How do we calculate the 42min time

[jsm6252]

if we made a hole from one side of the planet all the way through the Earth to the other side and jumped into it. it would take 42 minutes to get through to the other side before falling back down. my question is how do we caculate the magic 42 min time ?

 

[tiny-tim]

welcome to pf!

hi jsm6252! welcome to pf!

hint: g(r) is proportional to r (because it's proportional to the mass inside radius r, and to 1/r²)

 

[jsm6252]

does this mean that g on the serfice of the Earth is 9.8ms2 so 1/4 of the way down g is about 2.5ms2

 

[gneill]

does this mean that g on the serfice of the Earth is 9.8ms2 so 1/4 of the way down g is about 2.5ms2

Try 1/4 of the way up -- radius is measured from the center outwards.

 

[D H]

hi jsm6252! welcome to pf!

hint: g(r) is proportional to r (because it's proportional to the mass inside radius r, and to 1/r²)

That is assuming a constant density, and that is the assumption you need to make to get this result. (A non-constant density will yield a different answer.)

 

[jsm6252]

could you please give me an example.
physics is not my area of xpertise

 

[gneill]

could you please give me an example.
physics is not my area of xpertise

For the non-physicist:

http://www.physicscentral.com/explore/poster-earth.cfm"

 

[D H]

http://www.physicscentral.com/explore/poster-earth.cfm"

On the other hand, if all of the mass was concentrated at the center it would take only 29.9 minutes to make the same trip. Finally, with a realistic mass distribution the answer is 38.2 minutes. (See [thread=471524]this thread[/thread]).

 

[jsm6252]

how did you calculate the time

 

[gneill]

how did you calculate the time

Well, you could solve the differential equation representing the motion, or you could do something clever with the known result for the period of an elliptical orbit with respect to the length of its semimajor axis.

 

[jsm6252]

could you please give me an example.it's seems easier to follow with a few figures in equations
many thanks

 

[gneill]

could you please give me an example.it's seems easier to follow with a few figures in equations
many thanks

What class is this for? It is homework, right?

 

[jsm6252]

not for class just general interest as i saw a documentery about the universe

 

[gneill]

not for class just general interest as i saw a documentery about the universe

Ah. Well, this isn't quite the right forum for this inquiry. Nevertheless, for the case that D H mentioned, where all the mass is concentrated at a point at the center of the Earth, consider a body in an orbit around that mass. If the body were in a circular orbit with a radius equal to that of the radius of the Earth (so it just skimmed the surface), then its period would be given by the formula

$$T = \frac{2 \pi}{\sqrt{\mu}}\right) r^{3/2}$$

where $$\mu$$ is the gravitational parameter for the Earth, G*M.

Now, the same formula applies to an elliptical orbit, where r is replaced by the length of the semi-major axis of the ellipse.

This is where the clever bit comes in. It doesn't matter how broad or skinny the ellipse is, the period will remain the same if the length of the semimajor axis remains the same. So if we imagine that the orbit of the object is elliptical, and we let that ellipse become more and more eccentric (skinnier and skinnier), then eventually we will have an orbit that looks like a straight line, the object falling straight towards the center and bouncing back up again (the turn around the backside of the central mass point becoming arbitrarily close to the mass point). The semimajor axis of that "ellipse" is half the distance from the central point to the top of the "orbit", which is just half of the Earth's radius. This gives us for the trip to the center and back,

$$T = \frac{2 \pi}{\sqrt{G M}} \left( \frac{R_{earth}}{2}\right)^{3/2}$$

Plug in the numbers for M, G, and R_earth and see what you get.

For another pleasant surprise, plug in the values to obtain the period of the Earth-surface-skimming satellite. That is, the period of time it wold take a satellite to travel to the far side of the Earth and back again, traveling around, say, the equator (ignoring terrain issues and air resistance).

 

[jsm6252]

thank you