So in our talk about our thought experiment about drilling a tunnel through the earth I have this poser: When you go down into the earth, say in our tunnel, you will experience less and less force of gravity till at dead center, you would be in zero or microgravity. So considering that the spacetime curve that is the gravity field due to the mass of the earth, does it not stand to reason that at the center of the earth, the spacetime curve approaches flatness again and therefore has a dip in the curve from the max somewhere around the surface to a lessoning curve as you go underground to flat at the center, flat except we are still embedded in the spacetime curve of the sun, but locally zero?

Originally posted by sonhouse So in our talk about our thought experiment about drilling a tunnel through the earth I have this poser: When you go down into the earth, say in our tunnel, you will experience less and less force of gravity till at dead center, you would be in zero or microgravity. So considering that the spacetime curve that is the gravity field due to the mass of the ear ...[text shortened]... e center, flat except we are still embedded in the spacetime curve of the sun, but locally zero?

Originally posted by HandyAndy Now that's another story.

So here is a related question: the escape velocity of an object on the surface of the earth is about 11.2 Km/second. What about from a theoretical airless planet the same mass and size of the earth with a tunnel drilled all the way to the center of the planet. So suppose you launch from there, maybe you are afraid of aliens or something. So you launch your probe from dead center of the planet and you have magnets or something to keep you from contacting the sides of the tunnel and so forth, is the escape velocity still going to be exactly 11.2 Km/second or will it be higher?
Another way of posing the question is this: would the same amount of energy expended in the exact same rocket design be the same as if it were launched from the surface, would more energy be required to launch from the center of the planet?

Originally posted by sonhouse So here is a related question: the escape velocity of an object on the surface of the earth is about 11.2 Km/second. What about from a theoretical airless planet the same mass and size of the earth with a tunnel drilled all the way to the center of the planet. So suppose you launch from there, maybe you are afraid of aliens or something. So you launch your ...[text shortened]... aunched from the surface, would more energy be required to launch from the center of the planet?

More energy. Think about it this way: you need to get it from the centre to the surface. After which, it's an identical problem.

Or, your "escape velocity" at the centre will be the one that gives you a speed of 11.2 km/s when it reaches the surface.

It's the difference in potential energy that determines the escape velocity.

Originally posted by mtthw More energy. Think about it this way: you need to get it from the centre to the surface. After which, it's an identical problem.

Or, your "escape velocity" at the centre will be the one that gives you a speed of 11.2 km/s when it reaches the surface.

It's the difference in potential energy that determines the escape velocity.

Sure but wouldn't the energy required maybe even be LESS? Think of it this way: On the surface, you take off and you are accelerating say, three G's total. Well you feel three G's alright but your effective delta V is from TWO G's because you are overcoming the 1 G force of earth's gravity hole.
So when you are in the center of the planet, you are essentially weightless because of all the mass of the planet is distributed around you, which cancels out the gravity of the earth, at least in the center. Then you accelerate at 3 G's and you are increasing your velocity at a real 3 G rate. Of course the closer you get to the surface the more pull there is to slow you down but I think the net effect is you get to escape velocity with less energy required. So you might get to the surface with much more than 11.2 Km/sec.

Originally posted by sonhouse Sure but wouldn't the energy required maybe even be LESS? Think of it this way: On the surface, you take off and you are accelerating say, three G's total. Well you feel three G's alright but your effective delta V is from TWO G's because you are overcoming the 1 G force of earth's gravity hole.
So when you are in the center of the planet, you are essentia ith less energy required. So you might get to the surface with much more than 11.2 Km/sec.

No, that's not right.

Remember that the escape velocity is the velocity you need to be travelling at to escape the Earth's gravitational field without applying any further driving force (you can escape at a crawl if you can continuously apply the right force).

Launch from the centre at 11.2km/s, with no propulsion, and by the time you have reached the surface you will have slowed down.

A quick back of the envelope calculation suggests that the energy needed to lift from the centre to the surface is half the energy needed to lift from the surface to infinity. If that's right, then the escape velocity from the centre would be sqrt(1.5) of the escape velocity at the surface. Or about 22% higher.

Originally posted by mtthw A quick back of the envelope calculation suggests that the energy needed to lift from the centre to the surface is half the energy needed to lift from the surface to infinity. If that's right, then the escape velocity from the centre would be sqrt(1.5) of the escape velocity at the surface. Or about 22% higher.

Originally posted by sonhouse Where did you get the sqr root of 1.5 from?

Potential energy difference between the centre and surface = 1/2 of PE difference between surface and infinity.**

Therefore, escape energy from centre = 1.5 x escape energy from surface.

And energy = 1/2mv^2, so when comparing the escape velocities you get a square root coming in.

** This is obtained by integrating the size of the gravitational field (proportional to r inside the planet, and 1/r^2 outside it). I can't guarantee I made no errors there though.

Originally posted by mtthw Potential energy difference between the centre and surface = 1/2 of PE difference between surface and infinity.**

Therefore, escape energy from centre = 1.5 x escape energy from surface.

And energy = 1/2mv^2, so when comparing the escape velocities you get a square root coming in.

** This is obtained by integrating the size of the gravitational field ...[text shortened]... o r inside the planet, and 1/r^2 outside it). I can't guarantee I made no errors there though.

You mean the square root of 1.5 times surface escape velocity don't you?
1.2X and change. You said once before it was about 20% more energy to escape from the center than the surface. 13.7 Km/sec needed then.

Originally posted by mtthw ** This is obtained by integrating the size of the gravitational field (proportional to r inside the planet, and 1/r^2 outside it). I can't guarantee I made no errors there though.

Is this (propotional to r inside) a simplification that the Earth is homogenous, i.e. with a uniform density from the inner core and out to the surface?
If you calculate with a density as a function of distance from the center, what does this to the formula?

Originally posted by sonhouse You mean the square root of 1.5 times surface escape velocity don't you?
1.2X and change. You said once before it was about 20% more energy to escape from the center than the surface. 13.7 Km/sec needed then.

Originally posted by FabianFnas Is this (propotional to r inside) a simplification that the Earth is homogenous, i.e. with a uniform density from the inner core and out to the surface?
If you calculate with a density as a function of distance from the center, what does this to the formula?

True, that's exactly the simplication I used.

What does it do if you use a function? It makes it more complicated ðŸ™‚.

You need to integrate that function from 0 to r to get the gravitational force for r within the planet - you get something proportional to:

1/r^2 x Int{0 to r}(x^2 f(x) dx)

(which is proportional to r if f is constant)

You then need to integrate it again to get the potential energy difference.