Lets just say a huge asteroid hits the earth at the right speed and with the right amount of force and stops it's orbit around the sun. How long would it take for the earth to plummet into the sun?
Asked by: Mike Assad
There are two ways to go about this question...
One is, bite the bullet and solve the differential
equation, plug the values in, and get the answer.
The other is, use known quantities and some knowledge
The equation of motion for a body which has no angular
momentum about the sun, as described, is:
r'' = -GM/r2
[G is gravitational constant, M is mass of sun,
r is distance of body from the sun]
Where the primes denote differentiation with respect to
time. One would integrate this twice, the first
r' = - sqrt (2GM) * sqrt(1/r - 1/R)
Where R is the initial distance from the sun, here
it would be about the distance of the earth from
Second integration gets a little trickier, yet
still doable; and it yields the following equation:
Where t is the time measured from the time the body
started its motion at distance R. Now, we want to
find t when r goes to zero. It turns out easier than
it looks. The first term on the left vanishes, while
the arctangent's argument becomes infinite, and the
arctangent of infinity is Pi/2, we get, after some
t = sqrt(2*R3/(8*G*M))
Is the time it will take to fall into the sun, ignoring
the motion of the sun, and the size of the sun...
Now how do we evaluate this? The earth's period around
the sun is given by:
T = sqrt(4*2*R^3/(G*M))
And of course, it is a year.
Comparing the two, we see that...
t = T/(4*sqrt(2))
Which is roughly 64 days and 12 hours,
or just over two month's time to say goodbye to
the cruel world, however, it might get considerably hot
towards the end...
Now, the other, less math-more-physics method.
We know the period of any orbit is dependent only
on the semimajor axis; as long as it is an ellipse
with the sun at one focus.
We can take this problem to be one where the orbit
is an extremely flat ellipse. This puts the focus
at one end, while the body starts the motion at the
other end. So, what we need to calculate here is the
HALF of the period of an orbit for which the semimajor
axis is HALF of R, if you follow the argument.
So, the expression is:
t = (1/2)*sqrt(4*2*(R/2)3/(GM))
which reduces to:
t = (2*R3/(8*G*M))
The same expression as before.
Also note that the mass of the asteroid does not enter
into the equation, as long as it is not so huge
to make the mass of the earth after the collision
comparable to the mass of the sun, in which case
we'll have to consider the movement of the sun as
well in the derivation. (This can be taken care of
just by using the reduced mass, but that's another
Answered by: Yasar Safkan, Ph.D. M.I.T., Software Engineer, Istanbul, Turkey
Your proposed asteroid would need to have the same momentum as the mass of the Earth
multiplied by its orbital velocity, but in the opposite direction. One example would be
another Earth-sized object travelling at 18 miles/second. A lower mass with higher
velocity would also do, but you can easily imagine any such collision completely destroying
both objects with little left of the Earth to fall into the Sun.
Answered by: Paul Walorski, B.A., Part-time Physics Instructor
'A theory with mathematical beauty is more likely to be correct than an ugly one that fits some experimental data. God is a mathematician of a very high order, and He used very advanced mathematics in constructing the universe.'