# How strong is gravity in the center of the earth?

1. Dec 11, 2007

### Bjarne

How strong is gravity in the center of the earth?

2. Dec 11, 2007

### EL

At the center of the earth you would experince no gravtational force. (Since the force composants from the surrounding matter will cancel each other.)

3. Dec 11, 2007

### Bjarne

How strong is gravity then half way to the center of the earth ?

4. Dec 11, 2007

### jcsd

As a first order approximation (that is assuming that the Earth is a sphere of constant density, which isn't actually a bad approximation): half it's strength

Last edited: Dec 11, 2007
5. Dec 11, 2007

### AFG34

actually it would be a quarter of its strenght

g = Gm/ (.5r)^2
.5^2 is .25.....

6. Dec 11, 2007

### EL

According to you reasoning the force would actually be 4 times stronger.

The thing here is that the mass enclosed by a sphere with radius 0.5r is 0.5^3 times the mass enclosed by a sphere with radius r. (Assuming constant density of course.)

Since the enclosed mass goes as radius^3, the force will be proportional to the radius. Hence, as jcsd said, the gravitational force halfways to the center of the earth will be half of the force at the surface.

Last edited: Dec 11, 2007
7. Dec 12, 2007

### Chronos

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.

8. Dec 12, 2007

### jcsd

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.

9. Dec 12, 2007

### Chris Hillman

To be precise

Like everyone else, I'll assume we are discussing Newtonian gravitostatics (in the field theory version in which the field equation is Poisson's equation). If so, the potential of an isolated constant density ball of radius $r_0$ can be found from the conditions
$$\begin{array}{rcl} \lim_{r \to \infty} \phi & = & 0 \\ \Delta \phi & = & 4 \, \pi \, \rho \end{array}$$
(where I wrote the Poisson equation in Gaussian units, which are closest to standard notation in gtr, and where I set $G=1$, again for easy comparison with gtr). After setting $4/3 \pi \, r_0^3 \, \rho = M$ (the mass of the ball), the solution can be written
$$\phi = \left\{ \begin{array}{ll} \frac{M \, \left( r^2-3 \, r_0^2 \right)}{2 \, r_0^3}, & 0 < r < r_0 \\ \frac{-M}{r}, & r > r_0 \end{array} \right.$$
Differentiating gives the acceleration of static observers (comparable to the path curvature or magnitude of acceleration vector of world lines of static observers in gtr),
$$\frac{\partial \phi}{\partial r} = \left\{ \begin{array}{ll} \frac{M \, r}{r_0^3}, & 0 < r < r_0 \\ \frac{M}{r^2}, & r > r_0 \end{array} \right.$$
The acceleration vector of static observers points radially outward, showing that such observers have to fire their rocket engine radially inward in order to maintain their position (hence the possibly surprising sign convention I am using). As several posters stated, inside the ball, the acceleration increases linearly with radius; the potential has an inflection point at $r=r_0$ but is continuously differentiable. The attached jpg images (made using Maple) show (left) the potential and (right) the acceleration for the case $r_0 = \rho = 1$:

#### Attached Files:

File size:
2.8 KB
Views:
337
• ###### Force_Const_Density_Isol_Ball.jpg
File size:
3.1 KB
Views:
293
Last edited: Dec 12, 2007
10. Dec 12, 2007

### Chris Hillman

In gtr, the comparable model would be the Schwarzschild's constant density static spherically symmetric perfect fluid solution of the EFE, the first known solution of the nonvacuum equation found and the second known solution of any kind. I can't upload more figures until the mentors approve the first two so I'll wait to see expressions of interest to decide whether it would be helpful to discuss and compare this model with the Newtonian model just sketched.

Comparing the Poisson equation and EFE; note that the LHS results from applying a second order differential operator and the RHS describes the density of matter (and in gtr, of momentum), the source for the gravitational field. From this POV, the metric tensor is a kind of "potential" for the curvature tensor.

In past posts to PF, I have discussed what is known about the actual density profile inside the Earth, the Sun, and typical Neutron stars, and there is considerable variation! Typically, we wish to derive or assume some equation of state giving density as a function of pressure. As you would expect, constant density (as a function of pressure) turns out to be an overidealization in most astrophysically interesting cases! But all static spherically symmetric perfect fluid solutions of the EFE are known in various rather convenient forms, and one can even write down the relativistic polytrope solution in a fairly convenient form.

Last edited: Dec 12, 2007
11. Dec 19, 2007

### Bjarne

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.

Bjarne

12. Dec 20, 2007

### Janus

Staff Emeritus
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.

13. Jan 1, 2008

### Bjarne

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?

Bjarne

14. Jan 1, 2008

### yenchin

There is an interesting new mathematics book that deals with this issue and more, titled Hesiod's Anvil.

15. Jan 1, 2008

### Janus

Staff Emeritus
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)

However, the Earth isn't a constant density and the crust is less dense than the interior.

So, using an value of 2.7g/cm³ for the density of the Earth's crust, you get 9.797 m/s² or slightly more than that at the surface.

So, initially, as you begin to move down towards the center of the Earth, the g force will go increase, but as you continue down, it will begin to eventually decrease unitl it reaches zero at the center. (mainly because as you move closer to the center, the decreasing percentage of the Earth's mass contibuting to the force becomes a larger factor than your decreasing distance from the center.)

16. Jan 2, 2008

### Bjarne

Janus

Thank you

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”?

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?

Bjarne

17. Jan 2, 2008

### Kushal

18. Jan 2, 2008

### Janus

Staff Emeritus
I meant that the value of the acceleration would be less than the value of the acceleration at the surface.
This can be found by a=v^2/r.

So the adjustment would be 0.0337 m/s^2

19. Jan 2, 2008

### Bjarne

Janus

what was the values you was using to calculate 9.793 m/s² - at the surface at the earth
I mean M and r

Bjarne

20. Jan 2, 2008

### Janus

Staff Emeritus
M: 5.87e24 kg
r: 6.378e6 m.