Question

A geophysicist has drilled a hole through the earth, from the north pole to the south pole. When she is at the south pole she jumps in to the hole and a some time later she comes up at the south pole again. (there is no air-resistence in the hole)
a) How many seconds/minutes/ hours later does she come up again?
b) How big is her maxvelocity?
c) how big is her max acceleration?

I know it will become like a harmonic oscillator but I dont know how to calculate it.

thanks in advance.

Answer 1

@IrishBoy123

Answer 2

within the earth the gravity distribution changes from the inverse square law to one were g is zero at the dead centre and increases linearly to g at the surface.

so it would be just like being attached to a spring-like force with constant k where k = g/R where R is earth radius, with end of the relaxed spring being centred at earth's centre.

if x is distance from dead centre you can begin to sub in directly to the standard shm equation x = A cost ( wt + delta) where these all have the usual meanings, delta being phase shift which you might wish to use here if x = R (ie A in the shm equation) at t = 0.

in order for the person to re-appear at the S pole, they would have to go through 1 full cycle, is S to N to S pole.

you could also do it by deriving the shm equations from force (but why bother), or by comparing energy levels again by analogy to the spring system, but i'd personally just have a crack at this the obvious way.

Answer 3

here is a really good exploration of the idea, with pictures and links that go beyond the simple idea, including a derivation of the gravity profile within the earth.....

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

Answer 4

thanks @IrishBoy123

Answer 5

In that link why is the period of oscillation equation =

\[T=2\pi*\sqrt{R_e/g}\]

how can m=R_earth
and g=k?

@IrishBoy123

Answer 6

ah, nice spot!!

in simple horizontal (gravity independent) spring shm, we see m x'' + kx = 0. kx is measured in Newtons.

this leads to w = root (k/m), with k = spring constant, m = mass

here: m x'' + (mg/R) x = 0, where (mg/R) x is measured in Newtons

in this case, the m's cancel. which makes complete sense from the physical reality. the mass of the person jumping in should be irrelevant.

i think this was my error in my first post. so thank you.

to summarise, the equivalent of k in this example, lets call it k' is k'= mg/R.

thus, w = root ( k'/m) = root (g/R)

i hope that helps...

Answer 7

yeah thanks, but the maximum velocity? how do I derive that? and the max acceleration must be gravity at the earths surface? cause its just gets smaller the closer I get to earthcentre?

@IrishBoy123

Answer 8

have you gone with x = R cost (wt - π/2) ??????
where: R = rad earth & w = root (g/R)

if so

x' = - wR sint(wt - π/2)
x'' = - w^2 R cos(wt - π/2)

cos and sin oscillate between 1 and -1

so:

max/min vel = [+ or -] wR = root (g/R) * R = root (gR)
maxmin accel = [+ or -] w^2 R = (g/R) * R = g

Answer 9

No I dont think my math is that good to use derivate it like that. I did like this:

at the earth surface
\[F=mg\] and \[F_s=kR\]

\[F_s=F >mg=kR>k=mg/R\]

and a harmonic oscillator have the equations
\[\omega=\sqrt{k/m}\]
and
\[T=2\pi/\omega\]

So I first solved angular frequency:

\[\omega=\sqrt{mg/R/m}>\omega=\sqrt{g/R}>\omega=0.00125\]

then the frequancy in the Period equation.

\[T=2\pi/\omega>2\pi/0.00125>5060 s \approx 84 minutes\]

then for maximum speed,
\[F=mg.F=mv^2/r>F=F> mg=mv^2/R_j>v=\sqrt{g*R_j}=7909,7m/s\]

and maximum acceleration is 9.82m/s^2


Do you think i will get a passing grade for this assignment?
@irishboy123

Answer 10

@perl

Answer 11

it can be shown that as you move inside the Earth, the Force acting on you will reduce linearly and is proportional to radius. I think you had that
F = k*r

Answer 12

@pate16

i am happy to go through your approach. give me a bit so i can sit down and look.


i still think that from a physics perspective you should stick to well established shm principles, and that from a cursory glance your maths looks well up to what is needed to apply the basic, ephemeral physics shm paradigm.

i'll get back to you but perl might get you there in meantime. got to put my kids to bed.

Answer 13

(Assume that R_e is the earth's radius)
We know that for simple harmonic motion

$$
\Large {
T= 2\pi \sqrt{ \frac {m}{k}}

\\ \text{But } ~F = k\cdot R_e ~~\to k = \frac {F}{R_e}
\\ \therefore\\
T= 2\pi \sqrt{ \frac {m}{\frac {F}{R_e}}}
T= 2\pi \sqrt{ \frac {m}{\frac {mg}{R_e}}} = 2\pi \sqrt{ \frac {R_e}{g}}

}
$$

Answer 14

$$
\Large{ R_e = 6 378 100~ m\\
g = 9.81~ m/s^2\\
T = 2\pi \sqrt{ \frac{6 378 100}{9.81}} \approx5068.8856 ~sec
\\ 5068.8856 ~sec = 84.5~ min
}$$

Answer 15

I don't quite follow your reasoning for finding maximum velocity

Answer 16

No im not so sure of that either.. I saw it like a centripetal force but thats not quite true, is it?

Answer 17

F= mv^2/r is used for uniform speed circular motion

Answer 18

but the answer comes out approximately correct, so maybe you can use it

Answer 19

really? but how would you do it?

Answer 20

that approach would be used to find the speed of orbit for a satellite under the influence of gravity

Answer 21

but in our problem we are doing linear motion,
but I would have to ask @IrishBoy123 if this approach works.
F = mv^2/r
F =mg

mg = mv^2 / r ---> v = sqrt( gr)

maybe its a coincidence that it is the maximum velocity

another way to approach this would be to make a velocity equation at time t

Answer 22

I just checked this website, it is definitely correct.
The maximum velocity is sqrt( R_e * g)
http://www.quora.com/What-will-happen-when-a-ball-is-dropped-in-a-tunnel-that-passes-through-the-center-of-the-Earth

Answer 23

oh nice haha, But could you show how you would do?

Answer 24

here are some formulas
http://www.schoolphysics.co.uk/age16-19/General/text/Formulae_list/index.html

Answer 25

maximum velocity occurs when the derivative of velocity is zero , so when acceleration is equal to zero

Answer 26

we can use the equations for simple harmonic motion
$$
\Large{
\\ displacement = r \sin ( \omega t ) \\
velocity = \omega ~r \cos(\omega t)
\\
acceleration = -\omega^2r\sin(\omega t)

}$$

Answer 27

Let me give you an intuitive explanation, there are more general approaches that deal with it more generally.