How strong is gravity in the center of the earth?

[Bjarne]

How strong is gravity in the center of the earth?

 

[EL]

How strong is gravity in the center of the earth?

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

 

[Bjarne]

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

 

[jcsd]

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

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

 

[AFG34]

actually it would be a quarter of its strenght

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

 

[EL]

actually it would be a quarter of its strenght

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

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.

 

[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.

 

[jcsd]

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.

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.

 

[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 $r_0 = \rho = 1$:

 

[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.

 

[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

 

[Janus]

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. !

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.

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

 

[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

 

[yenchin]

There is an interesting new mathematics book that deals with this issue and more, titled https://www.amazon.com/dp/0883853361/?tag=pfamazon01-20.

 

[Janus]

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

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.)

 

[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

 

[Kushal]

some notes on gravitation:
http://science-student.com/category/course-notes/physics-1-notes

the answers to your questions are on it. i hope it helps.

 

[Janus]

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”?

I meant that the value of the acceleration would be less than the value of the acceleration at 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

This can be found by a=v^2/r.

So the adjustment would be 0.0337 m/s^2

 

[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

 

[Janus]

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

M: 5.87e24 kg
r: 6.378e6 m.

 

[Bjarne]

How deep inside the Earth approximate would gravity begin to decrease ?
Nearly halway?

Bjarne

 

[Kushal]

as soon as you get below Earth surface, an dthen move even deeper, it starts to decrease. it decreases proportionately to the distance from centre if you consider Earth as a perfect sphere and of uniform density.

 

[Andre]

So, initially, as you begin to move down towards the center of the Earth, the g force will go increase,...

Right or almost right, and all albout gravity inside the Earth in this thread:

https://www.physicsforums.com/showthread.php?t=207148

 

[epenguin]

So an object released in a tunnel through the centre of the Earth (with the obvious idealizations) would execute simple harmonic motion.?

Amusing.

 

[jsandow]

Due to increased in the density of the mantle, and the cores, gravity increases slightly as you descend to a depth of about 2000km, from 9.81 m/s² at the surface to about 10 m/s² at 2000km deep. it then increases faster to a maximum of about 10.7 m/s² at 3000km depth, then tapers roughly linearly to 0 at the very center.

While this general form may be followed for similar sized planets, it is important that the planetary body have the same approximate chemical makeup- iron core, silicate mantle/crust, etc. If the composition is different, an different density gradient will follow as depth increases to the core, and a somewhat different curve would follow.

Again, the reason for the deviation from the simple mathematical model curve is the varying densities of the core and mantle of the earth.

The actual curve, based on data from the CRC 94th ed., shows two peaks rather than one, before falling to zero. In this case, averaging the planet's density gives very inaccurate results.

A

 

[Dadface]

So an object released in a tunnel through the centre of the Earth (with the obvious idealizations) would execute simple harmonic motion.?

Amusing.

A cheap way to travel,just jump through the tunnel on one side of the planet and in about 25 minutes you are on the opposite side of the planet.I'm off to get my hammer drill.Does anyone want to invest in my new company?

 

[Dadface]

A cheap way to travel,just jump through the tunnel on one side of the planet and in about 25 minutes you are on the opposite side of the planet.I'm off to get my hammer drill.Does anyone want to invest in my new company?

Whoops the actual time is closer to 42 minutes


My calculator was playing up first time I did it(this is one of my favourite excuses please feel free to use it)

 

[jsandow]

I'll have to get more data points, and set up the series on a spreadsheet to get the actual time to accelerate based on the actual acceleration values as you approach the core. This seems like it would be significantly less than 21 minutes (to get to the center) you calculated, so it would be less than the orbital time.

The oscillation period in a fixed density is actually the orbital period for the altitude you start at. Of course, on Earth you either slow from atmosphere or burn up from 7900 meters per second velocity!

Up on the moon, however, you could theoretically orbit just above the surface (the highest points really) for practically ever.

 

[Dadface]

I'll have to get more data points, and set up the series on a spreadsheet to get the actual time to accelerate based on the actual acceleration values as you approach the core. This seems like it would be significantly less than 21 minutes (to get to the center) you calculated, so it would be less than the orbital time.

The oscillation period in a fixed density is actually the orbital period for the altitude you start at. Of course, on Earth you either slow from atmosphere or burn up from 7900 meters per second velocity!

Up on the moon, however, you could theoretically orbit just above the surface (the highest points really) for practically ever.

I agree jsandow, in my calculation I assumed a uniform density.Of course,because of density variations the motion will not be simple harmonic.

 

[jsandow]

the actual motion would equal a harmonic motion at four points in a single cycle: start, mid planet, the antipodal point where the direction would reverse, and mid planet again. but the curve would be a little different in between the points.

 

[mgb_phys]

So an object released in a tunnel through the centre of the Earth (with the obvious idealizations) would execute simple harmonic motion.?
Amusing.

It's also just a special case of a very elliptical orbit

 

[Chronos]

Interestingly enough, it takes about 42 minutes to free fall through the earth, no matter what angle the hole is pitched. Of course if the inclination angle relative to the center of the Earth is too shallow, you can expect some amount of 'rolling' time where friction will slow the journey.

 

[Bob S]

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?

The Earth's surface, and especially the oceans, are at a equipotential, including both the gravitational force and the centrifugal force. So the Earth is slightly prolate.

An equatorial bulge is a bulge which a planet may have around its equator, distorting it into an oblate spheroid. The Earth has an equatorial bulge of 42.72 km (26.5 miles) due to its rotation: its diameter measured across the equatorial plane (12756.28 km, 7,927 miles) is 42.72 km more than that measured between the poles (12713.56 km, 7,900 miles).

 

[Stoonroon]

The simple harmonic motion prediction for radial free-fall through a uniformly dense spherical mass (such as the idealized Earth discussed here) is very well known. Does anyone feel a desire (or need?) to verify this prediction empirically?

Though a high orbit satellite experiment would be the ideal way to do it, a laboratory version could also be done using a modified Cavendish balance.

Wouldn't it be nice to have all the theoretical discussion about this prediction backed up by a physical demonstration?

 

[Chronos]

Gravity probe B comes to mind.