, and the core of the atom, called the nucleus, is held together by the strongest force we know—strong nuclear force

      There are four fundamental forces in nature:

      1. Universal gravitation
      2. the electromagnetic interaction
      3. the weak nuclear force (which is now unified with electromagnetism -- we now speak of the electroweak interaction)
      4. the strong nuclear force, which binds protons and neutrons into a nucleus.
    1. The displacement vector Δ→rΔr→\Delta \mathbf{\overset{\to }{r}} is found by subtractin

      This is not the same as the three dogs + 1 example from the Ch. 2.3. That resultant vector was a SUM of the first three dogs' pull forces.

      This displacement is a DIFFERENCE of two vectors,

      $$\Delta\vec{r}=\vec{r}\left(t_2\right)-\vec{r}\left(t_1\right)$$

      using vector subtraction, which can be done by

      1. graphically drawing from the tip of the first vector to the tip of the second vector, as in this diagram, or
      2. subtracting component by component, e.g., \(x\left(t_2\right)-x\left(t_1\right)\)
    1. Thus, to describe the motion of any of the jets accurately, we must assign to each jet a unique position vector in three dimensions as well as a unique velocity and acceleration vector.

      We have already seen this in our study of free fall. To integrate the equations of motion (i.e., involving some acceleration, \(\ddot{s}\left(t\right)\)), one needs initial conditions -- initial position and initial velocity.

      Our entire program of integration means

      • we take each moment at which we know velocity and position
      • and use that information as initial conditions
      • to solve the equations of motion for position and velocity at the next moment of time.
    1. Kinematic Equations for Objects in Free Fall

      I hardly ever use the third equation. It is related to conservation of total mechanical energy, which is good, but difficult to use properly, in my opinion.

    1. The equation –v=v0+v2v–=v0+v2\overset{\text{–}}{v}=\frac{{v}_{0}+v}{2} reflects the fact that when acceleration is constant, v is just the simple average of the initial and final velocities.

      This is why it works, but only for constant acceleration systems

  7. Aug 2020
    1. If all else fails, bound it. For physical quantities for which you do not have a lot of intuition, sometimes the best you can do is think something like: Well, it must be bigger than this and smaller than that. For example, suppose you need to estimate the mass of a moose. Maybe you have a lot of experience with moose and know their average mass offhand. If so, great. But for most people, the best they can do is to think something like: It must be bigger than a person (of order 102 kg) and less than a car (of order 103 kg). If you need a single number for a subsequent calculation, you can take the geometric mean of the upper and lower bound—that is, you multiply them together and then take the square root. For the moose mass example, this would be

      Helpful when doing mental arithmetic. You get skilled with this, you can beat people on calculators!

    1. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take.

      "Checking the units" is a simple check on your derivations and solutions. Very important

    1. Check out a comprehensive source of information on SI units at the National Institute of Standards and Technology (NIST) Reference on Constants, Units, and Uncertainty.

      Well worth checking out.

    2. base units. All other physical quantities can then be expressed as algebraic combinations of the base quantities. Each of these physical quantities is then known as a derived quantity and each unit is called a derived unit.

      The distinction between base units and derived units is important.

    3. the units for some physical quantities must be defined through a measurement process

      This is significant.

      • The original methods of measuring the second involved the length of a day, a specific fraction, \(\frac{1}{86400}\) of a day.
      • The meter similarly was a specific fraction \(\frac{1}{10000000}\) of the distance on the line of longitude between the Equator and the North Pole that passes through Paris.
      • The current measurement method for meter and second use the frequency and wavelength of light emitted by krypton and cesium. E.g., a second is the amount of time taken by 9192631770 cycles of a specific microwave wavelength in the X-band uniquely emitted by a cesium atom.
      • These methods also mean that the speed of light is defined exactly, \(c=299 792 458\frac{m}{s}\). Very nice.
  8. May 2020
    1. For a star 500 light-years away, the light we detect tonight left 500 years ago and is carrying 500-year-old news.

      Electromagnetic waves carry energy, momentum and information!

  9. Apr 2020
    1. Given that maximum air displacements are possible at the open end and none at the closed end, other shorter wavelengths can resonate in the tube

      Boundary conditions are determined by the physical factors like length etc. You will find this all over physics!!

    1. ear. This wave function models the displacement of the medium of the resulting wave at each position along the x-axis. If two linear waves occupy the same medium, they are said to interfere.

      Interference!!!!

    2. to a linear restoring force of the medium

      Linear restoring force means porportional to displacement \(x\), like the spring force \(\vec{F}=-kx\left(\hat{i}\right)\).

      However, because the Taylor series exists, we can cobble together a linear wave equation from a nonlinear system, like the fluid we call air, for "small" displacements away from the equilibrium at ambient atmospheric pressure.

    3. the wave speed is constant and represents the speed of the wave as it propagates through the medium, not the speed of the particles that make up the medium

      Important distinction. The medium moves about equilibrium, but the wave energy and momentum move through the medium.

    4. A wave function is any function such that

      This is why there is a famous differential equation, the "one-way wave equation," viz. $$\frac{df}{dt}-v\frac{df}{dt}=0$$

    5. Whether the temporal term ωtωt\omega t is negative or positive depends on the direction of the wave.

      $$\frac{d}{dt}\varphi=\frac{d}{dt}\left(kx-\omega t\right)=k\frac{dx}{dt}-\omega$$

      $$kv-\omega=0\text{ since }\varphi\text{ is a constant}$$

      This means that the velocity of the phase \(\varphi\) is \(v=\frac{\omega}{k}=\lambda f\) since \(k=\frac{2\pi}{\lambda}\).

      Then consider \(\varphi=kx+\omega t\) instead. That would correspond to a leftward propagating wave, as the text mentions.

    1. The amplitude of the wave (A) is a measure of the maximum displacement of the medium from its equilibrium position.

      Or, in the case of sound waves, from its equilibrium pressure state -- the ambient atmospheric pressure.

    2. Matter waves are a central part of the branch of physics known as quantum mechanics.

      You might get to quantum waves in PHY2049 or PHY3101 or higher courses.

    3. For sound waves, the disturbance is a change in air pressure

      Very complex time evolution equations for pressure, but they boil down under certain conditions to linear wave equations for the pressure as a function of position and time, \(p\left(x,\, t\right)\)

    1. 34. One type of BB gun uses a spring-driven plunger to blow the BB from its barrel. (a) Calculate the force constant of its plunger’s spring if you must compress it 0.150 m to drive the 0.0500-kg plunger to a top speed of 20.0 m/s. (b) What force must be exerted to compress the spring?

      good workout involving max speed

    2. 32. Fish are hung on a spring scale to determine their mass. (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale?

      Good workout

    3. A suspension bridge oscillates with an effective force constant of 1.00×108N/m1.00×108N/m. (a) How much energy is needed to make it oscillate with an amplitude of 0.100 m? (b) If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1.00×104J1.00×104J of energy each second, how long does it take for the bridge’s oscillations to go from 0.100 m to 0.500 m amplitude.

      Nice workout

    1. if you are familiar with

      Another way to think about this approximation is the Taylor series for a function \(f\left(x\right)\) near its minimum at \(x = a\).

      $$f\left(x\right)\approx f\left(a\right)+f^{\prime}\left(a\right) \left(x-a\right)+\frac{1}{2}f^{\prime\prime}\left(a\right)\left(x-a\right)^2$$

      At the minimum, the derivative is zero, so \(f^{\prime}\left(a\right)=0\). Therefore, the first order term drops out; the zeroth order term, \(f\left( a\right)\) and the second order term \(\frac{1}{2}f^{\prime \prime}\left(a\right)\left(x-a\right)^2\) remain, i.e.,

      $$f\left(x\right)\approx f\left(a\right)+\frac{1}{2}f^{\prime\prime}\left(a\right)\left(x-a\right)^2$$

      If the function \(f\) is your potential, then you now have an oscillator potential for this small neighborhood of \(x=a\). We know the solutions for \(x\left(t\right)\) in this neighborhood, as we discussed in remote session 26.

      Many physical systems can be studied profitably from this standpoint, the Taylor series expansion of the potential near one of its minima.

    2. energy proportional to the amplitude squared

      In the electromagnetic field of electric field \(\vec{E}\) and magnetic field \(\vec{B}\), the energy density of the field, \(u\), is also proportional to the squares of the fields: $$u=\frac{1}{2}\epsilon_0E^2+\frac{1}{2}\frac{1}{\mu_0}B^2$$

      The constants \(\epsilon_0\) and \(\mu_0\) are related to the speed of light \(c\) :

      $$c^2=\frac{1}{\epsilon_0 \mu_0}$$

    3. A practical application of the concept of stable equilibrium points is the force between two neutral atoms in a molecule.

      Entree to current research areas!

    4. is proportional to the square of the amplitude

      Because, when \(x=\pm A\), the kinetic energy is zero, no matter what your initial conditions were, and all the energy \(E\) is potential.

    1. can be modeled with the following equations of motion:

      Alternate model uses the generic solution but with the phase \(\phi\) converted into a sine term. E.g.,

      $$x\left(t\right)=a\, \cos\left(\omega t\right)+b\, \sin\left(\omega t\right)$$

      The constants \(a\) and \(b\) look like components of a vector whose magnitude is \(A\), viz. $$a=A\cos\left(\varphi\right) \ b=-A\sin\left(\varphi\right)$$

    2. pulling it out to x=+0.02m.x=+0.02m.x=+0.02\,\text{m}\text{.} The block is released from rest

      Initial conditions easiest to use: $$x\left(0\right)=0.02\, m \ \dot{x}\left(0\right)=0.0\frac{m}{s}$$

      Result: \(\varphi = 0\). Nice.

    3. Figure 15.6

      The velocity is leftward, derivative is negative, $$\dot{x}<0$$ during the first half of the motion, then rightward, positive derivative, $$\dot{x}>0$$ in the second half of the cycle... from THESE initial conditions $$x\left(0\right)=A\ \dot{x}\left(0\right)=0$$

    4. For one thing, the period T and frequency f of a simple harmonic oscillator are independent of amplitude.

      I think we demonstrated this with variation of the swing angle back when we used pendula in the classroom.

    5. The stiffer the spring is, the smaller the period T. The greater the mass of the object is, the greater the period T.

      We already know this from earlier in the semester: $$\omega^2=\frac{k}{m}$$

    6. The mass now begins to accelerate in the positive x-direction

      Incorrect. Its acceleration is rightward as soon as it crosses equilibrium, \(x<0\), i.e., in (c).

    7. The mass is displaced to a position x=Ax=Ax=A and released from rest.

      Initial conditions: $$\vec{F}\left(0\right)=kA\left(-\hat{i}\right)$$ $$x\left(0\right)=A$$ $$\vec{v}\left(0\right)=0$$

    8. In simple harmonic motion, the acceleration of the system, and therefore the net force, is proportional to the displacement and acts in the opposite direction of the displacement.

      These words imply, as we have already seen, that the total energy \(E\) is proportional to the square of the displacement, \(q\), using a generic coordinate \(q\) and to the square of the coordinate velocity, \(\dot{q}\). Or, as the math department might say, the function \(E\) is "quadratic" in \(q\) and \(\dot{q}\).

    9. much highe

      $$\frac{\text{human}}{\text{ultra}}=\left(2\times 10^4\right)/ \left(2.5\times 10^6\right)\approx 0.01$$

      So that is "small" in the sense of calculus shortcuts.

      This also means the wavelengths are much smaller for ultrasound.

    1. for objects within a few kilometers of Earth’s surface, we can take

      "a few kilometers"

      I.e., from the top of Mt. Everest (elevation \(\approx 9\, km\)) to the bottom of the Mariana Trench (depth \(\approx 11\, km\). Cf., NASA planetary fact sheet.

      E.g., $$\text{Everest} \longrightarrow\frac{1}{6380^2}=\frac{1}{40704400}\approx 2.457\times 10^{-8}$$ $$\text{sea level}\longrightarrow\frac{1}{6371^2}=\frac{1}{40589641}\approx 2.464\times 10^{-8}$$ $$\text{Challenger Deep}\longrightarrow\frac{1}{6360^2}=\frac{1}{40449600}\approx 2.472\times 10^{-8}$$

    2. the radius of Earth

      Not true for the Earth-moon system, as discussed, but good for most terrestrial objects like the oceans and for human objects like stadia and great pyramids.

      Earth-moon system:

      Earth and the average depth of the ocean (blue line):

      Spectrum stadium, a blip next to Earth itself:

    1. lever arm

      Be sure to work this out with some careful sketching. You can even define orbital angular momentum \(\vec{l}\) for a straight line trajectory.

    2. Angular Momentum of a Particle

      Orbital angular momentum as opposed to spin angular momentum. It is amazing to think that, on a complex gravitational trajectory like an ellipse, parabola or hyperbola, the vector product of \(\vec{r}\) and \(\vec{p}\) is conserved. The magnitudes of momentum and radial distance and their geometric relationship are controlled by gravity to lock the object into its elliptical, parabolic or hyperbolic orbit.

  10. Mar 2020