From: Ronald Wong (ronwong@inreach.com)
Date: Mon Jan 10 2000 - 15:13:18 PST
Date: Mon, 10 Jan 2000 15:13:18 -0800 (PST) Message-Id: <l03102800b49b83995fce@[209.209.19.11]> From: Ronald Wong <ronwong@inreach.com> Subject: Re: Astronomical help needed
On Wed, 05 Jan, Tim said:
> We are in the middle of our astronomy section of my physics class
>and one
>of my students asked how long it would take to fall from the earth orbital
>radius to the sun ( the straight-line solution). I have figured how to do
>it iteratively with a spread sheet, but that solution lacks the elegance of
>a closed solution. The fact that the acceleration changes as you get closer
>to the sun has stymied my integration attempts. Anyone remember how to do
>that problem?
Tim:
I don't remember having to do this problem before but I do remember that a
frontal assault on the problem of planetary orbits using Newton's laws
involved some pretty heavy slogging as well as a little imagination.
One approach to solving your problem is to recognize that an object that is
falling radially towards the sun is a special case in the study of
planetary orbits. It's special because this object will run into the sun
and self-destruct before it can complete its "orbit" (probably the reason
why the problem wasn't studied). If somehow it was able to pass through the
sun - through a tunnel that passes completely through the center of the sun
for instance - it would be able to oscillate back and forth in a periodic
fashion with the sun at the center of its "orbit".
Basically what we have is a planet that happens to have no tangential velocity.
One quarter of the period of this object's "orbit" is nothing more than the
time it would take to travel from its initial position, when it was at
rest, to the center of the sun. Basically, this would be the answer to your
student's question.
Like a planet, the square of the period of this object is proportional to
the cube of its mean distance from the sun - Kepler's third law.
In the case of the sun, T^2 = 4*PI*PI*R^3/(G*Ms).
Where T^2 means T squared and R^3 means R cubed. G is the universal
gravitational constant (6.67X10^-11 Nm^2/kg^2) and Ms is the mass of the
sun (1.99x10^30 kg).
You can derive this relationship by assuming that a planet's orbit is a
circle centered on the sun and equating the centripetal force acting on the
planet to the gravitational force. When you take advantage of the fact that
the tangential velocity, v, is equal to 2*PI*R/T you will find yourself
looking at Kepler's Third Law as mathematically described above.
What determines the planet's mean distance from the sun is it's total
energy. Interestingly enough, the relationship is a simple inverse
proportionality where R, the planet's mean distance from the sun, is
inversely proportional to (-) 2*E. Where E is the total energy/mass
(working the problem out in terms of energy/mass simplifies matters). The
constant of proportionality is just G*Ms again.
Mathematically: (1) R = G*Ms/(-2*E).
Now the total energy/mass at any time, E, is equal to the sum of the
(Kinetic Energy)/mass (K) and the (Potential Energy)/mass (U).
That is: (2) E = K + U
Since the object isn't moving initially, K = 0 and therefore E = 0 + U -
Where U is the initial (potential energy)/mass.
For our particular problem the value of U is determined by the following:
(3) U = (-) G*Ms/Re
The negative sign has to do with the fact that the gravitational force is
an attractive force and that the gravitational potential (U) is defined so
as to be zero at infinity. Re is the mean distance from the earth to the
sun (the distance your object is going to fall after it has been released -
1.49X10^11 m).
Since E = 0 + U,
(4) E = (-) G*Ms/Re.
That E is negative is important to the problem. Only objects with values of
E less than zero orbit the sun. If E = zero, than the path of the object is
a parabola (this is how you calculate the escape velocity) and if E > zero
the path is a hyperbola. In these last two cases, the planet has sufficient
energy to escape the gravitational influence of the sun.
Substituting eq.(4) into eq. (1) you arrive at:
(4) R = Re/2.
As a result, T^2 = 4*PI*PI*R^3/(G*Ms) becomes T^2 = PI*PI*Re^3/(2*G*Ms).
Putting the appropriate values into the last equation and taking the square
root gives you the value for the period:
(5) T = about 128 days.
One quarter of this is the time it takes to reach the center of the sun
after falling freely from a distance equal to the earth's mean distance
from the sun.
Voila!!: About 32 days.
Of course, in the actual case, the object will hit the surface of the sun
first.
The difference between the 32 days and the time taken to just reach the
surface of the sun can be estimated by considering the following:
>From the conservation of energy law, you can determine the speed of the
object after it has fallen from the earth's orbit to the surface of the
sun. The kinetic energy it has, just as it strikes the surface of the sun,
equals the change in it's potential energy.
The speed turns out to be about 616 km/sec. If, at this moment, it entered
a tunnel that went to the center of the sun and continued traveling at this
speed, it would take a little less than 20 more minutes to reach the sun's
center.
The fact is that it would continue to accelerate as it fell towards the
center so the actual time would be less than 20 minutes. Thus the
difference in time between falling towards and slamming into the surface of
the sun as opposed to reaching it's center would be 0.04% at most. A
pretty trivial difference.
Sorry for being so long winded but you did ask for a solution to the problem.
You could have solved the problem by consulting a math mavin and having the
mavin slog through the second order differential; or make use of the
subtle, mathematical alternatives introduced by Lagrange and Hamilton a
little less than 200 years ago (the solution is more mechanical and
straight forward and focuses on the energy of the system rather than the
forces involved but it's still a slog for the mathematically challenged). I
think that the above solution which looks at the problem from a different
point of view is a little more straight forward. You do have to know that
the mean distance is an inverse function of the total energy of the body
but, if you do, the solution is pretty "straight forward".
The only questions are: "Is it a correct?" and "Are the numbers right?"
Well, I'm pretty sure I'll hear about it if the answer turns out to be "NO!"
ron
P.S. There is a fourth solution which is a lot simpler and more elegant but
I didn't offer it here since the reader would be justified in asking for
the proof of it's validity. The proof is in the above solution. I'll leave
it to the physicists in this mail-list to figure out what the fourth
solution is.
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