MR_T

so i figure ---- it we got people posting about std's HMT might save the day. I got this physics assigment due tomorow and cant figure out this problem. Its a new chapter and I'm busting my *** trying to get my ---- done before 8am. Somebody take a crack at this and save a nignog from dropping out.

heres the problem

"Suppose we dig a hole through Earth that extends all the way through the center forming a straight shaft. Show that if we drop an object through the shaft, it will oscillate between the two far ends exhibiting simple harmonic motion; in other words, show that the gravitational force on the object has the form of Hooke’s Law F = kx, and derive the equivalent expression of the k constant. You will have to use Newton’s law of gravitation:

F = Gmom/r^2 where mo is the object’s mass and m is the mass enclosed within a spherical volume of radius r and the same r is the distance of the object from the center of Earth.

Assume that Earth has uniform density ñ, mass M, radius R and that the shaft is a complete vacuum."

thanks HMT

buk9tp

LOL HE SAID SHAFT

MikeJ-2009

If Joseph Davis was here he would come up with a bunch of big words, and even your teacher would be tricked into thinking it was the right answer.

buk9tp

heres some info that might be useful

http://www.school-for-champions.com/...l_equation.htm

MR_T

thanks buck... haha i was laughing first time i read the problem too, had people look at me like i was stupid for laughing at the shaft. anyway i am good with newton's laws, i just dont have a clue how to relate them with hooks law. what i am really wondering is if g varies at different as you go deeper, come on i know there are some physics gurus out here, somebody bust out an equation

RacingTiger

Im going to take a stab at it but wouldnt the equation to determine the force they apply be F=g(m1 x m2)/r^2 where m1 and m2 are the object mass respectively and g is newton's constant 6.67 x 10- 11 N m 2 /kg 2. and r is the distance between the two. I would assume you'd need to double this since newtons third law states they'ed be attracting each other.

Hook's law f=-kx
since K is the is the force constant it would be replaced with G Newton's constant since thats the spring
since x is the spring distance stretched it would be r


Thats my thoughts but hey this is the internet after all

NIGn0g

That suck that I'm drunk, I like that kind of problem. I know what the problem want you to prove, but my brain is stuck in alchool and weed.

USS

This was on my Physics final freshman year... I still have mine

for some r<= Rearth

F(r) = -GmM(r)/r^2

Volume of sphere, we know M(r) = ((4*pi*r^3)/3)p p = density

so, F(r) = (-Gm/r^2) * (4*pi*p*r^3)/3

Or in the form you need

F(r) = -(4*pi*G*mp*1/3)*r

So for you comparison the -4... blah blah blah thing is similar to your linear spring constant and your "deflection" term is r.

rsmith2786

I hate retarded questions like that. I skipped over a lot of the retarded classes by studying abroad. I got A's, drank way too much, traveled, and didn't learn anything. It screwed me for some of my later classes but that just forced me to learn the important stuff from the previous classes.

t_cel_t

do you have to factor the small amount that the earth moves towards the object as it falls?