Time taken to free fall to centre of earth - if allowed?

AI Thread Summary
The discussion revolves around the time it would take to free fall to the Earth's center, with estimates suggesting it could be around 20 to 32 hours when accounting for air resistance and the decreasing gravitational force. Initial calculations indicate a lower limit of approximately 20 hours if a constant terminal velocity of 320 km/h is assumed. However, as the fall progresses, both air density and gravitational pull change, which would likely increase the total time needed. A simplified model without air resistance suggests a much shorter time of around 21 minutes, based on harmonic oscillation principles. The conversation also touches on the feasibility of such a fall on other celestial bodies like the Moon or Mars.
cool_ranjit
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Hi,
Some time back I was saw a discovery documentary, can't remember the name, but in that somewhere they told that under free fall if possible to centre of earth, it will take around 32 Hrs or so.. No calculations or details were given.

So I wonder will it take so much time or how they came to that calculation, they might be wrong ..

So my question is how much time will it take to reach the centre of earth. Following assumptions are taken into account:

1) Person weight : 60 KG
2) Waist diameter : 34 cm ( to determine the cross section if required )
3) Air drag or friction has to be taken into account
4) We know that g ( 9.8 m/s 2 ) reduces towards we approach the centre of earth. The graph is linear. So if possible this can be accounted.
5) Radius of Earth : 6300 KM
 
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Welcome to the forum, cool_ranjit - cool thread ! :smile:

Though I can't provide a complete solution to you, some ideas might be useful for a start (they might be wrong, I'm just a hobby physicist. Maybe some kind soul would like to comment on their validity ?)

First, calculating the exact air resistance is difficult, even with the weight and waist diameter you provided. Just one point: Would the person jump head first or with a horizontal orientation ? Suggestion: Just consider the range of velocities that a parachutist can have during his free-fall phase:

"They reach terminal velocity (around 120 mph (190 km/h) for belly to Earth orientations, 150-200 mph (240-320 km/h) for head down orientations)" http://en.wikipedia.org/wiki/Parachuting

So if a person has a velocity of 190-320 km/h (depending on the circumstances) there will be an equilibrium between air resistance and gravity. Just for fun, let's take the 320 Km/h and see how long it would take to reach the centre of the Earth if this velocity would be constant all the time: 70875 seconds, nearly 20 hours.

Now its easy to see that these 20 hours are a lower limit for the time he would actually need: As you correctly stated, gravity will decrease the closer we come to the centre. At the same time, air pressure and air density will continuosly increase. Both effects together mean that the equiliibrium velocity of the jumper will continuosly decrease from the maximum value of 320 km/h that I assumed at the surface of the earth. So the guy will actually need more time.

So we have a lower limit of 20 hours - the 32 hours you qouted are still credible.

To get the real time, we would have to calculate air density as a fuction of radius r. Hmm... I once knew how to derive air density near the surface of the earth, but this assumes that g is constant... I wonder whether there is still a nice analytic expression if g is proportional to r - perhaps we have to go numerical.

Anyway, when we have air density as a function of r, we can calculate the equilibrium velocity for every r. Doing an integration will then give us the total time needed. (It's cheating a litlle bit - at the moment the guy jumps into the hole, he will not have the equilibrium velocity of 320 km/h, but instead 0 km/h. But he will accelerate to equilibrium velocity on the small time scale of 10 seconds, so it's not too bad an approximation).

EDIT: Just for completeness: If you would be willing to drop the condition that air resistance has to be taken into account, then it would be quite easy. The linear dependence of gravitiy on r would result in a harmonic oscillation (the position as a function of time would be a sinus curve) which is quite easy to calculate. But the resultant time would be much smaller than 32 hours. And it would not be as much fun as your problem. :wink:
 
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Pity the core is too hot and high-pressure to drill (and shore up) a hole. That would be a fun ride. I'd pay to do that. And it's safe because the sinusoidal oscillation ensures no abrupt jerking as you gently settle down to the center. Then I guess they have to hoist you back up (Jesus I'd hate to have to take the stairs).

Maybe on a less geologically active planet? Can we do it in the moon? Or Mars? I guess it would be less fun because of the smaller gravity, but I'd still pay to do it.
 
Xezlec said:
Pity the core is too hot and high-pressure to drill (and shore up) a hole. That would be a fun ride. I'd pay to do that. And it's safe because the sinusoidal oscillation ensures no abrupt jerking as you gently settle down to the center. Then I guess they have to hoist you back up (Jesus I'd hate to have to take the stairs).

Maybe on a less geologically active planet? Can we do it in the moon? Or Mars? I guess it would be less fun because of the smaller gravity, but I'd still pay to do it.

I'd pay for it too, but maybe not as much as you'd pay. Say you could do it on the moon. How much exactly is it worth to you?

Name the right price and Disney will colonize the moon in under twenty years. Get enough people to prepay and they'll do it in ten.

And no doubt impress their logo into one corner as seen from earth. Political pressure would never let them reform the whole surface into Micky Mouse ears.
 
Because the strength of gravity grows linearly (for a sphere of constant density) with distance from the center, this is a Hooke's Law problem - neglecting air (which is probably a good thing: you don't want the atmosphere falling down your hole!). The spring constant k/m is 9.8 m/s^2 /6300 km or 1.56 x 10^-6 /s^2. The period is 2pi/sqrt(k/m) = 83 minutes. The time to the center is 1/4 that, or 21 minutes.
 
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