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Orbital Long

How long a time brain needs to develop!

I want to write this article first in star language. I f You do not understand my article I´ll publish my article in english very soon!

E´ electrecircui des Papez sa nucleanfissiones des Sunno, alfaCenD!

De synkrotroneradiate comes to Eugen fra Stare sa Sunno sa bestamen de´ funten des Humanos. De Electrecircui des Papez hafte funkten uith de´ Synkritroneradiate. Nog hafte is ei seconde contenuos. Uir conten de Energos des Gravite aproxomat  10 to power of 33 J is de´Planeten des Sunno. Is Tellus es aproxomat 6 x 10 to power of 9 inhabes sa de´mass des Tellus es 5.974 x 10 to power of 24 kg encludes inhabes mass.

De Sunno hafte impulse 76/60 1/s = 1.266666667 1/s (ore Tellus geh is Orbite throug de Magnetosfäre). Whe protones fissiones hafte samen amonten berdo is de samen Timeno ore minder sa de protonen geh to de Centre des Sunno vie konvektiones, followes 1.26666667 1/s aproxomat 10 tp power of 33 J aproxomat 167392032 J/kgs. Ege des Humanos es aproxomat 80 yere befor NU-birde, 80 yere aproxomat 2522880000 s aproxomat 1673920321 J/kgs. 25220000 s x 80 kg aproxomat 3.37848008 x 10 to power of 20 J denes energos fur de lefen Humanos 80 yere.

De Brane gewices aproxomat 4 kg sa energos hafte be 1673920321 x 4 kg = 6695681284 J/s sa es 4 kg es 2.391453289 x 10 to power of 27 pc protone, aproxomate wis denien 2.799837796 x 10 to power of -18 J/ protone is Papez circui (aproxomat 1.60217733 x 10 to power of -19 J corretes 1.60217733 x 10 to power of -19J/mp. De amontes´ des electrones hafte be 2 x 2.391453281 x 10 to power of 27 pc followes 2.391453281 x 10 to power of 27 pc/1AU fra Sunno endest, = 1.594302187 x 10 tp power of 16 pc/m followes contes electrese-poverte sa 1.594302187 x 10 to power of 16 pc/1s ore Timeunites, followes 5.074821477 pc x 10 to power of 16/m2 is de Eugen followes 0.002554354 J/m followes 1.60217733 x 10 to power of -19 J = 6.241506363 x 10 to power of 18 eV followes 1.594302187 x 10 to power of 16 pc aproxomat 9.950847245 x 10 to power of 34 eV = 9.950847245 x 10 to power of 28 MeV followes:

o.511 MeV/me- = 9.950847245 x 10 to power of 28 MeV/sigma me-

sigma me- = 9.950847245 x 10 to power of MeV x me-/0.511 MeV

sigma me- = 0.177389717 kg

4 kg/sigma me- = 22.54922138 (yere)

es de utveces des Branes.

(?) Tellus movides is Orbit des Sunno sa mo electrones circuiten to Tellus followes de Birde wai formes af de circuiten electrones sa emisione.

That´s all for this time!

Antti Johannes Vaalama

 

About the Author

If an object fell toward the Sun from the Earth's distance, with no orbital motion, how long would it take...?

to reach the Sun?

answer: 64.6 days (changed later to 64.560 days)

This is a little tricker than I first thought. Intuitively, I figured the answer would be something of the order of half a year.

The potential energy of an object of mass m in the gravitational field of a large body of mass M is
U = -(GMm)/r
where r = distance between bodies.

M is the mass of the sun, m the mass of the small object.

Let
R1 = distance from sun to earth
R2 = radius of sun

We can use conservation of energy and the definition of kinetic energy to say the following:

mv^2 /2 - (GMm/r) = -(GMm / R1)

Therefore, as the object falls from a distance R1 from the sun to R2, it has velocity
v = sqrt (2 ((GM/r) - (GM / R1)))

So we know the velocity v as a function of r, the distance from the sun. We need to integrate this to get the position as a function of time, and then calculate the answer.

Unfortunately, I'm rusty at integration, so I wrote a little program to do a numerical integration instead. (For the very earliest part of the trip to the sun, I simply assumed the normal physics of a falling object in a constant gravitational field. This was necessary because the initial velocity is 0, so we have to "kickstart" the integration from a point where the velocity is non-zero.)

The result of this -- and the answer to your question -- is
64.6 days.

I think I did this correctly, but would be interested in hearing any other answers, particularly from people who see how to do the integration analytically.

-- added a few minutes later:
It's easy to calculate an upper limit to this time. Simply assume that the gravitational force between here and the sun matches that found at earth's distance from the sun. (The gravitational force actually increases as we approach the sun; that's why this calculation gives an upper limit.)

We first calculate the acceleration of gravity from the sun at the earth's distance:
a = GM / R1^2

We then use the standard formula for distance versus time in a constant gravitational field
s = (1/2) * a * t^2
to calculate the time. The result is 82.2 days. Thus, the correct answer *must* be less than 82.2 days. That gives me more confidence that the number I gave above (64.6 days) is correct.

------
Here's one more tip that the answer above is good: If you project the earth's orbit into a single axis, it makes the trip from a far point to the sun in a quarter of the year, or 91.31 days. The ratio of this time to the answer given above is 1.41. This is the square root of two, and this number shows up in physics when we talk about escape velocity versus circular velocity. So now I'm very sure that the above answer is right, but would still be interested in seeing an analytical solution from someone who's better at differential equations.

-- final edit:
I found a web page that solves this problem analytically:

http://www.physlink.com/Education/AskExperts/ae226.cfm

That page says that the answer is a quarter of a year divided by the square root of two. That quantity works out to 64.57 days, which matches the answer I found from numerical integration. (But this is a slightly incorrect approximation for reasons given below.)

Keith P's answer above is an elegant easy way of solving the problem (but, in fairness, he gave a grossly incorrect answer, and then posted the correct answer long after I posted mine). Also, technically speaking, he's solving a slightly different problem. You asked for the time to reach the sun, and I assumed you meant the surface, not the center. According to my numerical integration, the object would fall past the surface of the sun at least 12.6 minutes before it reaches the center, so the correct answer is the following:
T <= (0.5 * (0.5)^(1.5)) years - 13 minutes
so
T <= 64.56 days

The reason this is an upper limit is that I assumed the gravitational potential of a point source. Because the sun is an extended body, the acceleration of the particle through the sun (if you imagine a particle that could survive the heat) will be less than was assumed in my integration, so the travel time from surface to center will be somewhat greater than 13 minutes. I don't feel like doing this calculation now.

-- and finally, the exact approach
The web site above gives the following relation between time and distance from sun:
r*R*sqrt(1/r - 1/R) + (R^(3/2))*arctan(sqrt(R/r -1)) = sqrt(2GM)*t
where
r = distance from sun (which we set to the solar radius)
R = distance of earth from sun (1 AU)
M = mass of sun
G = gravitational constant
t = time

Plug in the numbers, and you get
t = 64.560 days
and that's my final answer! It is about 0.01 days less than Keith P's answer because it is the time to fall to the surface of the real sun, not the center of a hypothetical point sun. His approach wins the "most elegant" award, but the ugly formula above wins the "most precise" award.

In summary: The time to the surface is a bit less than Keith P's answer. The time to the center is a bit more than his answer because once the object is inside the sun, you can no longer treat the sun as a point object of fixed mass. The differences are very small, of the order of 0.01 days.

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The Russian Orbital Segment of the International Space Station $124.99 The Russian Orbital Segment of the International Space Station - Wall Mural

Orbital Floor Lamp by Foscarini $2026 The Foscarini Orbital Floor Lamp pairs mid-century modern shapes with soft illumination to create a playful lamp for bold, contemporary interiors. The Oribtal Floor Lamp, a 1992 design by Ferruccio Laviani, features 5 glass shapes with large screw details. Glass shapes are individually backlit. Foscarini, headquartered in Italy, offers superb contemporary lighting products in playful and inspiring designs. In partnership with both established designers and emerging designers, Foscarini has created an exciting collection of products for residential and commercial applications. The Foscarini Orbital Floor Lamp is available with the following: The Foscarini Orbital Floor Lamp is available with the following: Details: Silkscreen Printed Glass shadesLacquered metal baseNot UL/ETL ListedDesigned by Ferruccio LavianiMade in Italy Options: Color: Multi Colored, or White.Lighting: Five 40 Watt 120 Volt Incandescent lamps (not included). Shipping: This item usually ships in two weeks. Dimensions: Fixture: Height 67 In., Diameter 20.75 In.

Orbital Sciences Corporation $89.22 High Quality Content by WIKIPEDIA articles Orbital Sciences Corporation (OSC, though commonly referred to as Orbital) is a company which specializes in satellite launch and manufacture. Its Launch Systems Group is heavily involved with missile defense launch systems. Orbital formerly owned ORBIMAGE (now GeoEye) and the Magellan line of GPS receivers, though they are now divested (the latter to Thales). Orbitals NYSE ticker symbol is ORB. It has its headquarters in the Dulles area of unincorporated Loudoun County, Virginia, United States. Author: Surhone, Lambert M./ Timpledon, Miriam T./ Marseken, Susan F. Binding Type: Paperback Number of Pages: 144 Publication Date: 2010/05/19 Language: English Dimensions: 5.98 x 9.01 x 0.33 inches

The Fragment Molecular Orbital Method $129.95 Considered a promising alternative for modeling large molecules of biochemical interest, the fragment molecular orbital method (FMO) shows great promise when applied to any number of biochemical problems. This book demonstrates applications to biochemical studies that emphasize practical aspects of performing actual simulations.

Plastic and Orbital Surgery $147.95 This textbook covers the whole of ophthalmic plastic and reconstructive surgery, i.e., eyelid, lacrimal, orbital, and socket surgery, at a basic level. It is aimed at ophthalmologists worldwide. It covers what any fully trained ophthalmologist should know about the subject, to enable him/her to practise with competence and confidence.

Specific Orbital Energy $60.54 PMHigh Quality Content by WIKIPEDIA articles In astrodynamics the specific orbital energy epsilon, (or visviva energy) of an orbiting body traveling through space under standard assumptions is the sum of its potential energy (epsilon_p, ) and kinetic energy (epsilon_k, ) per unit mass. According to the orbital energy conservation equation (also referred to as visviva equation) it is the same at all points of the trajectory: epsilon=epsilon_k+epsilon_p={v 2over{2}}{muover{r}} ={1over{2}}{mu 2over{h 2}}left(1e 2right) where: v, is the orbital speed of the orbiting body; r, is the orbital distance of the orbiting body; mu = GM, is the standard gravitational parameter of the primary body; h, is the specific relative angular momentum of the orbiting body; e, is the orbit eccentricity. It is expressed in J/kg = m s2 or MJ/kg = km s Author: Surhone, Lambert M./ Timpledon, Miriam T./ Marseken, Susan F. Binding Type: Paperback Number of Pages: 70 Publication Date: 2010/07/31 Language: English Dimensions: 6.00 x 9.02 x 0.17 inches

Molecular Orbital Theory $101.96 High Quality Content by WIKIPEDIA articles In chemistry, molecular orbital theory (MO theory) is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. In this theory, each molecule has a set of molecular orbitals, in which it is assumed that the molecular orbital wave function f may be written as a simple weighted sum of the n constituent atomic orbitals i, according to the following equation: Author: Miller, Frederic P./ Vandome, Agnes F./ McBrewster, John Binding Type: Paperback Number of Pages: 152 Publication Date: 2010/07/31 Language: English Dimensions: 5.98 x 9.01 x 0.35 inches

Superior Orbital Fissure $81.25 High Quality Content by WIKIPEDIA articles High Quality Content by WIKIPEDIA articles The superior orbital fissure is a foramen in the skull, although strictly it is more of a cleft, lying between the lesser and greater wings of the sphenoid bone. A number of important anatomical structures pass through the fissure, and these can be damaged in orbital trauma, particularly blowout fractures through the floor of the orbit into the maxillary sinus. Author: Surhone, Lambert M./ Timpledon, Miriam T./ Marseken, Susan F. Binding Type: Paperback Number of Pages: 104 Publication Date: 2010/07/19 Language: English Dimensions: 6.00 x 9.02 x 0.25 inches

Factory Reconditioned Skil 4290-02-RT 4.5 Amp Variable Speed Orbital Jigsaw $20.99 4.5 Amp variable Speed Orbital Jigsaw - 4290-02-RT

Factory Reconditioned Skil 9225-01-RT 9.5 Amp Orbital Reciprocating Saw $47.99 9.5 Amp Orbital Reciprocating Saw - 9225-RT

Factory Reconditioned Skil 4490-01-RT 5 Amp Orbital Cut Scrolling Jigsaw $41.16 5 Amp Orbital Cut Scrolling Jigsaw - 4490-01-RT

Factory Reconditioned Skil 4495-01-RT 6 Amp Orbital Jigsaw $34.99 6 Amp Orbital Jigsaw - 4495-01-RT. Carrying Bag

Antidoto Orbital $14.99 Track Listing: 1. Revolver - Eufemia Tufano, 2. Dixie - Dildo, 3. No Te Necesito - Cohete, 4. Febrero 31 - Vedette, 5. Am?rica Latina - Dragon Zaga, 6. Viajar en la Luz - Nukler, 7. Solo Otra Vez - Hydro, 8. Wake Up - Gula (Remix, remix), 9. Kyoto - Vedette, 10. D?jame - Cohete, 11. Noches de Vinil - Dildo, 12. 3 A.M. - Eufemia Tufano, 13. Volar - Hydro, 14. Hoy - Nukler, 15. Oyeme Jos? - Dragon Zaga

Orbital Velocity $5.99 With a mandate to combat terror, the ultra-covert action team called Stony Man works outside official channels. Elite field commandos operate with real-time intelligence from master cybernetics experts. Grit and tactical brilliance are Stony Man's best weapons—and America's best chance when terror strikes from the sky to threaten the globe. An American neo-Nazi has declared war on the governments of the world. His group, Fist of Heaven, controls six weapons platforms in deep space and has launched high-shock kinetic missiles at major cities. The deadly spears have struck London, Moscow, Los Angeles and Tokyo. The death toll is staggering. America is losing the battle to save the planet from the hands of a madman. Grim and determined, Stony Man's teams prepare for the worst as they unleash their own brand of righteous retribution against the Fist of Heaven.

The Orbital Hit $11.62 A world class hit man, the ultimate Professional fulfills contracts under the alias of "GOLGO13." Without allegiance to any group, cause, or nation, he follows his own personal code when choosing "jobs." And once he accepts a contract, it is always honored. ?A true professional, GOLGO13 is an expert marksman familiar with most of the world's military hardware and master of several of the more lethal Oriental matial arts. Fluent in at least thirteen languages, GOLGO13 is able to slip in and out of even hostile nations without detection. Victims, they never know what hit them.

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