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,

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.

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.

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.

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.

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.