# How it Feels to Fall through the Earth

I was just watching a program on tv about gravity.

The showed one example of a tube through the Earth from one point on the Earth to another point on the Earth's surface.

This shows how we would accelerate relative to the Earth until we reached the centre, and then decelerate until we get to the second point, and then return in the same way on the way back in a continuous back and forth yoyo.

What they didn't mention and I think is true - but please correct me - is that if you boarded a sealed capsule which had one end up and some nice pictures to show the original up direction - that when the sealed capsule was dropped you would not ever be able to tell where you were by feel. You would feel weightless the entire journey and - apart from the pictures - lose all concept of up and down. There would be no feeling of acceleration or deceleration and there would be no feeling of changing direction of motion at anytime.

Is that correct?

DrGreg
Gold Member
Is that correct?
Yes.

sophiecentaur
Gold Member
2020 Award
If, however, the container were big (long) enough, you could detect the slight difference in g force between the parts which were nearest and furthest from the Earth's centre centre. The whole container would be falling due to the average forces (it would be rigid) but there would be a detectable microgravity within the capsule (a small compression force, aamof) You could calculate where you were on your journey from this.

Cool. Thanks people. Isn't that awesome hey! I think I read a sci-fi book that got that wrong..

Hi Sophie. I understand what you are saying I think. I think you are saying that the gravitational difference between head and toe will change throughout the travel.

I'm just wondering if that is correct within the bounds of the planet. I thought gravity changed at a linear rate through the depth of a planet unlike it decreases at approximately a squared rate from the centre at altitude above the planet surface? If that is so then the gravitational difference between the head and toe wouldn't change throughout the travel.

I just thought I heard that about the linear change. Not that it matters but I just want to be accurate. Which one is correct?

sophiecentaur
Gold Member
2020 Award
The attractive force towards the centre is proportional to the distance from the centre.
Easy to prove.
The there is no net force due to any of the Earth at a bigger radius than where you are so it's only the mass in the sphere beneath your feet that counts - that is proportional to the cube of the distance from the centre. The gravitational effect of that mass is proportional to one over the radius squared. Taken together, that gives the attractive force as proportional to the distance from the centre (cube/square).
You might notice that this force variation produces simple harmonic (oscillatory) motion and the period is the same for all amplitudes of oscillation. It's also the same period as for a Low Earth orbit.
It seems, btw, that the microgravity when you're falling through the Earth tends to compress objects whereas the microgravity when you're in orbit is tending to stretch things. Strange.

It was my understanding that the gravitational force decreased at a linear rate specifically due to the geometrical attributes of spheres that you described Sophie but being that of the mass overhead offseting the mass underneath to reduce change to a linear proportionate rate. But I can stand to be corrected. Can you point me to somewhere that specifically explains what you have just explained to me.

Can I infer from what you are saying that if you stand inside a massive spherical shell of any depth that you will be weightless on the inside surface and further that you will be weightless at all points within that shell at any anti-altitude above the inside surface right up(down) to the centre?

Last edited:
sophiecentaur
Gold Member
2020 Award
Try googling Gravitational Potential inside Spherical Shell. That should take care of the effect of all the 'overhead' parts of the planet. (of course this all assumes spherical symmetry).

Hi Sophie. I love what you are saying. Thanks. So effectively a Dyson shell could never work because insiders would float towards the Sun. The shell itself would have to be able to withstand the pull exerted from the Sun.

Regarding the earlier question I came across the following on Newton's Shell theorem. It says:

1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
2. If the body is a spherically symmetric shell (i.e. a hollow ball), no gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
3. Inside a solid sphere of constant density the gravitational force varies linearly with distance from the centre, becoming zero at the centre of mass.

The last one seems to agree with what I originally said if I am reading it right. Of course the Earth isn't of constant density and I was making that presumption in my example that it was. Is what I said okay in line with a constant density?

sophiecentaur
Gold Member
2020 Award
This question has an interesting knock-on. Because the restoring force to the centre is just proportional to density and distance from the centre the motion will be sinusoidal with time (SHM) and independent of the size of the planet. This means that, although the experiment is a no no with the Earth, because of the fantastic temperatures and pressures, you could do this thing for real if you could get hold of a near-spherical asteroid made of Earth-like rock and a solid, cool core. If you dropped a golfball / pebble down the hole. it would oscillate back and forth with an identical time period (about 90 minutes). You'd just have to choose the hole to be along the axis of rotation or the pebble would hit the sides.

A very expensive thing to do and it wouldn't advance the sum of human knowledge but what a fun project that everyone would appreciate.

Yeh, its cool hey. It fascinates me because in a perfect density sphere you would tend to expect to feel the change in direction but you actually feel nothing at all. Thanks for answering my questions Sophie. Shows how fascinating gravity is.

The tunnel must not have air in it :)

sophiecentaur
Gold Member
2020 Award
Yeh, its cool hey. It fascinates me because in a perfect density sphere you would tend to expect to feel the change in direction but you actually feel nothing at all. Thanks for answering my questions Sophie. Shows how fascinating gravity is.

But you would feel something if you are not a point mass. The motion of the object relates to the centre of gravity (this is not necessarily the same as the centre of mass, where the g field is not uniform) but the actual g field is not the same over the whole extent of a finite object. Hence, there will be tiny 'differential' forces acting on different parts of that object. Possibly the forces would not be 'feelable' but they would be detectable, in principle.
By measuring the microgravity, you could know where you were with respect to the centre of the planet. Hence, you could detect your motion and 'know' when you change direction.
The same goes for all 'free fall' situations in non-uniform g fields (which is everywhere, actually).

K^2
If there is an actual tube, it does not run between poles, and we actually account for rotation, the picture changes quite dramatically. You will not experience any acceleration along the tube, but you will experience a sideways acceleration due to Coriolis Effect, which means you'll be very far from "weightless" experience.

sophiecentaur
Gold Member
2020 Award
Why do you say the coriolis is 'intead of', rather than 'in addition to'? Depending on the speed of rotation, one or the other may dominate but it would be case-specific.

Yeh true K^2. So it would have to be non-spinning as well as even density throughout.

But Sophie if the gravitational change through the planet - in the above non-realistic but perfect scenario - is linear as Newton's Shell theorem indicates then the gravitational difference between head and toe of the capsule (or indeed the person) would always be the same - irregardless of if they are a point or not - would it not? In this respect the microgravity would remain consistent throughout the journey would it not also?

K^2
Why do you say the coriolis is 'intead of', rather than 'in addition to'? Depending on the speed of rotation, one or the other may dominate but it would be case-specific.
In the rotating frame, you'll experience gravitational, centrifugal, and Coriolis forces. However, only Coriolis force is perpendicular to the direction of the tube. The forces along the tube still act on the falling capsule, but because of free-fall in that direction, you won't experience these forces. Tube prevents the capsule from freely accelerating in direction perpendicular to the tube, however, so the Coriolis force will push the capsule against the tube wall, and anyone inside the capsule to the walls of the capsule. It will be the only force actually experienced by anyone inside.

sophiecentaur
Gold Member
2020 Award
Yeh true K^2. So it would have to be non-spinning as well as even density throughout.

But Sophie if the gravitational change through the planet - in the above non-realistic but perfect scenario - is linear as Newton's Shell theorem indicates then the gravitational difference between head and toe of the capsule (or indeed the person) would always be the same - irregardless of if they are a point or not - would it not? In this respect the microgravity would remain consistent throughout the journey would it not also?

If the forces are proportional to the distance from the centre then so is the Difference between the forces. That's what the theorem tells you. You could tell exactly where you were by measuring this difference.
Edit. I just re-read that. It's rubbish and you must be right!!!!

Last edited: