Gravity Question. (more)?

Question

The concept of gravity has always boggled my mind. But here's my hypothetical question: If you could did a hole to within ten feet of the center of the earth then how much gravity would there be? If I was at the exact center of the earth would I be weightless? I know the earth's inards are molten. This is purely hypotheticle.

Answer 1

If you were right at the center then the symmetry of the situation would mean that you'd be weightless since the gravitational force would be balanced all around.

Similar situation if you were just a little off the center. Most of it would balance except for a little bit. Think about starting at the earth's surface and moving to the center. You'd start at your weight and become zero at the center. Assuming continuity and noting the nice behaviour of gravity then it must decrease in a smooth fashion. If you continued through your hole to the opposite side you'd weigh the same but in the opposite direction, always pointing towards the center.

Answer 2

If you remember g=9.8 m/s^2 is valid only not very far off the Earth’s surface.
In common case f=k*M*m/x^2=m*g, f is strength of attraction of two objects M & m at distance x between them, that is g(x)=k*M/x^2. This is true when objects are outside of each other! But you ask what will be when you are inside the Earth.
Let average density of rock be d, then a ball of rock with radius x has the mass M=d*4*x^3/3.
If x is less than radius of the Earth, then g(x)=d*4*x^3/3*k/x^2=(4d*k/3)*x. Now you see that the deeper you are the less you will weigh & in the very center when x=0 you’ll weigh nothing!
The outer rock sphere does not effect on you when you are inside of the sphere no matter where exactly you are. Why so? How can I prove it? The proof you’ll learn in a college!

Answer 3

yes, at the very center you'd be weightless

if you were a bit off center, you'd be brought back to the center

if you made a hole going through the Earth, and jumped in from the surface, you'd accelerate and accelerate but the acceleration would diminish and reach zero at the center (weightlessness) while speed would be at a maximum, and then you'd continue and slow down more and more, until you'd reach the surface on the other side, and then fall right back in.

and if there was no friction, you'd oscillate like this forever. But if there was some friction, then you'd end up stabilising at the center of the Earth.

at which point I guess you'd want to come out, which would be easy at first, simply walking, and at some point would require climbing. And there would be 6'400 kilometers, or nearly 4'000 miles of it.

good luck!

Answer 4

Call 10 feet as 3 metres.

As others have said, the symmetry of the earth cancels out all the mass in the shell of earth above you.
You would experience your weight as being due to a sphere of radius 3 metres and with the same density as the earth (fluid made of nickel iron alloy at that point ).

When at the centre, a sphere of radius 0 metres will attract you and cause your weight to be 0 Newtons, whilst your mass will remain at about 40kg,


The following is an example, but I have used the STP density of NiFe (as about 6000 kg/m3) so it will be inaccurate.

At 3 metres from the centre:
The volume of the sphere is: 113 cubic metres
The mass is 670000kg (or 670 tons)

The force on you would begiven by: F = G*m1*m2/R*R
The force on you would be: 0.0002 Newtons
as opposed to 400Newtons on the surface of the earth.

Where G = 6.673exp(--11) (cubicmetres/kg*squaresecond)
I haven't checked the calculation. Somebody correct me if it is wrong!

Answer 5

you would feel weightless. The gravitational pull from the surrounding earth would cancel itself out and you would feel weightless

Answer 6

gravity 0.Yes

Answer 7

i dont know