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# Free Fall to the Sun

theotex
In Off on a Comet, a novel written by Jules Verne, at one point some characters are wondering what would happen to the earth if suddendly stopped on its circular path around the sun.
Well, supposing that our planet survives the heat generated by kinetic ernergy, it will begin begin to fall toward the sun. And how long would this fall take ? Lieutenant Procope, one of the personnages, says : 64 days.

A few lines of integral calculus can confirm this number reported by Jules Verne (who was not a scientist). But there is a much easy way to find it, without paper and pencil.

Who can explain the trick, linked to a famous astronomer born in 1571 ?
Bikerman
LOL..that is a mathematicians 'suppose' rather than a physicists, I'll wager.. :=)
Normally only mathematicians can casually chuck that much energy around with a single word. Physicists normally act dumb-struct and look shifty at such a suggestion

Earth would, of course, not only 'not survive intact' it would certainly melt and I suspect it might even sublime. Rather than tackle the interesting question - I'll leave that to others, even though it concerns my favourite scientist of the era and a man so much greater than Copernicus that it is embarrassing when people know the latter and not this chap.

Instead of solving this nice little problem (I'm thinking a Mars a day might help to shape the problem) I'll consider the energy of earth stopping and see if we can deduce anything from that.

Kinetic energy uses an m^2 term so it's gonna be huge....
In fact it calculates to 2.6488 * 10^33 kg m^2 s^-2 Joules of energy.

SHC for Basalt is 0.84. Melting at about 1000C
energy to melt 1kg basalt is therefore
mxc x delta T = 1 x 0.84 x 1000 Joules = 840 Joules
Divide one by t'other and we get about 3x10^30kg of rock that this energy could melt.
The actual mass of rock is about 6x10^24 kg so we have an excess of magnitude 6

I've got stuff to do now but I'll try to find time to do the calc for actually evaporating instead of melting and see if there is enough energy for that - later dude.
C.
Dennise
Bikerman,

The OP is asking how long it would take the Earth to 'fall' into the sun if the Earth's orbital motion were zero. It appears he is ignoring energy considerations related to dissipation of Earth's kinetic energy from the loss of orbital velocity. I.e. his initial condition is the Earth is simply a large stationary mass 93 million miles from a much larger mass.

I'd be interested to learn what the 'trick' is that avoids a paper and pencil calculation of (I think) this equation:

distance = 1/2 × acceleration × time2 + intial speed × time

solved for time.
theotex
Here is the trick :

You have to apply the third law of Kepler : squares of periods are proportional to cubes of major axis.

A fall to the sun can be considered as a flat ellipse which major axis is the distance from earth to the sun, that is half of the major axis of the ellipse described by the earth around the sun. Then the cube of the major axis of this flat ellipse is 8 times less than the one of the earth orbit. So its period is equal to a year divided by square root of 8, or 2X sqrt(2). But the fall to the sun represents just half of the flat period ellipse (no return), so its duration is a year divided by 4X sqrt(2), or 3 months divided by 1.4, which after a bit of mental calculus gives around 64 days.
ocalhoun
 Dennise wrote: I'd be interested to learn what the 'trick' is that avoids a paper and pencil calculation of (I think) this equation: distance = 1/2 × acceleration × time2 + intial speed × time

The acceleration is not constant because gravity is stronger as you move closer, and while this effect is negligible on the surface of the earth, on the scales we're talking about here, that's a big difference.
That will add some significant complications: you'll need to replace the 'acceleration' variable with an equation that relates the gravitational force to the distance. I'd look it up if I weren't so lazy.
theotex
It is not very hard. You write the increase of kinetic equals to the decrease of potential energy

1/2mv²=integral(GmM/x²)dx=GM(1/x-1/x°)

so dx/dt=sqrt[2GM(1/x-1/x°)]

then

dt=dx/sqrt[2GM(1/x-1/x°)]

this integral is classical and can be solved with a trignonometric function.

Yet, Kepler way is much more elegant.
asnani04
This is a very interesting question and I'm a bit lost. I think energy conservation should do it though. Equating Earth's potential energy (current) with its Kinetic Energy when it reaches the Sun will give an expression for its velocity, which can be written as dx/dt, and further integrated after transposing dt on the other side to find the time. If earth's melting is ignored, it would be like a point charge falling through an electric field centered at a fixed point charge (the sun) and the equations would follow suit.
theotex
The formal equation of an electric charge "falling" toward and other one of an opposite sign would be the same for Coulomb law is in squared inversed like gravitation, but the ratio of their masses is far from the sun/earth one (1836 at the most for a proton/electron) ; so you would have to rather choose your orgin at their barycenter.

An other problem would be that the accelerated will emit an electromagnetic radiation and kinetic ernegy won't be anymore equal to potential energy decrease. This complicated situation has already been studied end 19th century/beginning 20 th.
kelseymh
 theotex wrote: The formal equation of an electric charge "falling" toward and other one of an opposite sign would be the same for Coulomb law is in squared inversed like gravitation, but the ratio of their masses is far from the sun/earth one (1836 at the most for a proton/electron) ; so you would have to rather choose your orgin at their barycenter.

Nope. For charged particles, you don't use their mass -- you use their relative electric charge, and the analogue to the barycenter is the aveage location of all the charge in the system.

 Quote: An other problem would be that the accelerated will emit an electromagnetic radiation and kinetic ernegy won't be anymore equal to potential energy decrease. This complicated situation has already been studied end 19th century/beginning 20 th.

Of course, that's also true of graviating systems (look up "gravitational waves" or "gravitational radiation"). The difference is merely in the relative magnitudes and distances involved.
theotex
 Quote: Nope. For charged particles, you don't use their mass -

Nope. To establish the movement equation or two charged particles you have to use their mass as well as their charges.The mass barycenter is the most convenient origin because the two particles will move toward it, while for the sun/earth couple, the barycenter or the two bodies is very near from the sun center.

 Quote: Of course, that's also true of graviating system

The speed of a celestial body falling on the sun (around 600 km/s) never reaches a magnitude where you have to take into account gravitational waves. While for an electron or a proton, they radiate electromagnetic waves everyday. That's why you have radio, tv, satellite communications, and why it is so difficult to prove gravitational waves existence.