## Thursday, January 21, 2010

### Fermi Gravity (FPOTW)

Here is a Fermi problem with a (perhaps) surprising answer, and several methods of attack.
The question is simply stated thus:

Where is gravitational acceleration greater, on the Earths surface, or 100 km underground?

There is a very easy way to solve this, a Fermi way that will answer the question. There is also a back of the envelope method, which will give a numerical answer.

For a hint I give you the following picture:

Enjoy!

1. You made it too easy with that picture.
However, I was surprised. It was not as "easy" as you said to answer this question even without a number.
How did you do it?

2. The acceleration is greater 100 KM below the surface. 100 KM through the boyant crust leaves the vast majority of the gravitational mass at the earth's interior ... primarily the cores and mantle. Since gravity follows the inverse square law, the affect if gravity increases by this amount less the small bit in the 100 KM above. Also, gravitational fields cause time dialation (not very large in the case of the earth) but nonetheless the clock runs slower at 100 KM below and that means greater acceleration since it takes less time to cover a certain distance.

3. Ahh! Something fun!

I will try to solve a slightly different question at first:

Given only gravitational acceleration and density of a substance of the Earth at the Earth's surface and the Earth's radius estimate maximal value of gravitational acceleration in Earth's interior.

Instead of a Fermi way i choose another, rather a long path because it involves some neat calculus.

Solution:

As a first approximation lets suppose that the density inside the Earth is a linear function of the distance r from the center of the Earth:

d(r)=d(0)-[d(0)-d(R)]*(r/R)

d(0) is a density at the centre (r=0) of the Earth
d(R) is a density at the surface (r=R) of the Earth

Now, gravitational acceleration inside the Earth, at distance r from the centre is:

g(r)=G*M(r)/r^2

where M(r) is a mass of the solid sphere with radius r and center of which coincides with the center of the Earth.
G is the gravitational constant.

M(r) = integral (z=0 to z=r) 4*pi*d(z)*z^2

M(R)=M is the mass of the Earth

I skip calculus part of this and write down a final formula for the gravitational acceleration :

g(r)=4*pi*G*r*{d(0)/3-[d(0)-d(R)]*r/(4*R)}

In this formula, density at the centre (r=0) of the Earth, d(0) is unknown. But g(R) is given.

From this we have:

d(0)=3*[g(R)/(pi*G*R)-d(R)]

Using a formula for the gravitational acceleration at the surface of the Earth:

g(R)=G*M/R^2

we can eliminate G:

d(0)=3*[M/(pi*R^3)-d(R)]

It is the time to get a first numerical result now.

Namely, an estimation of the density d(0) at the centre of the Earth.

So, given:

Radius of the Earth: R=6,4*10^6 m

Mass of the Earth: M=6*10^24 kg

Density at the surface of the Earth: d(R)= 3 g/cm^3
..........................................

We have d(0)=13 g/cm^3
............................................

The next step is to find maximum of g(r):

g(r)=4*pi*G*r*{d(0)/3-[d(0)-d(R)]*r/(4*R)}

Jumping over the part of the mathematical we get that g(r) has a maximum at:

r=2/3*R/[1-d(R)/d(0)]

Plugging numbers into this formula we get the second numerical result:

.............................................
r=0.87*R or maximal value of gravitational acceleration in Earth's interior is h=R-0.87*R = 850 km away from the surface.
.............................................

But what is the max. value of gravitational acceleration?

Jumping again over the math i got such expression:

max[g(r)]/g(R)=(2*x)^2/(4*x-1)

where x=1-pi*d(R)*R^3/M

..................................
max[g(r)]/g(R)=1.02
..................................

A nice problem! Was time well spent. Thanks Michael.

4. Thanks Martin!

I love problems like this, as you can explore the solution to whatever depth you wish. (Pun intended!)

For instance, the simplest Fermi type solution I came up with answers the question of which is greater, without using numbers. The solution leads to some further exploration.

My first approach was to work from the surface of the earth and outwards to an altitude of 100 km. We know from Gauss's law that the gravitational field of a symmetric (or uniform mass distribution) spherical shell enclosing you is zero. So, the thin shell of Earths atmosphere contributes nothing to the gravitational field you feel at the surface. Also, the mass below you can be considered to be a point mass.

If you could do a Mary Poppins, and sit on a cloud 100 km in altitude, via g = GM/r^2 your gravitational acceleration would decrease, since the additional mass of the air is a small fraction of that of the Earth.

Likewise, going to a depth of 100 km below the Earth, the thin shell of the crust above you does not contribute to the gravitational field and represents a small percentage of the total mass of the earth. By the inverse square law we find a greater gravitational acceleration 100 km below the surface.

However, this is making an assumption that the crust is less dense than that of the average density the Earth and is at least 100 km thick.

How much less dense should the crust be compared to the average density of the Earth?

Using scaling laws and how density vs surface area changes you can determine that g(r) increases as you approach the center so long as the density of the crust (or what is above you) is not more than 2/3 the average density of the Earth. I will leave that as an exercises.

However, one can use your solution Martin, to calculate this directly, with calculus. In other words, that factor of 2/3 in your rmax is pretty telling! =)

5. Hi Gary,

The "small bit [of gravity] in the 100 KM above" actually is zero.
Gauss's law states that the gravitational field inside a spherical shell of uniform mass distribution is zero.

With respect to time dilation, you are correct. (Try calculating the effect for the Surface of the Earth).
However, you should state that because the gravitational field is greater 100 km below the surface, clocks run slower, not the other way around as you stated. One does not imply the other, as you can have time dilation via velocity or gravity (space-time curvature), but given a time dilation you cannot assert from that alone what is causing it.

Speaking of clocks, can you calculate the period of a pendulum anchored at the surface of the Earth with a length of 100 km below the surface? How about of pendulum of length L first at the surface of the Earth, then taken down to 100 km?

6. Hi Michael -

This was fun. I just stumbled on your site and I think it is cool that you try and get people to put some real scientific thought processes to use!

I appologize for not being clear about the clock running slower 100 KM below the surface. I thought I said that but I guess I made it confusing.

At any rate, great site and keep up the interesting challenges!

7. @gary

Thanks! Have a great week!