# What is the derivation for decrease in g with depth?

1. Apr 22, 2011

### Quantum Mind

1. The problem statement, all variables and given/known data

We know that 'g' decreases with height and the derivation for the formula is straight enough, but how is the formula for decrease of 'g' with depth derived?

3. The attempt at a solution

If an object is taken to a height 'h' above the surface of the earth and assuming the radius of the earth as 'R', then acceleration due to gravity at a height 'h' is given by the formula

gh = GM/(R+h)2

After adding R2 to both denominator and numerator, we get the formula

gh = g[R/(R+h)]2

However, the same formula will not work for depth because if (R-h) is substituted for (R+h) in the above formula, then the value of 'gh' will actually increase, which is not the case. I know that the correct formula to be used here is gh = g[(R-h)/R], but how is that derived ? I know this is pretty basic but I am not able to get the hang of it.

Would appreciate if somebody can please explain.

Last edited: Apr 22, 2011
2. Apr 22, 2011

### kewljerk

It is somewhat similar as we derive electric field inside a dielectric sphere.

=>1st approach
use gauss law for finding out the electric field inside a dielectric sphere. it will come to be Qx/4*pi*epsilon*R^3
simply replace 4(pi)esilon by G and Q by M-mass of earth, you will get the formula of g inside the earth.Mind it here x is distance from the center.

=>2nd approach
g=GM/r^2 for a spherical shell where r=distance from the center

so, here take a spherical shell at a distance x from center, so g due to that shell=G(p*4pi*r^2)dr/x^2
where p=density of the earth=M/(4/3&pi*R^3)
integrate above differential equation from 0 to x and put the value of p. If you want answer in terms of distance from the surface of the earth simply replace x by R-h.

3. Apr 22, 2011

### ashishsinghal

4. Apr 22, 2011

### AlexChandler

Yes the basic idea is this: picture the point where you want to calculate g. Now imagine a sphere with the center of the earth as its center, that has this point on its surface. Gauss' law says that only the mass contained within this sphere counts toward the gravitational field at that point. So really we can find the g value here simply as

$$\vec g = - \frac{V'}{V} \frac{GM}{r^2} \hat r$$

Where V' is the volume of the imaginary sphere and V is the volume of the earth. M is the mass of the entire earth, and r is the distance from the center of the earth to the point in question.
Of course this relies on the assumption that the earth is uniform in its density which is not quite true, but this is a decent approximation.

And actually this is just

$$\vec g = - \frac{r^3}{R^3} \frac{GM}{r^2} \hat r$$

where R is the radius of the earth. or we can say

$$\vec g = - \frac{GMr}{R^3} \hat r$$

This is quite interesting because if we were to drill a hole all the way through the earth and drop an object in, it would feel a restoring force proportional to the distance r from the center, which would act to make the object oscillate exactly like a mass and spring system.
Comparing to the mass spring system we see the object will have a period:
$$T = 2 \pi \sqrt{ \frac{R^3}{GM} }$$

Last edited: Apr 22, 2011
5. Apr 22, 2011

### AlexChandler

Just to clarify, I arrived at the first equation by considering the density of the earth

$$\rho = M/V$$

and then the mass contained within the sphere is

$$M' = \rho V'$$

so we have

$$\vec g = - \frac{GM'}{r^2} \hat r$$

6. Apr 22, 2011

### AlexChandler

And if you are curious, here is gauss' law:

$$\oint_S \vec g \cdot \hat n dA = - 4 \pi G \oint_V \rho dV$$

Does this make much sense to you? I can elaborate if you like.

7. Apr 22, 2011

### AlexChandler

Or alternatively (in differential form):

$$\nabla \cdot \vec g = - 4 \pi \rho G$$

8. Apr 24, 2011

### Quantum Mind

Thanks all. I think I am beginning to understand. But can I have the derivation for gh = [(R-h)/R] please ?

9. Apr 24, 2011

### AlexChandler

What is this formula for? I dont think its right... the left side has dimensions of acceleration and the right side is dimensionless.

10. Apr 24, 2011

### AlexChandler

Oh i see what you meant.

Yes it is as follows. As you see, a result from the equations above is,

$$\vec g = - \frac{GM'}{r^2} \hat r$$

Following your convention let us call the gravitation field vector at this location g sub h
Then we can write that the magnitude of g sub h is

$$g_h = \frac{GM'}{(R-h)^2}$$

where i have substituted r= R-h

Now we also have, as explained above,

$$M' = \frac{r^3}{R^3} M$$

So we have

$$g_h = \frac{GM}{R^2} \frac{r^3}{R} \frac{1}{r^2}$$

and therefore, substituting r=R-h and g=GM/R^2

$$g_h = g \frac{R-h}{R}$$

11. Apr 24, 2011

### Quantum Mind

Thank You, I get it now.