### Centripetal acceleration & centrifugal force

Centripetal acceleration is directed inward towards the center of curvature of a trajectory.  Centrifugal force is the inertial resistance of a mass to this acceleration, and is directed outward.  It's easy to get wrapped around the axle with the terms centripetal and centrifugal.  One way of avoiding confusion while keeping the physics straight is to only use centripetal to refer to the inward acceleration, and reserve the use of centrifugal to refer to the outward force, which is the inertial resistance of a mass to inward centripetal acceleration.

e.g. when a mass on the end of a string is swung in a circle, the inward centripetal acceleration causes an outward centrifugal force on the mass which causes a matching tensile force in the string.  Of course, it is the tensile force of the string which causes the curvature of the trajectory in the first place, and thus, the centripetal acceleration.  (Attempts to unravel causality often lead to this chicken/egg circular logic.)

For a circular trajectory with constant tangential velocity, the equation for centripetal acceleration is:

$\ddot{R}=-\frac{v^{2}}{R}$

Designating centripetal acceleration as R double-dot does not mean the radius length is changing - it just indicates that the direction of the acceleration is always along the radius vector.  The minus sign indicates the inward, or minus R direction.  If the radius is constant, centrifugal force does no work.  Force arises from one of two sources: an energy gradient, or a time rate of change of momentum.  In the case of centrifugal force, the momentum vector is changing -- its tip moving in the direction of the center of curvature.

The equation for centripetal acceleration can be derived from the following vector diagram.

Let R1 and V1 be the radius and velocity vectors at some time, t1.  (V1 is a free vector, i.e. it may be drawn anywhere as long as its direction and magnitude is preserved.)  Assume a circular trajectory with constant tangential velocity and radius of curvature.  Some short duration of time, dt, later, the radius and velocity vectors have the directions of R2 and V2.  Since R1 equals R2 and V1 equals V2, the two triangles are isosceles.  Since V1 is perpendicular to R1, and V2 is perpendicular to R2, the acute angles of both triangles are equal.  Since the two triangles are similar, the ratios of corresponding sides are equal.

$\frac{dV}{V}=\frac{Vdt}{R}$

The subscripts have been dropped since the magnitudes of the vectors are not changing.  The above equation can be rearranged:

$\frac{dV}{dt}=\frac{V^{2}}{R}$

Note that if the velocity vector triangle were multiplied by the scalar, mass, it would become a momentum vector triangle.

The centripetal acceleration can also be expressed as a function of the rate of change in direction.  Letting d-Theta represent the small angle (in radians) that the radius vector travels through in differential time, dt.

$d\Theta =\frac{Vdt}{R}$

The angular velocity is represented by Omega.

$\Omega =\frac{\mathrm{d}\Theta }{\mathrm{d} t}=\frac{V}{R}$

Squaring:

$\Omega ^{2}=\frac{V^{2}}{R^{2}}$

Multiplying both sides by R yields the two equivalent expressions for centripetal acceleration.

$\Omega ^{2}R=\frac{V^{2}}{R}$

The above expressions for centripetal acceleration are adequate for circular trajectories with constant velocity.

### Aerodynamic lift from centripetal acceleration

Explaining aerodynamic lift to students, whether aeronautical engineers or pilots, presents a dilemma for instructors.  All of the popular explanations invoke at some point the Bernoulli equation.

One problem with the Bernoulli equation (1738) is that it’s wrong.  Bernoulli equates the pressure of a gas to the internal kinetic energy density.  It wasn’t until the kinetic theory of gasses was fully developed a hundred years later that it became known that pressure is equal to only 2/3 the internal kinetic energy density.

$p=\frac{2}{3}\left [ \frac{1}{2}\rho v_{i}^{2} \right ]=\frac{2}{3}u_{i}$

or,

$u_{i}=\frac{3}{2}p$

For an idealized gas flow with constant density, (incompressible), no friction, (inviscid), and no heat transfer, (adiabatic) the total energy density must remain constant, since no work is being done on or by the gas, and no heat is being added or lost. Total energy density is the sum of the random internal kinetic energy density and the kinetic energy density of the flow, known as dynamic pressure, q.

$q=\frac{1}{2}\rho V_{flow}^{2}$

Total energy density is:

$u_{t}=u_{i}+q$

or,

$\frac{3}{2}p_{0}=\frac{3}{2}p+q$

Giving the correct Bernoulli equation as:

$p_{0}=p+\frac{2}{3}q$

Another problem with Bernoulli is that, even if we attribute all of the translational kinetic energy density to a pressure drop, we still can’t come up with enough lift.  Even NASA acknowledges that all of these Bernoulli based theories of lift are incorrect. NASA does not present a correct theory, other than suggesting that if you want to understand lift, you must be proficient with Euler’s equations.  (With a little Navier-Stokes thrown in for viscose effects.)

Let’s take a step back and think of the airfoil as a device that induces a curved path in the airflow both above and below the wing.

Consider the flow above the wing.  The downward centripetal acceleration of the airflow at some height, h, above the wing is:

$a_{d}=\frac{V^{2}}{R(R_{0 },\alpha ,h)}$

Where the radius of curvature is a function of camber, (R0), angle of attack, and height above the wing.  The radius of curvature decreases with either increasing camber or increasing angle of attack.

For a given combination of camber and angle of attack the radius of curvature is a complex function of height above the wing.

The vertical pressure gradient at each height is equal to the density of the air times the downward acceleration:

$\frac{\mathrm{d} P}{\mathrm{d} h}=\rho \frac{V^{2}}{R(h)}$

The above pressure gradient is analogous to the tension in a string attached to a swinging mass - it is the pressure gradient required to maintain the flow curvature.  You can think of the pressure gradient as causing the flow curvature, or the flow curvature as causing the pressure gradient through centripetal acceleration -  ishkabibble.

The two effects are in equilibrium.  The upward centrifugal pressure gradient is balanced by the product of density and downward centripetal acceleration.

Since the curvature of the flow is convex on top, the pressure gradient must be positive, i.e. increasing pressure with increasing height.  At some height above the wing, the radius becomes infinite for all practical purposes, and the pressure is equal to ambient.  (The flow becomes straight.)

At the top surface of the wing, the total pressure drop from ambient becomes:

$\Delta P_{top}=-\rho V^{2}\int_{0}^{h_{0} }\frac{dh}{R(h)}$

The same argument can be made for the flow beneath the wing,  The radius of curvature function will be different, however, there will still be some finite distance from the wing at which the influence of the wing is negligible, the flow becomes straight, and the pressure is equal to ambient.

For the bottom surface, the equation for the total pressure difference from ambient has the same form as above except for the sign, i.e. the pressure at the wing is greater than ambient.

$\Delta P_{bottom}=\rho V^{2}\int_{-h_{0}}^{0 }\frac{dh}{R(h)}$

The total pressure difference between the bottom of the wing and the top of the wing is:
$\Delta P_{wing}=\Delta P_{bottom}-\Delta P_{top}=\rho V^{2}\left [ \int_{-h_{0}}^{0 }\frac{dh}{R(h)}+\int_{0}^{h_{0} }\frac{dh}{R(h)} \right ]$

We have insufficient knowledge of the radius of curvature distributions to allow solving the non-dimensional integrals inside the brackets; instead we rely on wind-tunnel test data.  The dimensionless value of the bracketed expression is, by convention, evaluated as one half the coefficient of lift.

$\Delta P_{wing}=\frac{1}{2}\rho V^{2}C_{L}=qC_{L}$

Total lift on a wing of area, S is:

$L=\Delta P_{wing}S=qC_{L}S$

The radius of flow curvature, R, decreases with increasing camber or increasing angle of attack, which explains why increasing camber (flaps) increases lift for a given angle of attack and why increasing angle of attack increases lift for a given camber.

The above equation gives us all the practical information about lift that we need.  For a given airfoil, at a given angle attack, lift is proportional to dynamic pressure.  At constant dynamic pressure and camber, lift increases proportionate to angle of attack.  (Up to a point.)

For any airfoil there is some minimum radius of flow curvature which defines the maximum angle of attack; beyond which, the flow separates from the top surface, curvature of flow is lost, and the wing stalls.

### Centrifugal force on the hydrogen electron

Modern physics has evolved its explanation of the atomic structure well beyond the simplistic and naive idea of classical electron orbits.  It is of historical interest, however, to occasionally revisit this original intuitive model, since many of its results are still valid today.

The outward centrifugal force for a circular orbit is the product of centripetal acceleration and mass.

$F_{out}=m_{e}\frac{v^{2}}{r}$

The inward Coulomb force of attraction to the proton nucleus is inversely proportional to radius squared.  From the definition of the classical electron radius, re, it can be written as:

$F_{in}=\frac{r_{e}m_{e}c^{2}}{r^{2}}$

Setting these two forces in equilibrium:

$\frac{r_{e}}{r}=\frac{v^{2}}{c^{2}}$

(Divide both forces by mec2, and multiply by r.)

One of the great mysteries of quantum physics is the experimental evidence that the ratio v/c of the hydrogen electron is always an integer fraction of the constant alpha, originally known as Sommerfeld's constant.  (Approximately 1/137)

$\frac{v}{c}=\frac{\alpha }{N};N=1,2,3...$

Substituting for v/c reveals that the equilibrium radius increases by the square of the integer sequence.

$r=\frac{r_{e}}{\alpha ^{2}}N^{2}; N^{2}=1,4,9...$

$a_{0}=\frac{r_{e}}{\alpha ^{2}}$

Some down to Earth examples of centrifugal force.

#### How much lighter is an object at the Equator Vs. a pole?

Only the centripetal acceleration of a point on the Equator must be considered.

The determination of the Earth's circumference facilitates the determination of both the tangential velocity at the Equator and the radius of the Earth.

$C_{e}=360^{deg}\times 60^{nm/deg}\times 6076^{ft/nm}=1.3124\times 10^{8}ft$

WGS-84 established the circumference of the Earth at the Equator to a larger value of 131,479,713.54 ft.  We'll use the WGS-84 value, rounded to 1.3148 X 108 ft.

Tangential velocity of the Equator is:

$v=\frac{1.3148\times 10^{8}}{24\times 60\times 60}=1522 ft/sec$

The radius of the Earth at the equator is:

$R=\frac{1.3148\times 10^{8}}{2\pi }=20,925,692 ft$

Centripetal acceleration at the Equator is:

$\ddot{R}=-\frac{1522^{2}}{20,925,692}=-0.1107ft/sec^{2}$

The centrifugal force acts outward (up), partially canceling gravitational weight.

$\frac{\Delta W}{W} =\frac{0.1107}{32.2}=0.003438$

(As a point of interest, the above ratio is approximately the same as the flattening ratio of the Earth - 0.003353.)

A fully loaded 747-400 (875,000lb) on the ground in Singapore weighs approximately 3,000lb less than it would in Thule Greenland (N76.5o) with the same mass.  (Both centrifugal force and its vertical component decrease with the cosine of latitude.  The vertical component of centrifugal force, therefore, decreases by the cosine of latitude squared.)

#### How much lighter is an airplane flying east at 600 knots over the Equator?

This is a problem of the third kind.  The aircraft is flying along an arc which has the same radius of curvature as the Earth.

The total centripetal acceleration can be found by adding the aircraft's ground speed to the tangential velocity of Earth.

The ground speed of the aircraft in feet per second is:

$v=\frac{600\times 6076}{60\times 60}=1013ft/sec$

The total tangential velocity of the aircraft is, therefore, 2535 ft/sec.  Total centripetal acceleration is:

$\ddot{R}=-\frac{2535^{2}}{20,925,692}=0.3071ft/sec^{2}$
The fractional reduction in weight is:

$\frac{\Delta W}{W}=\frac{0.307}{32.2}=0.009534$

A 747-400 at maximum gross weight, (875,000lb), flying east over the equator at 600 knots (ground speed) is approximately four tons lighter than its one-g gravitational weight.

#### What is the altitude of a geostationary satellite?

For a satellite to remain stationary relative to the surface of the earth, it must be in a circular equatorial orbit with an angular rate matching that of the earth.  The centripetal acceleration must also equal the acceleration of gravity at the orbit radius.

$g=R\Omega _{e}^{2}$

Since the acceleration of gravity is inversely proportional to radius-squared, the product of radius-squared and acceleration of gravity is constant.  Using the known equatorial radius of the earth, Re, and the known acceleration of gravity at the Equator, ge:

$R^{2}g=R_{e}^{2}g_{e}$

Substituting from the first equation for g:

$R^{3}\Omega _{e}^{2}=R_{e}^{2}g_{e}$

$R=\left ( \frac{R_{e}^{2}g_{e}}{\Omega _{e}^{2}} \right )^{\frac{1}{3}}$

Now it's just a matter of plugging in the following known quantities on the rhs.

The radius of the earth at the Equator:

$R_{e}=6.37\times 10^{6}m$

The acceleration of gravity at the Equator:

$g_{e}=9.8ms^{-2}$

The rotation rate of the earth relative to the inertial reference frame:

$\Omega _{e}=7.3\times 10^{-5}rad/s$

$R=\left ( \frac{6.37^{2}\times 10^{12}}{7.3^{2}\times 10^{-10}}\times 9.8 \right )^{\frac{1}{3}}=42,170km$

This is the radius of the geosynchronous orbit from the center of the earth.  For the altitude, we must subtract the radius of the earth.

$h=35,800km$

#### Will PANSTARRS C/2011 L4 ever return?

If only the gravitational field of the Sun is considered, the answer appears to be no, since the velocity of the comet at perihelion exceeded the escape velocity of the Sun.

The velocity at perihelion has been reported to be 172,000 mph.

$V_{p}=172,000mph=\frac{0.172\times 10^{6}\frac{mi}{hr}\times 24\frac{hr}{day}\times 365\frac{day}{yr}}{92.96\times 10^{6}\frac{mi}{au}}=16.2\frac{au}{yr}$

The specific kinetic energy (kinetic energy per unit mass) is merely one half velocity squared.

$u_{k}=\frac{1}{2}V_{p}^{2}=131.22\frac{au^{2}}{yr^{2}}$

Dimensioning specific energy as velocity squared is perfectly valid regardless of the units of distance and time.  For example, we could choose an arbitrary unit of mass as one earth mass.  We might then define some force (I'll call it "sum" - short for "some force".) as the force required to accelerate one earth mass at the rate of one astronomical unit per year per year.  A natural unit of energy would then be "au-sum".  (Note that the above equation has the equivalent dimension of "au-sum per earth mass".)

Specific potential energy can also be expressed in au2/yr2.

Since Earth's orbit is nearly circular, the Sun's gravitational acceleration at 1au can be estimated by setting it equal to the centripetal acceleration of the earth.
$g_{e}=\omega _{e}^{2}R_{e}=\left ( 2\pi \frac{rad}{yr} \right )^{2}\left ( 1au \right )=4\pi ^{2}\frac{au}{yr^{2}}$
( V2/R would yield the same result.)

The gravitational acceleration is also equal to the product of the universal gravitational constant, G, and the mass of the Sun, Ms, divided by radius squared.

$g_{e}=\frac{GM_{s}}{(1au)^{2}}=4\pi ^{2}\frac{au}{yr^{2}}$

Therefore, the product , GMs, can be defined as:

$GM_{s}=4\pi ^{2}\frac{au^{3}}{yr^{2}}$

Gravitational force per unit mass is equal to gravitational acceleration.  With radius in astronomical units:

$\frac{F}{m}=\frac{4\pi ^{2}}{R^{2}}\frac{au}{yr^{2}}$

The specific potential energy of the sun at any radius can be found by integrating the force per unit mass from the radius to infinity.  Since it takes positive work to completely remove a mass, the potential energy is considered negative.

$u_{p}=-\int_{R}^{\infty }\frac{F}{m}dR=-\int_{R}^{\infty }\frac{4\pi ^{2}}{R^{2}}dR=-\frac{4\pi ^{2}}{R}\frac{au^{2}}{yr^{2}}$

The comets radius from the Sun at perihelion was:

$R_{p}=0.301609au$

$\therefore u_{p}=-\frac{4\pi ^{2}}{0.301609}=-130.89\frac{au^{2}}{yr^{2}}$

$u_{total}=131.22-130.89=0.33\frac{au^{2}}{yr^{2}}$

Since the total energy per unit mass is greater than zero, the kinetic energy is sufficient to escape the gravitational field of the Sun.  This agrees with the JPL data which lists the eccentricity of the comet's trajectory as greater than one.  (An eccentricity of zero is a circle, less than one an ellipse, one a parabola, and greater than one a hyperbola.)

Since the Sun contains over 99.8% of the entire mass of the solar system, it appears that the comet would escape even if the entire mass of the solar system were concentrated in the Sun.  (Dividing the specific potential energy by 0.998 gives a potential energy for the entire solar system of -131.15au2/yr3.)

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