Question

Lets say we can drill a tunnel straight through north- and southpole. Describe the movement of a rock that falls from northpole. Lets say the earth has a homogenous density and discard friction. Calculate after how much time the stone reaches the mid-point of the earth.

Answer 1

i know i should use : K = GmMr/R^3

Answer 2

radius of the earth: 6371 km, mass: 5,9737.10^24

Answer 3

Just a suggestion:

\[v=\frac{dy}{dt}\]

Find an expression that defines the kinetic energy and thus speed at every point, then do an integral for y=R to 0. You should get an expression for time.

Answer 4

find an expression the defines kinetic energy..

Answer 5

you just mean the formula for kinet. en.?

Answer 6

yes.

Answer 7

why? i have the answer of one of my classmates and it doesn't look like he used kinetic energy formula anywhere

Answer 8

i see: \[\frac{ dm/dr*r }{ R^2 }\] and \[-kx=ma\]

Answer 9

and the first equation i posted

Answer 10

I have no idea what \[\frac{ dm/dr*r }{ R^2 }\] is supposed to do, but if you want there is another way using -kx=ma.

Recall what g is in terms of G, M and R. \[g=\frac{MG_{E}}{R^{2}}\]
Thus, Force becomes: \[F=-mg\frac{r}{R}\]
This looks similar to hooke's law:\[F=-kx\]
Which you will find that:\[k=\frac{mg}{R}\]

For spring oscillations,angular frequency is:
\[\omega=\sqrt{\frac{k}{m}}\]\[T=2\pi\sqrt{\frac{mR}{mg}}\]

This expression is for 1 compelte oscillation, so half of it is:\[t=\pi\sqrt{\frac{R}{g}}\]

Answer 11

wow, alright

Answer 12

I would do it like this:
\[dy=\frac{ adt^2 }{ 2 }=\frac{ GMdt^2 }{ 2y^2 }\]
\[2dyy^2=GMdt^2 \rightarrow 2 \int\limits_{0}^{R}y^2dy=GM \int\limits_{0}^{t}dt^2\]
\[\frac{ 2R^3 }{ 3 }=GMt^2 \rightarrow t=\sqrt{\frac{ 2R^3 }{3GM }}=657,65 s\]

Answer 13

could you explain hooke's law? what is k and x?

Answer 14

Well, you threw out the idea of :−kx=ma...
Hooke's law states that the restoring force is proportional to the displacement x. this proportionality constant is k.

@Fifciol i'll try to make sense of that :P

Answer 15

restoring force..proportionality constant... alright...

Answer 16

@Fifciol you used the kinematic equation for dy?

Answer 17

then dt^2 so it's a function of time...?

Answer 18

I used kinematic equation and dt^2 is differential

Answer 19

alright

Answer 20

and a = GM/y^2 ???

Answer 21

yes

Answer 22

how'd you think of that?

Answer 23

F=ma=mg

Answer 24

F=GMm/y^2=ma

Answer 25

y is the radius, right?

Answer 26

distance

Answer 27

R would be at surface of the earth

Answer 28

oooh ok

Answer 29

so you described the movement using kinematic equations, then integrated to know the time?

Answer 30

yes

Answer 31

got it, i think it's correct. Unless @Festinger has a different answer?

Answer 32

I'm not sure the first equation I wrote down

Answer 33

Even my first suggestion gave me a different expression.

Answer 34

@Mashy any thoughts?

Answer 35

@amistre64 is that correct:
\[dy=\frac{ adt^2 }{ 2 }\]
?

Answer 36

if a is constant, it is right?

Answer 37

i havent really learned to model differential equations

Answer 38

wouldnt this act like a pendulum tho? or are we trying to determine when itll first reach the center?

Answer 39

when it will reach the center

Answer 40

I am unsure if \[\int_{0}^{t}dt^{2}=t^{2}\] is a legal move....

Answer 41

and there is some relation to the inverse of a square of a distance right?

Answer 42

Fificiol is wrong.. cause the mass of the earth is not a constant as you go down the earth..

Answer 43

i mean the effective mass that provides the gravitational force :P

Answer 44

doesnt assuming homogenous density and whatnot mean that we are simplifying all the complexity from it?

Answer 45

yesyes

Answer 46

yes.. so u need to get mass as a function of distance..

use the densitiy as a constant.. and get a function for mass in terms of M and R

Answer 47

it has homogenous density, and when you deal with gravity and planets you can treat the mass as if all of it were at center

Answer 48

While true that:\[\int d(t^{2})=t^{2}+c\] is true, but in this case it is actually: \[\int (dt)^{2}\]

Answer 49

@fificiol.. as you go inside the earth, the gravitational force decreases..

in your equation u should modify M

m = My^3/R^3...

Answer 50

because according to your equation, the force keeps increasing as you go inside the earth.. so you get a huge error.. u ll get acceleration to be increasing.. infact the acceleration is decreasing! ^_^

Answer 51

i think.. everything else holds..!!

Answer 52

Mashy is right.

Answer 53

i'm seeing this equation from my classmate: \[m.a=\frac{ G.m.M }{ R^3 }\]

Answer 54

times r

Answer 55

yes, Mashy you're right

Answer 56

\[m.a=\frac{ G.m.M }{ R^3 } . r\]

Answer 57

so...what now?

Answer 58

follow what i posted for now until someone comes with a better solution.

i am uncomfortable with the pi appearing...

Answer 59

ok

Answer 60

but i would really suggest you go back to look at hooke's law. your friend seems to know hooke's law.

Answer 61

i don't understand how it applies to this

Answer 62

the expression that you posted that your friend wrote ma=-kx gave me the idea to link hooke's law and gravity in this case.

the equation of force shows that it is proportional to r, like the x that appears in hooke's law.

after that i just rode on the results from hooke's law with spring oscillations.

Answer 63

i see, you used it for the proportionality

Answer 64

My new way of thinking on this problem:

\[ma=-kx \rightarrow x''+\frac{ k }{m }x=0\]
\[\frac{ GMm }{ R^2 }=kR \rightarrow k=\frac{ GMm }{ R^3 }\]
\[x''+\frac{ GM }{ R^3 }x=0 \rightarrow T=2\pi \sqrt{\frac{ R^3 }{ GM }}\]
\[t=\frac{T }{ 4 }=\frac{ \pi }{ 2 }\sqrt{\frac{ R^3 }{ GM }}\]

Answer 65

Yes, you are right. It should be divided by 4 instead of 2. My bad.

Answer 66

why?

Answer 67

One oscillation is from one end > center > another end > center > original end

when i divided by 2 it is the solution to: one end > center > another end

Answer 68

and you need, one end>center ?