How strong is gravity in the center of the earth?
According to you reasoning the force would actually be 4 times stronger.actually it would be a quarter of its strenght
g = Gm/ (.5r)^2
.5^2 is .25.....
Actually when I think about it constant density is a pretty poor approxiamtion despite what I said even in this context as a 1.1-th order 'guess' I'd say the strenght at half the Eart's radius is probably closer to the same strenght at the Earth's surface than it is to half the strength when varying density is factored in.The center of the earth is easy - 0. Fractional distances are more complex because the density of earth increases as you approach the core. EL's explanation is correct for constant density.
As you drop below the surface of the Earth, that portion of the Earth's mass which is "above you"(further from the center than you are) doesn't contribute to the force of gravity that you experience. Thus the above formula doesn't give the right answer as it assumes that the M responsible for the acceleration remains constant as r decreases, when in fact M decreases to zero as r decreases to zero.Chris (and others)
Sorry. I am not highly mathematically educated..
Above is mention that gravity should be (at the strongest at the surface of the earth) and from here decreasing to 0 at the centre.
When I use the formula g = G*M/r^2 - We will have 4E14 m/s^2 in the centre. !!
Now I am confused. - Does that mean that gravity increases or decreases towards the direction of the centre of the earth?,
I mean, is the answer that gravity is all just being cancelling out to 0 of opposite corposants in the centre?
How fast would acceleration of gravity be (approximated) half way to the centre, above we have different suggestions?
What about the theories of relativity, does that also conclude that gravity is 0 in the center and agrees to this.
If you assume a constant density for the Earth, you would get a value of 9.786 m/s², a little under that at the surface. (9.793 m/s²)Imaging measuring the acceleration of gravity by equator, in a 5000meter deep mine, or 5000 Meter below the surface of the ocean.
M = 5.97E24Kg
r = 6378 Km (by ækvator) – 5 km = 6373 Meter
What would the acceleration of gravity (g) be here?
I meant that the value of the acceleration would be less than the value of the acceleration at the surface.Janus
I am just a little confused.
I understand you answer that 5000 Meter below the surface of the earth, - in a mine, or in the ocean the acc. of gravity will be 9.786 m/s² (adjusted for density it will be 9.797 m/s) correct?
But what do you mean with: “a little under that at the surface. (9.793 m/s²)” ?
How deep is “a little under the surface”?
This can be found by a=v^2/r.PS!
If we also adjust for the speed of the earth’s rotation at equator?
Radius 6378km – 5km = 6373km - circumference = 40.058 Km = 40058000Meter / 86400s = 463,63 m/s
Do you know the influence of this?
Right or almost right, and all albout gravity inside the earth in this thread:So, initially, as you begin to move down towards the center of the Earth, the g force will go increase,.....