Suppose the Earth suddenly stopped in its orbit.

physics/fall.into.sun New

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Problem

Suppose the Earth suddenly stopped in its orbit.  How long would it
take to fall into the Sun?  Assume the orbit is circular, the Sun is
much heavier than the Earth, both are small compared with the orbit,
gravity between them is the only force, and there are no relativistic
effects.  More generally, how long does it take to fall to distance r
from the Sun?

Show solution

I must admit that when I first came across this problem a while ago I
just bashed out the integration, too. However, you've inspired me to
think about it again and discuss it with some colleagues, and we found
that it's actually quite easy:

I'm not sure Kepler's Laws apply to bodies which collide, even if
they're point masses, so give the Earth a small push sideways as it
starts to fall - it will follow a long narrow elliptical orbit passing
very close to the Sun - and then let the size of the push -> 0. If the
Earth's initial distance from the sun is 1 then the semi-major axis of
the new orbit -> 0.5. By Kepler's ? Law the period of an orbit is
proportional to the 3/2 power of the semi-major axis, so the period of
the new orbit -> 0.5^(3/2) years, or 1/(2*sqrt(2)) years. The time
taken for the fall is thus half this time, or 1/(4*sqrt(2)) years, or
about 64.6 days.

Fortunately this is the answer I got by integration, too.  Of course I
never doubted it would be, you understand 8-)

-- Thomas D Bending <tdb@maths.qmw.ac.uk>

obPuzzle:
Suppose the Earth suddenly stopped.
How long time will it take to fall to distance r from the Sun ?

I found the following (closed) expression for the fall time, t, to
distance r from the sun:

     t = 1/(sqrt(2)*pi) * (pi/4 - 1/2*arctan(x) + x/(2*(x*x+1))) year
where
     x = sqrt(r/(R-r)), r = [0, R[
and
     R is the original Earth orbit radius.

-- Jens Joergen Nielsen <jjn@login.dknet.dk>