Dig a hole halfway to the center on the earth what would you weigh?

[Alan McDougall]

Discounting the real problems of digging such a hole, such as huge air magma, pressure, heat etc!

 

 

If you could dig a hypothetical hole half way down to the center of the earth and your weight was 100 kilograms on the surface, what would you weigh at the bottom of the hole?

[imatfaal]

At the centre nothing - both Newton and Gauss showed that anywhere within a symmetric shell of matter no attraction from the shell is felt in any direction; in a very narrow (compared to size of earth) mineshaft you would only be affected by the sphere of matter closer to the centre than you. When you got to the centre of the earth no mass would be in a sphere closer to the centre than you and also you would feel no attraction from the "shell" further away from the centre.

 

DH wrote a great post on the varying weight of a body through the earth - it's not simple rate of change (linear) as would be expected in an ideal case but varies due to the changing density of the earth. I will look for the post and make a link.

 

http://www.sciencefo...post__p__660350

and

http://www.sciencefo...post__p__652185

 

BTW - 100 kg is a mass - your mass will not vary. Your weight is what changes

[Alan McDougall]

At the centre nothing - both Newton and Gauss showed that anywhere within a symmetric shell of matter no attraction from the shell is felt in any direction; in a very narrow (compared to size of earth) mineshaft you would only be affected by the sphere of matter closer to the centre than you. When you got to the centre of the earth no mass would be in a sphere closer to the centre than you and also you would feel no attraction from the "shell" further away from the centre.

 

DH wrote a great post on the varying weight of a body through the earth - it's not simple rate of change (linear) as would be expected in an ideal case but varies due to the changing density of the earth. I will look for the post and make a link.

 

http://www.sciencefo...post__p__660350

and

http://www.sciencefo...post__p__652185

 

BTW - 100 kg is a mass - your mass will not vary. Your weight is what changes

 

I am referring to weight, not mass I know that at the very center you would weigh nothing, and that your mass would remain the same. I will go to the link thanks for that!

[Phi for All]

Another couple of threads on this might help:

 

http://www.scienceforums.net/topic/64539-earth-hole/page__p__660289__hl__%2Bearth+%2Bhole__fromsearch__1#entry660289

 

http://www.scienceforums.net/topic/20566-why-wouldnt-hole-through-earth-work/page__hl__%2Bhole+%2Bhalfway+%2Bthrough+%2Bthe+%2Bearth

[between3and26characterslon]

Assuming a uniform density approximately 1/8th you would weigh at the surface. You only need consider the radius of the sphere beneath your feet so if you were standing on a planet the same density as Earth but half the diameter you would weigh 1/8th what you do on Earth.

 

The volume of a sphere has the term r³ in it, if r is twice as big r³ is 8 times bigger

[Alan McDougall]

Assuming a uniform density approximately 1/8th you would weigh at the surface. You only need consider the radius of the sphere beneath your feet so if you were standing on a planet the same density as Earth but half the diameter you would weigh 1/8th what you do on Earth.

 

The volume of a sphere has the term r³ in it, if r is twice as big r³ is 8 times bigger

 

 

That is correct, some people think the huge amount of mass above the bottom of the hole must somehow effect the equation that gives the correct answer.

 

The density of the earth varies as you get nearer the center. So in reality the sphere beneath your feet would equate to a smaller denser planet, with greater relative gravity. (Although less than on the surface) Thus ;in reality you would weigh more than 1/8th than you would have on the surface.

[uncool]

So far, people seem to have forgotten to take into account the changing radius. While the apparent mass of the object attracting you is 1/8th that when you are on the surface, its apparent center is also half as far as it would be if you were on the surface, which quadruples the effect of gravity. Therefore, you would weight half what you do on the surface. The force decreases linearly.

=Uncool-

[Greg H.]

[math]g = \frac{4\pi}{3}G pr[/math]

 

where p = density of the object.

 

If we assume p remains constant (which isn't accurate, but let's make the assumption for the moment), then we can eliminate all the constants values from the equation, then all we're left with is the surface gravity being directly proportional to the radius. As the radius from the center decreases so does the gravity, and thus the weight.

 

In truth, it's probably not quite exactly half, since the mean density of the planet goes up as you approach the iron core (the mean density of the Earth is roughly 5.54 g/cm3, while iron has a density of 7.87 g/cm3). Assuming you split the difference on the density at 6.701 g/cm3, then the actual force of gravity would be

g_midpoint = .604g_surface or there abouts (as always, if I did my math right). In other words, you would mass the same, and weight about 60% as much as you do on the surface.

[D H]

[math]g = \frac{4\pi}{3}G pr[/math]

 

where p = density of the object.

 

If we assume p remains constant (which isn't accurate, but let's make the assumption for the moment), then we can eliminate all the constants values from the equation, then all we're left with is the surface gravity being directly proportional to the radius. As the radius from the center decreases so does the gravity, and thus the weight.

 

In truth, it's probably not quite exactly half, since the mean density of the planet goes up as you approach the iron core (the mean density of the Earth is roughly 5.54 g/cm3, while iron has a density of 7.87 g/cm3). Assuming you split the difference on the density at 6.701 g/cm3, then the actual force of gravity would be

g_midpoint = .604g_surface or there abouts (as always, if I did my math right). In other words, you would mass the same, and weight about 60% as much as you do on the surface.

 

You had the first part correct (halfway down you would weigh half as much as your surface weight if the Earth did have a uniform density) but you were well of the mark on the second part. Inside our real Earth, you would weigh more halfway down, not less.

[Greg H.]

You had the first part correct (halfway down you would weigh half as much as your surface weight if the Earth did have a uniform density) but you were well of the mark on the second part. Inside our real Earth, you would weigh more halfway down, not less.

 

Ok, as that makes no sense to me, I'll illustrate my math, so you can show me where I went wrong in my calculations (I'm assuming I did something wrong, rather than you being mistaken DH.)

Edit to clarify some notations

g_s = acceleration due to gravity at the mean surface of the earth.

g_m = acceleration due to gravity at a point half way between the point center of the earth and the point of g_s

p = mean density of the earth at the point that g_n is calculated.

r = the distance between the point g_n is calculated and the point center of the earth. [math]r_{g_{m}} = \frac{1}{2}r_{g_{s}}[/math]

 

Starting with

[math]g = \frac{4\pi}{3}G pr[/math]

 

Following from that (Edit: sorry, hit the wrong button )

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8} cm^3}{g\times s^2} \times \frac{5.54g}{cm^3} \times 637810000 cm[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 5.54 \times 637810000 cm[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 3.533\times10^{9}cm[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10\times 3.533 cm}{s^2}[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{235.792 cm}{s^2}[/math]

 

[math]g_{s} = \frac{314.39\pi cm}{s^2}[/math]

 

 

I am not going to bother taking the pi out. We've done enough to make a comparison.

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8} cm^3}{g\times s^2} \times \frac{6.701g}{cm^3} \times 318905000 cm[/math]

 

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 6.701 \times 318905000 cm[/math]

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 2.136\times10^{9}cm[/math]

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10\times 2.136 cm}{s^2}[/math]

[math]g_{m} = \frac{4\pi}{3}\times\frac{142.557 cm}{s^2}[/math]

 

[math]g_{m} = \frac{190.075\pi cm}{s^2}[/math]

 

This shows, again assuming my maths are right, that [math]g_{m} = .605g_{s}[/math].

 

Is there a flaw in my math?

Edited August 11, 2012 by Greg H.

[Alan McDougall]

Ok, as that makes no sense to me, I'll illustrate my math, so you can show me where I went wrong in my calculations (I'm assuming I did something wrong, rather than you being mistaken DH.)

Edit to clarify some notations

g_s = acceleration due to gravity at the mean surface of the earth.

g_m = acceleration due to gravity at a point half way between the point center of the earth and the point of g_s

p = mean density of the earth at the point that g_n is calculated.

r = the distance between the point g_n is calculated and the point center of the earth. [math]r_{g_{m}} = \frac{1}{2}r_{g_{s}}[/math]

 

Starting with

[math]g = \frac{4\pi}{3}G pr[/math]

 

Following from that (Edit: sorry, hit the wrong button )

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8} cm^3}{g\times s^2} \times \frac{5.54g}{cm^3} \times 637810000 cm[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 5.54 \times 637810000 cm[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 3.533\times10^{9}cm[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{6.674\times 10\times 3.533 cm}{s^2}[/math]

[math]g_{s} = \frac{4\pi}{3}\times\frac{235.792 cm}{s^2}[/math]

 

[math]g_{s} = \frac{314.39\pi cm}{s^2}[/math]

 

 

I am not going to bother taking the pi out. We've done enough to make a comparison.

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8} cm^3}{g\times s^2} \times \frac{6.701g}{cm^3} \times 318905000 cm[/math]

 

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 6.701 \times 318905000 cm[/math]

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10^{-8}}{s^2} \times 2.136\times10^{9}cm[/math]

[math]g_{m} = \frac{4\pi}{3}\times\frac{6.674\times 10\times 2.136 cm}{s^2}[/math]

[math]g_{m} = \frac{4\pi}{3}\times\frac{142.557 cm}{s^2}[/math]

 

[math]g_{m} = \frac{190.075\pi cm}{s^2}[/math]

 

This shows, again assuming my maths are right, that [math]g_{m} = .605g_{s}[/math].

 

Is there a flaw in my math?

 

I like your maths, I did a web search and was very suprised by how different the answers to this question are, even from science forums?

[D H]

Is there a flaw in my math?

A couple of things are wrong.

 

First off, you inserted numbers way too early in the game. The math game is best played if one stays symbolic as long as possible. Do that and you will find that the ratio of the gravitational acceleration at two different distances r₁ and r₂ from the center of the Earth is

 

[math]\frac{g(r_1)}{g(r_2)} =

\frac {\frac 4 3 \pi G \bar{\rho}(r_1) r_1}{\frac 4 3 \pi G \bar{\rho}(r_2) r_2} =

\frac {\bar{\rho}(r_1)}{\bar{\rho}(r_2)}\,\frac{r_1}{r_2}[/math]

 

Notice you the common factor [imath]4/3\pi G[/imath] nicely drops out. Now substitute [math]r_2=R_E=6371\,\text{km}, r_1 = r_2/2[/math] and you get

 

[math]\frac{g(R_E/2)}{g(R_E)}\frac 1 2\,\frac {\bar{\rho}(R_E/2)}{\bar{\rho}(R_E)}[/math]

 

Using your 6.70 gm/cc³ as the average density for the halfway point does lead to your 60%.

 

 

This leads to the second problem. That density value of 6.70 gm/cm³ is completely wrong. Inside a spherical body, it's only the matter below that contributes to gravitational acceleration. All the stuff that is at a distance from the center of the body greater than that of the point in question contributes nothing. Halfway down to the center is about 300 km into the Earth's outer core. The Earth's core is mostly iron, compressed iron at that. We call liquids and solids incompressible, but that is a misnomer. The density of the material at this halfway down point is about 10.35 gm/cm³. Density builds up to 13.088 gm/cm³ at the very center of the Earth.

[Greg H.]

A couple of things are wrong.

 

First off, you inserted numbers way too early in the game. The math game is best played if one stays symbolic as long as possible. Do that and you will find that the ratio of the gravitational acceleration at two different distances r₁ and r₂ from the center of the Earth is

 

[math]\frac{g(r_1)}{g(r_2)} =\frac {\frac 4 3 \pi G \bar{\rho}(r_1) r_1}{\frac 4 3 \pi G \bar{\rho}(r_2) r_2} =\frac {\bar{\rho}(r_1)}{\bar{\rho}(r_2)}\,\frac{r_1}{r_2}[/math]

 

Notice you the common factor [imath]4/3\pi G[/imath] nicely drops out. Now substitute [math]r_2=R_E=6371\,\text{km}, r_1 = r_2/2[/math] and you get

 

[math]\frac{g(R_E/2)}{g(R_E)}\frac 1 2\,\frac {\bar{\rho}(R_E/2)}{\bar{\rho}(R_E)}[/math]

 

Using your 6.70 gm/cc³ as the average density for the halfway point does lead to your 60%.

 

 

This leads to the second problem. That density value of 6.70 gm/cm³ is completely wrong. Inside a spherical body, it's only the matter below that contributes to gravitational acceleration. All the stuff that is at a distance from the center of the body greater than that of the point in question contributes nothing. Halfway down to the center is about 300 km into the Earth's outer core. The Earth's core is mostly iron, compressed iron at that. We call liquids and solids incompressible, but that is a misnomer. The density of the material at this halfway down point is about 10.35 gm/cm³. Density builds up to 13.088 gm/cm³ at the very center of the Earth.

 

Ok, I see where you're going. I was kind of afraid my assumptions would come back and bite me, but I was too lazy to go look up any kind of accurate numbers for the mean density at specific depths. That'll teach me to be lazy.

 

Thanks for the explanation DH, I appreciate it.