Particle Falling to Earth: Solving w/ MultiV Calc

[Trying]

Ok. I'm not sure if this belongs in this section. This is a problem relying on physics 1 concepts, but I do believe it requires MultiV Calc (which I haven't taken) to solve.

1. The problem

How long would it take a particle of distance D away from the center of the Earth to fall to the surface of the Earth (from rest)? This is of course assuming that the only force acting upon the object is the Earth's gravitational force.

Homework Equations

Fg = GmM/d^2 (M is mass of Earth and m is mass of object. G is universal gravitational constant and d is the distance from the center of the earth.

The Attempt at a Solution

This was way beyond my current skill level, but I tried it anyway.
from the equation for Fg, I derived acceleration as a function of distance from the earth, so a(d) = GM/d^2.
but to make this in terms of x, starting from position "zero": a(x) = GM/(D-x)^2

Then came my attempt to incorporate the time:
a = dv/dt; v = dx/dt → dx/v = dt → a = v*dv/dx = GM/(D-x)^2 → ∫vdv = ∫GM/(D-x)^2dx

v^2/2 = GM/(D-x) + C , but v initial is zero so C = 0. → v = √(2GM/(D-x))

v = dx/dt = √(2GM)/√(D-x) → √(D-x)dx = √(2GM)dt → -2/3(D-x)^(3/2) + C = √(2GM)t
Since we know at t = 0, x = 0, C = 2/3D^(3/2)
so → -2/3(D-x)^(3/2) + 2/3(D)^(3/2) = √(2GM)t
t = 2/3/√(2GM)(D^(3/2)-(D-x)^(3/2))

Conclusion:
As I was typing this up, I started to see things and made some educated guesses and progressed along. But being a novice at calculus, I don't know if the process I took was acceptable.

I would truly appreciate any comments, corrections, or hints as to what direction I should be taking. (FYI: just to plug in some values, using the avg distance from the center of the moon to the center of Earth, it came out to take a little over 49 hrs for the moon to first touch the Earth surface to surface.)

 

[ehild]

a(x) = GM/(D-x)^2

Then came my attempt to incorporate the time:
a = dv/dt; v = dx/dt → dx/v = dt → a = v*dv/dx = GM/(D-x)^2 → ∫vdv = ∫GM/(D-x)^2dx

v^2/2 = GM/(D-x) + C , but v initial is zero so C = 0. → v = √(2GM/(D-x))

∫vdv = ∫GM/(D-x)^2dx=>v^2/2=GM/(D-x)+C

It is correct so far, but you used the initial condition incorrectly. At t=0, x=0, so C has nonzero value to fulfil the initial condition

v(0)^2/2=GM/(D-X(0))+C--------> 0=GM/D+C

ehild

 

[Trying]

Thanks a lot ehild for your help.

It's good to know I was atleast using the right approach. I can fix the rest now.

Thanks again!

 

[Trying]

Ok I take back what I said before. I'm stuck.

v^2/2 = GM/(D-x) + C
so v = √(2GM/(D-x)) - √(2GM/D)
v = dx/dt= √(2GM)(√D-√(D-x))/(√(D-x)√D)
dx(√(D-x)√D)/)(√D-√(D-x)) = √(2GM)dt

TI-89'd and walframalpha'd the left side to get:
D[-√DLn(x) + 2√DLn(√(D-x)-√D) + 2√(D-x)] + D^(3/2)Ln(x) -√(D)x + C = √(2GM)t
2D^(3/2)Ln(√(D-x)-√D) + 2D√(D-X)-√(D)x + C = √(2GM)t

Two problems with this:
No matter what x is, I end up taking the LN of a negative number. I "fixed" this my moving an even multiple coefficient to the power of the inside of the LN. (not sure that's legal)

The other problem is when I attempt to solve for C, noting the initial condition t=0,x=0, I end up taking LN(√(D-0)-√D) = LN(0) which is undefined.

Am I making illegal manuevers somewhere?
I am stuck at this point. Any pointers would be greatly appreciated
Thanks ahead again!

 

[D H]

Don't rely on wolframalpha. Yes, it's there, and you can use it as a crutch, but it is a crutch. Use it as a hint if you must. So, you've seen the (goofy) solution at wolframalpha. wolframalpha is not intelligent and is not aware that x is decreasing.

- What did wolframalpha do, and how would you do it differently?
- What does the fact that x is decreasing mean with regard to velocity?

 

[gDavidov]

1st step...
v^2 = A + B doesn't exactly imply v = sqrt(A) + sqrt(B) :B

 

[Trying]

gDavidov, I agree.
But I just skipped showing a few steps. v^2/2 = GM/(D-x) + C1
I just ignored C for the moment as it's not important I solve for it here: v = √(2GM/(D-x)) + C2
plugging in initial conditions, x=0, v=0, C2 = -√(2GM/D)

And to D H
I've been trying to make sense of what you said for a while. Could you give me some more hints?
And what do you mean x is decreasing? I'm probably not understanding you correctly, but the way I am setting up the problem, the variable x is increasing and that distance is being subtracted from the total distance for the position relative to the Earth's core. I mean it does make sense that at t=0, x=0, I would have a problem. But how do I fix this?

Thank you both for responding so quickly btw. I'm one of those people who can't get any sleep in when I can't figure out a problem I've set out to do.

 

[gDavidov]

you can't just ignore C1. Maybe you could if it was something like a differential equation or integral equation (in terms of derivatives and integrals) but this is only an algebraic equation.

v^2/2 = GM/(D-x)-GM/D
v^2 = 2GM/(D-x)-2GM/D
v = sqrt(2GM/(D-x)-2GM/D)
v = sqrt(2GM) * sqrt(1/(D-x)-1/D)
v = sqrt(2GM) * sqrt((D - D + x)/(D(D-x)))
v = sqrt(2GM) * sqrt(x/((D(D-x)) ) = dx/dt...

 

[Trying]

Thank you gDavidov, I will re-evaluate from there and see what troubles I run into next!

 

[ehild]

@gDavidov: It looks more clear in tex:

v^2/2 = \frac{GM}{D-x}-\frac{GM}{D}
v = \sqrt{2GM}\sqrt{ (\frac{1}{D-x}-\frac{1}{D})}

@Trying: Remember: you must nut ignore the integration constant.

You can make the expression under the square root a bit easier to handle:

\frac{1}{D-x}-\frac{1}{D}=\frac{x}{D(D-x)}
v=\frac{dx}{dt}=\sqrt{\frac{2GM}{D}}\sqrt{\frac{x}{D-x}}

this is a separable first-order differential equation, leading to the integral equation

\int{\sqrt{\frac{D-x}{x}}dx}=\sqrt{\frac{2GM}{D}}\int dt=\sqrt{\frac{2GM}{D}}t

Wolframalpha is a gorgeous help, but you need to check what you typed in, and the solution too, especially when the result is very complicated. You can try, type in int(sqrt((D-x)/x)dx). It looks awful, so I try to make the integral a bit simpler with the substitution
u=\sqrt{\frac{x}{D-x}}
x=D\frac{u^2}{1+u^2}
\frac{dx}{du}=D\frac{2u}{(1+u^2)^2}

The substitution leads to the integral

\int{\sqrt{\frac{D-x}{x}}dx}=D\int{\frac{1}{u} \frac{2u}{(1+u^2)^2}du}=D\int{\frac{2}{(1+u^2)^2}du}

Type in int(2/(1+u^2)^2)du) into Wolframalpha : it gives a much nicer result than before. Do not forget that
u=\sqrt{\frac{x}{D-x}},
and u=0 if x=0, so you can find the constant of integration with the initial condition u=0 at t=0. Then find the time when x=(original distance D - the radius of Earth R).

Just for fun, I tried the integration without Wolframalpha just by hand, as I had to do all in my life. It took some time while I could find out the method.

***\int{\frac{2}{(1+u^2)^2}du}=\int{\frac{2(1+u^2-u^2)}{(1+u^2)^2}du}=\int{\frac{2}{(1+u^2)}}du-\int{\frac{2u^2}{(1+u^2)^2}du}
The first integral is simply 2arctan(u). I integrate the second term by parts. :

\int{\left(\frac{2u}{(1+u^2)^2}\right) u du}

I integrate the expression in parentheses:

\int{\frac{2u}{(1+u^2)^2}du}=-\frac{1}{1+u^2}

and differentiate the second one, so

\int{\frac{2u^2}{(1+u^2)^2}du}=-\frac{u}{1+u^2}+\int{\frac{1}{1+u^2}du}=-\frac{u}{1+u^2}+\arctan(u)

Now I collect all terms in ***:
\int{\frac{2}{(1+u^2)^2}du}=2 \arctan(u)-\left(-\frac{u}{1+u^2}+\arctan(u)\right)=\frac{u}{1+u^2}+\arctan(u)+C

Wolframalpha is correct!ehild

 

[Trying]

@ehild and gDavidov

yes! I will never ignore that C again. Fixing that solved all my problems.

Thanks ehild for such an in-dept explanation.
And thanks everyone else for your input. I'll definitely frequent this forums from now on.

 

[ehild]

You are welcome. I hope seeing you soon again. That was a challenging problem .

ehild