What comprises the mass of the earth?

[Jsimonetti]

My question is as follows:
Do we (and all stationary objects on the surface of the earth) make up for the overall mass of the earth? If not why?

In an attempt to apply my understanding of this matter, I would wonder if I stand outside and jump, are my legs creating force on the earth, pushing the Earth downwards (even a small amount), or since I am part of the mass of the earth, does the Earth simply absorb the energy from my legs?

 

[Pengwuino]

Yes, you help make up the mass of the Earth. When you jump, you do push the Earth downwards.

 

[Jsimonetti]

Thank you!
I can then assume that If I lay on the ground, and rest a 1kg weight on my chest, that the 1kg weight has only 1 force pulling it down, the force of the gravity of the earth, and not two forces, the gravitational force of the earth, and its gravitational attraction to the matter in my body. Is this also correct?

 

[Pengwuino]

If you consider the mass of the Earth and hte mass of your body as 2 different entities, the weight feels a gravitational force from the Earth AND you since you create your own gravitational field.

 

[Jsimonetti]

But gravity is determined by the mass of an object. If I am part of the mass of the earth, than surely, you would have to agree that the weight is acted upon by one force, the gravity of the earth.

 

[Pengwuino]

Well, ok, let's just settle this. The Earth is $$m_1$$, you are $$m_2$$. If all there was in the universe was you, the earth, and the moon, the force of attraction on the moon is because of the Earth, $$m_1$$, and you, $$m_2$$. If the universe just encompassed you and the Earth, the gravitational field you would feel would be solely from the Earth, $$m_1$$.

I believe I might have tossed some confusing info out there due to the original question.

 

[D H]

But gravity is determined by the mass of an object. If I am part of the mass of the earth, than surely, you would have to agree that the weight is acted upon by one force, the gravity of the earth.

Jsimonetti, you appear to have a common misperception of the gravitational force between two objects, namely that

$$F=\frac{GmM}{r^2}$$

where F is the magnitude of the gravitational force, G is the universal gravitational constant, m and M are the masses of the two objects, and r is the distance between the centers of mass of the two objects. The above equation is of course Newton's law of universal gravitation.

Newton's law of gravitation applies to point masses only. For a real object, one has to break things down to infinitesimally small objects to which Newton's law does apply. The gravitational force between two non-point masses is the vector sum of the gravitational forces amongst all pairs of infinitesimally small objects in the objects. In short, you have to do a double volume integral.

It turns out that the gravitational force between a point mass and a mass with a spherical mass distribution is the same as that given by Newton's law of gravity. Newton himself developed the proof of this; it is called Newton's shell theorem. The planets in our solar system are very close to spherical objects, particularly when viewed from far, far away. Approximating them as point masses yields a very good model of the solar system.

How about the problem at hand? On one hand you could laboriously compute the gravitational force by the earth+human system on some other object by means of this double integral approach. On the other hand, suppose you individually compute the gravitational force on the object by the Earth and by the human. Add these two forces together and you will get the exact same answer as the laborious approach. Gravitational forces, like forces in general, are additive.

 

[monty37]

well this just crossed my brain.. the mass of Earth as such without considering
all the people on it would be less than its mass with people on it,
but will the difference be negligible..dont you
think if you remove Earth of its whole population its mass would considerably reduce?

 

[Jsimonetti]

Thank you for all the replies, they are enlightening.
However this brings me to one more question:

We are then attracted to the Earth by the force equal to the sum of all the infinitesimally small particles that make up the earth. If I were to dig out a hole (lets say a labor intensive 7 foot cube) and stack this dirt next to the hole, the force of gravity pulling me down in a perpendicular direction would be less if I was standing at the bottom of the hole, than on the top of the pile, correct?

 

[Dale]

dont you
think if you remove Earth of its whole population its mass would considerably reduce?

This is ridiculous, the mass of the Earth is about 6E24 kg there are about 6E9 people each weighing on average far less than 100 kg for a total of less than 6E11 kg of people. So the Earth is roughly 1E13 times greater mass than the mass of all of the people on the earth. In other words, that is about 0.00000000001% of the mass of the earth.

The gravitational constant G is only known to a precision of about 0.0014% (http://www.aip.org/pnu/2000/split/pnu482-1.htm" ) and the mass of the Earth is only known to about the same precision. So the mass of all of the people on Earth could be added or subtracted millions of times over without any detectable difference to our most precise measurements.

 

[DaveC426913]

Thank you for all the replies, they are enlightening.
However this brings me to one more question:

We are then attracted to the Earth by the force equal to the sum of all the infinitesimally small particles that make up the earth. If I were to dig out a hole (lets say a labor intensive 7 foot cube) and stack this dirt next to the hole, the force of gravity pulling me down in a perpendicular direction would be less if I was standing at the bottom of the hole, than on the top of the pile, correct?

Yes.

What you want to explore is called Newton's Shell theorem.

Let's simplify the setup.

Pretend the Earth is a perfect sphere, 4,000,000 metres in radius. Three examples:

a] Standing on its surface (0 metres depth), you will feel the full effect of a 4,000,000 metre radius sphere under you.

b] In a small hole 2 metres below the surface of the Earth, you would feel the effect of a perfect sphere 3,999,998 metres in radius. You would feel no net gravitational pull from the 2 metre thick spherical shell that constitutes the entire surface layer of the Earth.

c] In a small hole 1,000,000 metres below the surface of the Earth, you would feel the effect of a perfect sphere 3,000,000 metres in radius. You would feel no net gravitational effect from the 1,000,000 metre thick spherical shell that constitutes the upper layers of the Earth.

d] In a small hole 3,999,998 metres below the surface (only 2 metres from the centre), you would feel the g-pull of a sphere only 2 metres in radius. You would feel no net gravitational effect from the 3,999,998 metre spherical shell above you.

 

[D H]

If I were to dig out a hole (lets say a labor intensive 7 foot cube) and stack this dirt next to the hole, the force of gravity pulling me down in a perpendicular direction would be less if I was standing at the bottom of the hole, than on the top of the pile, correct?

Yes.

What you want to explore is called Newton's Shell theorem.

No!

DaveC, you implicitly assumed the Earth is of a constant density throughout. This absolutely is not the case for the Earth. The Earth's core is far denser than the material above it. This simplifying assumption leads to exactly the wrong answer for this specific yes/no question.

A much better approach is to assume a radial mass distribution. I don't have time to go through the derivation now. I will do so later this evening if someone else hasn't already filled in the gaps. The bottom line is that gravitational acceleration will decrease with depth if

$$\rho_s > \frac 2 3 \,\bar{\rho}$$

In English, the condition for gravitational acceleration to decrease with depth is that the density of the surface material must be more than 2/3 of the mean density of the Earth.

The mean density of the Earth is 5515 kg/m³. 2/3 of this is 3677 kg/m³. Dirt varies in density from about 1200 to 2000 kg/m³, far less than the requisite 3677 kg/m³. Solid granite has a density of 2691 kg/m³, so gravitational acceleration will increase with depth even if you dig through solid granite.

 

[mgb_phys]

If I were to dig out a hole (lets say a labor intensive 7 foot cube) and stack this dirt next to the hole, the force of gravity pulling me down in a perpendicular direction would be less if I was standing at the bottom of the hole, than on the top of the pile, correct?

This is even used practically.
Searching for oil you measure the force of gravity on a weight, when you are over a buried oil reserve the force will be slightly less (because oil is less dense than rock).

When you are building a telescope you also have to be careful about the difference between level as measured by a spirit level or plumb bob and straight up to the stars. Local gravity might be pulled off sightly to one side because of a nearby mountain or denser rock - like your pile of dirt - affecting the spirit level.

 

[DaveC426913]

No!

DaveC, you implicitly assumed the Earth is of a constant density throughout.

Yes. I should have made it explicit. The poster wants to understand the principles involved. I see no reason to complicate the principles with messy reality, at least until the poster undestands the principles.

But if you do want to be picky about reality, it's still perfectly true. It is still Newton's Shell Theorem. It's just that you need to factor in the varying density.

Additonally, the poster's hole is only a few metres deep. While your factoring of the varying density will indeed change the numerical answer (which was not asked for), it will not change the principle of the answer. The person in the hole will still weigh less than the person standing above the hole.

So, really, I'm not sure where you're coming from.

 

[Dale]

While your factoring of the varying density will indeed change the numerical answer (which was not asked for), it will not change the principle of the answer. The person in the hole will still weigh less than the person standing above the hole.

That is just the point, the person in the hole will weigh more, not less. There are two competing factors here, the person is closer to the center of the Earth which tends to increase the gravitational attraction, and there is a smaller amount of matter in the sphere below the person which tends to decrease the gravitational attraction. If the density of the outer shell is less than 2/3 the average density in the sphere below (as is the case on the surface of the earth) then the closer distance effect wins and the gravity increases.

 

[KingNothing]

We like to think of things as point masses when it comes to gravity. However, in reality, one could imagine the gravitational force of each individual atom as being a separate force.

 

[D H]

But if you do want to be picky about reality, it's still perfectly true. It is still Newton's Shell Theorem. It's just that you need to factor in the varying density.

Additonally, the poster's hole is only a few metres deep. While your factoring of the varying density will indeed change the numerical answer (which was not asked for), it will not change the principle of the answer. The person in the hole will still weigh less than the person standing above the hole.

So, really, I'm not sure where you're coming from.

Where I am coming from is that you are wrong. The person will weigh more at the bottom of the hole.

You are right that the shell theorem still applies. Ignoring the rotation of the Earth and assuming the Earth has a spherical mass distribution (density depends only on distance from the center of the Earth), the gravitational acceleration experienced at some distance r from the center of the Earth is exactly that given by Newton's law of gravitation,

$$a(r) = \frac {Gm(r)}{r^2}$$

where m(r) is the mass of that portion of the Earth inside a sphere of radius r centered at the center of the Earth.

Suppose the Earth comprises an inner shell of mass surrounded by an outer shell. The inner shell has spherical mass distribution, mass m₀, and radius r₀. The outer shell is a spherical shell of constant density ρ_s. The mass of the part of the Earth within a sphere of radius r, R_e>r>r₀ is

$$m(r) = m_0 + \frac 4 3 \pi \rho_s \left(r^3-r_0^{\,3}\right)$$

Denote the difference between the mass of the inner shell and a mass of the same volume but the density of the outer shell as Δm,

$$\Delta m \equiv m_0 - \frac 4 3 \pi \rho_s r_0^{\,3}$$

With this, the mass m(r) becomes
$$m(r) = \frac 4 3 \pi \rho_s r^3 + \Delta m$$

The gravitational acceleration at some point r within this outer shell is thus

$$a(r) = \frac {Gm(r)}{r^2} = \frac 4 3 \pi G \rho_s r + \frac {G\Delta m}{r^2}$$

The gravitational acceleration will decrease/increase with depth if the derivative of the acceleration with respect to r is positive/negative. Differentiating,

$$\frac{\partial a(r)}{\partial r} = \frac 4 3 \pi G \rho_s - 2 \frac {G\Delta m}{r^3}$$

The conditions for an decrease gravitational acceleration with respect to depth is

$$\rho_s > 2 \frac {\Delta m}{4/3 \pi r^3}$$

Substituting $\Delta m = \left(m(r) - 4/3 \pi \rho_s r^3$ and simplifying,

$$\rho_s > 2 \frac {\left(m(r)}{4/3 \pi r^3} - 2 \rho_s$$

The first term on the right hand side, $m(r)/(4/3\pi r^3)$ is the mean density of of that portion of the Earth inside a sphere of radius r. With this,

$$\rho_s > \frac 2 3 \bar{\rho}$$

Two thirds of the mean density of the Earth is significantly greater than the density of granity (2.7 grams/cc) or even basalt (3.0 grams/cc). Gravitational acceleration increases with depth near the surface of the Earth.

In fact, gravitational acceleration increases all the way down to the mantle/crust boundary. It then decreases a bit with increasing depth but then starts increasing again, reaching a maximum at the mantle/outer core boundary. The gravitational acceleration at the mantle/outer core boundary, 2,890 km below the surface, is greater than the gravitational acceleration at the Earth's surface.

 

[DaveC426913]

Got it. Conceded.