cpumasterwv":9ywszni9 said:
EUREKA! I wasted an entire weekend trying to work out the whole variable mass theory. So then help me out with this. If you were working out the amount of time it would take to reach the center, then said object would actually decelerate as you approched the center due to a lower amount of gravity? (Were assuming air resistance equal to sea level)
And if this is the case you get into longer and longer decimals to the point of would you ever reach the true center? I guess thats all up to scale. .000000625 in doesn't really figure into anything when the person... I mean... object is 6 ft tall.
WARNING: Math ahead!
This is actually in general a very non-trivial problem, since it's essentially a fairly difficult second-order differential equation (acceleration, the second time derivative of distance, as a function of distance^-2), although adding in the fact that it's not a point source can make things more or less difficult, depending on your assumptions. So let's look at the problem:
Newton's second law: F = ma, relates force to acceleration
Newton's gravity law: F = -GmM/r^2, relates distance, mass of planet, etc., to force.
Put them together and cancel out m (your mass), you have:
a = r'' = GM/r^2 (equation 1)
Where ' denotes a time derivative. This problem is rather subtle even assuming that M is constant (as you did - a point source). But we're doing a different problem, where (assuming we're already beneath the Earth's surface), M is a function of r - call it M(r). That's the mathematical difference. Let's rewrite the equation in terms of the density of the Earth, since that makes life easier - density ρ is equal to mass divided by volume, or M/V=M/(4/3 pi r^3) = 3/4pi M/r^3 (assuming the Earth is spherically symmetric, or that a thin shell of Earth at any given radius looks uniform). Let's absorb the 3/4pi into our constants (so far there's just -G), call our constants -k^2 (the reason will be clear in a moment), and we can rewrite equation 1 as:
r'' = -k^2 ρ(r) r
So the solution of our equation depends on our choice of ρ(r), the density of the Earth as a function of distance from the center. I have no idea what this is in real life, though I'm sure it's not pretty, but let's (for fun) assume that ρ(r) is constant - in other words, every point inside the Earth has the same density as any other point. If that were the case, then we absorb ρ into our constants and we have a familiar equation:
r'' = -k^2 r
This is the equation for simple harmonic motion. Now you see why we called our constants -k^2 (instead of k) - we need to make it clear that the constants are negative (due to the minus sign in Newton's gravity law), because if they were positive, we'd have exponential behavior - clearly wrong! So the general solution to this is:
r(t) = A sin(kt) + B cos(kt)
Where k is some combination of positive constants. So the motion is harmonic, nice periodic trig motion. With initial conditions we can knock this down a bit further. Let's assume we start from rest at the surface of the Earth (call it capital R). Then we have:
r(t=0) = R = A sin(0) + B cos(0) = B
r'(t=0) = 0 = k A cos(0) + k B sin(0) = k A
So A is equal to 0, and B is equal to R. The final equation of motion is:
r(t) = R cos(kt)
I just saw dangineer's post - he's entirely right, of course
Air resistance is a schlep, and I won't bother; not to mention we're dealing with the inside of the Earth's surface, where air resistance is hardly a concern!
Apologies for thread drift, MeteorWayne.
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