I Forgot My Calculus >.<

[CanadaAotS]

Apt title... hehe

 

Anyways, I was trying to do some calculations based on the "Tunnel Through Earth" thread.

 

I wanted to actually find out how fast an object, that was dropped from earths surface to the center, would be going.

 

This involves Physics, but its more math intensive, so I put it in Math. If that rubs anyone the wrong way, then move it

 

Earths Radius: 6378100 m

Earth Mass: 5.9742 x 10^24 kg

Object (Sphere with app. human mass): 70 kg

 

Now, I'm thinking that throughout the fall the force of gravity is constantly changing.

 

[math]F_g = \frac{G*m_e*m_o}{{r_e}^2}[/math]

 

So I need to find out what [math]\frac{dF_g}{dt}[/math] is. (Force of gravity changing over time)

 

[math]\frac{dF_g}{dt} = \frac{({r_e}^2)*(G*m_e*m_o)' - (G*m_e*m_o)*({r_e}^2)'}{{r_e}^4}[/math]

[math]\frac{dF_g}{dt} = \frac{({r_e}^2)*(G*(\frac{dm_e}{dt}*m_o + \frac{dm_o}{dt}*m_e)) - (G*m_e*m_o)*2r_e*\frac{dr_e}{dt}}{{r_e}^4}[/math]

 

Now I think the change in masses are all 0 (mass never changes). I can probably simplify alot with that because it cancels out [math]\frac{dm_e}{dt}*m_o + \frac{dm_o}{dt}*m_e[/math] completely. I'll try fixing it up a bit...

 

[math]\frac{dF_g}{dt} = \frac{-2Gm_om_e\frac{dr_e}{dt}}{{r_e}^3}[/math]

 

Well, that worked out better then I thought it would

Not sure how to proceed from here...

 

I think I need acceleration or something so that I can take the integral to get velocity. But I'm in the dark... never took Physics Calculus, just regular Calculus lol

 

Pointing me in the right direction would be great

[CanadaAotS]

Well I've been told that my whole change in gravity is wrong lol.

 

Apparently gravity decreases linearly from the surface to the center

so...

 

[math]\frac{dF_g}{dt} = \frac{F_g}{r_e} = \frac{686N}{6378100m}[/math]

 

So force of gravity is dropping at -1.0756 x 10^-4 N/s

 

Acceleration = Force / mass

[math]a = \frac{F_g}{m_o}[/math]

[math]\frac{da}{dt} = \frac{\frac{dm_o}{dt}*{F_g} - \frac{dF_g}{dt}*m_o}{{m_o}^2}[/math]

 

I think I'm still right with change in mass being 0. plus I can substitute 1.0756 x 10^-4 N/s for [math]\frac{dF_g}{dt}[/math]

 

[math]\frac{da}{dt} = \frac{1.0756\cdot 10^{-4} Ns^{-1}}{m_o}[/math]

 

Again things simplify nicely the [math]m_o[/math]'s cancel out leaving me with a nice equation. When I put 70 kg in for mass of object I get da/dt equaling [math]1.5365*10^{-6} ms^{-3}[/math]

 

And now I'm stuck again lol... I know I have to take an integral somewhere here because I have to get from acceleration equation to velocity.

[insane_alien]

-1.0756 x 10^-4 N/s

 

this should be N/m not N/s

[CanadaAotS]

 

yah I realize that now that I look. result is the same though, just screwed up the units

[CPL.Luke]

I forgot the formula for gravity inside of the earth, (it shouldn't be linear) but if you find it all you have to do is integrate with respect to distance use a proper integral and then set whatever number you get out of that equal to 1/2mv^2

 

I may be able to derive the equation for the gravity in a minute, if I do then i'll post it

[CPL.Luke]

woops I guess it is linear, the formula is G Mx/R^3

 

where G is the gravitational constant

x is the distance from the center of the earth to the point where your measuring the field

R is the radius of the earth

 

integrating with respect to x from R to 0 will give you

 

(GM/2R^3) x^2

 

(taken between the limits it becomes)

 

(GM/2R^3)(-R)

 

which becomes GM/2R^2 multiply by m to get the potential energy and you'll have

 

-GMm/2R^2 get rid of the negative because were talking about speeds in a 1 dimensional instance, and set the equation equal to 1/2mv^2

 

(GM/R^2)^1/2=v or the velocity when you reach the center of the earth, neglecting friction

[CanadaAotS]

Yay! thanks

 

so...

[math]v = \sqrt{\frac{GM_e}{{r_e}^2}}[/math]

 

I wasnt THAT far off.

[math]v = \sqrt{\frac{(6.673 \mbox{ x } 10^{-11})\cdot (5.9742 \mbox{ x } 10^{24})}{6378100^2}}[/math]

[math]v = 3.13 ms^{-1}[/math]

 

wtf?!?

Is this right? I thought you'd be going ALOT faster...

[Sisyphus]

Ahhhhh! Your math is waaaay unnecessarily complicated! And it's wrong!

 

Here's a tip. The force of gravity decreases linearly by distance, since g decreases by the square of the radius and the amount of matter underfoot increases by the cube of the radius.

 

So, force varies as distance. Or, F = kx, where F is force, k is some constant, and x is the radius from the center of the Earth. Look familiar? That's right, it's Hooke's law. You can treat the whole thing as if it's a gigantic, very weak, ideal elastic spring.

[CanadaAotS]

That unfortanetly doesnt help me -_-'

 

Some math showing what I did wrong in get ~3m/s would be good lol

[Sisyphus]

Ok, look.

 

F = kx

ma = kx

 

at surface:

(1kg)*(9.8m/s^2) = k*(6400000m)

Therefore, k = 1.53*10^-6

 

Ok, now, since m=1, we can ignore it in calculations. Therefore:

a = (1.53*10^-6)x

This is acceleration as a function of radius.

 

Now, you just have to integrate to get velocity as a function of radius:

V = [(1.53*10^-6)x^2]/2 + p, where p is some constant. Simplify to:

V = (7.66*10^-7)x^2 + p

 

To complete the equation we need to know what p is. Luckily, we already know what V is at a given x. Namely, that your velocity is zero at the surface:

0 = (7.66*10^-7)6400000^2 + p

Hence, once solved, p = 31375360

 

Now, we have a complete function for velocity as a function of distance from the center:

V = (7.66*10^-7)x^2 + 31,375,360

 

Of course, at the center, x=0, therefore.............

 

V=31,375,360 meters per second

 

Or roughly 112,000,000 km per hour.

Or about 0.1c, meaning we've pretty much left Newtonian mechanics altogether.

 

 

As for your actual question, it looks like you just misplaced a decimal somewhere. I'm too lazy to look for it.

[CanadaAotS]

DAMN did I misplace a decimal lmao.

 

now thats a bit better! Can you imagine that? .1 c by the time you reached the center? lmao!

[CPL.Luke]

yeah I accidentally dropped an R^2 along the line, it should have been

 

[MATH]sqrt {\frac{GM_e}{r_e}}[/MATH]

 

sysyphus your math doesn't work, you can integrate to acceleration with respect to time in order to get velocity, but not acceleration with respect to distance. Just look at the units

[CPL.Luke]

or 7905.96 m/s

[Sisyphus]

 

sysyphus your math doesn't work' date=' you can integrate to acceleration with respect to time in order to get velocity, but not acceleration with respect to distance. Just look at the units[/quote']

 

Yeah, I realized that later. I was hoping nobody would notice.

[insane_alien]

lets look at this from the energy perspective.

 

at the centre you will have 0J of gravitational potential energy

at the surface you will have

(-6.6742*10-11*5.9742*10^24*70)/6278100

=-4376084406 J of potential energy

KE = 0.5*m*v^2

v^2=2*KE/m

v=sqrt(2*KE/m)

v=11181.7 m s^-1

 

which just so happens to be escape velocity as well.