How did Cavendish find the density of the Earth?

  • #1
marc873a
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Homework Statement
Explain how Cavendish found the density of the earth
Relevant Equations
P_earth = 3g/4pi*r_earth*G
I'm trying to find how Cavendish got the density of the earth to 5,48 times the density of water. In all of the YouTube videos and webpages I have seen, they mention different formulas where the mass of the earth or the gravitational constant, G, is included. But as far as I understand, these values were found later using the density that Cavendish found. How did he calculate the density without the other two values?

Thanks ins advance :)
 
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  • #2
According to my 3 seconds of research on the Internet, he found G and THEN found the density from that.

EDIT: Hm ... I did a few more seconds and now I'm getting the same results you got. Very confusing.

Well, someone here will know, I'm sure.

2nd EDIT: Numerous more reports agree that he found G.
 
  • #4
marc873a said:
I have found a few articles that say the opposite. Very confusing indeed
I do not understand. There is no distinction to be made between measuring G and measuring the density of the Earth. They are simply two ways to express the same experimental result.
 
  • #5
Well, using the formulas I have seen so far, the density is calculated using the mass of the earth or G, and G is estimated using the density or mass of the earth. I have not yet found any formulas where the result is found without using the other values which is what confuses me
 
  • #7
marc873a said:
Well, using the formulas I have seen so far, the density is calculated using the mass of the earth or G, and G is estimated using the density or mass of the earth. I have not yet found any formulas where the result is found without using the other values which is what confuses me
Most equations can be rearranged to put a particular unknown quantity alone on the left while the right hand side is filled with quantities that you know.

It is called algebra. Or "solving an equation" for an unknown.

Without even looking at the original explanation from Cavendish, we know a few things. He knew the radius of the Earth, ##r_\text{Earth}##. He knew how strongly a mass would accelerate in the Earth's gravitational field, ##g##.

The Cavendish experiment can be seen as establishing a traceable chain of comparisons between the attraction of a known mass to the Earth and the attraction of a known mass [pair] to a known pair of test bodies. That is, it amounts to a comparison between the active gravitational mass of the Earth and the active gravitational mass of a known pair of test bodies.

With that comparison in hand and knowing ##g## and ##r_\text{Earth}## and the mass of one's chosen test bodies (##158 \text{ kg}##), one can choose to report the experimental result as the mass of the Earth, as the density of the Earth or as the value of ##G##. It is just a matter of what one chooses to solve for. One can solve for any or all of them.
 
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  • #8
phinds said:
According to my 3 seconds of research on the Internet, he found G and THEN found the density from that.

EDIT: Hm ... I did a few more seconds and now I'm getting the same results you got. Very confusing.

Well, someone here will know, I'm sure.

2nd EDIT: Numerous more reports agree that he found G.
I wonder if it would make sense to add a sort of standard deviation to the 5.5 g/cm^3 , or is it mostly uniform at that value?

And his wife is reported to have said Henry never even came close to finding her G.
 
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  • #9
Oh, so he just found the ratio between attraction between the lead balls and the attraction between the earth and another object? But I'm still pretty confused on he calculated that attraction. I know of F = GMm/r^2, but that can't be it because G wasn't invented, right? What other formula calculates the attractive force, that he could have used?
 
  • #10
marc873a said:
Oh, so he just found the ratio between attraction between the lead balls and the attraction between the earth and another object? But I'm still pretty confused on he calculated that attraction. I know of F = GMm/r^2, but that can't be it because G wasn't invented, right? What other formula calculates the attractive force, that he could have used?
You do know that Newton's universal law of gravitation predates the Cavendish experiment by over a century, right?

It does not matter whether the constant of proportionality in the law had a standard name such as ##G##. The proportionality is the bit that matters. A constant of proportionality follows from that, named or not.

Personally, I find the algebra easier to do if the constant has a name. But one can work directly with proportionalities instead. Our ancestors knew how to do algebra, perhaps better than us.
 
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  • #11
WWGD said:
I wonder if it would make sense to add a sort of standard deviation to the 5.5 g/cm^3 , or is it mostly uniform at that value?
IMG_0057.jpeg


marc873a said:
Oh, so he just found the ratio between attraction between the lead balls and the attraction between the earth and another object? But I'm still pretty confused on he calculated that attraction. I know of F = GMm/r^2, but that can't be it because G wasn't invented, right? What other formula calculates the attractive force, that he could have used?
IMG_0058.jpeg
IMG_0059.jpeg
 
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  • #12
  • #13
It was Archimedes.
"Give me a rope to hang my soap and a bathtub large enough and I will determine the density of the Earth."
 
  • #14
WWGD said:
Thanks, Frabjous, what I meant was whether the density at different points in the ( interior of the ) Earth is overall uniform. Are there points where it's value is , say 6 gm/cc, other points where the density is, say 5 gm/cc?
The density increases with depth or am I misunderstanding your question.

IMG_0060.jpeg
 
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  • #15
WWGD said:
Thanks, Frabjous, what I meant was whether the density at different points in the ( interior of the ) Earth is overall uniform. Are there points where it's value is , say 6 gm/cc, other points where the density is, say 5 gm/cc?
There is no way to discern that through purely gravitational measurements at the Earth's surface. Any object which has a spherically symmetric mass distribution and the same size and mass as the Earth will gravitate identically with respect to external measurements.

Other experiments lead to a model for the Earth's density which then leads to the graph of gravitational acceleration versus depth shown here.
https://en.wikipedia.org/wiki/Preliminary_reference_Earth_model said:
1710458431555.png
The iron core is, of course, far more dense than 6 gm/cc.
 
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  • #16
WWGD said:
Thanks, Frabjous, what I meant was whether the density at different points in the ( interior of the ) Earth is overall uniform. Are there points where it's value is , say 6 gm/cc, other points where the density is, say 5 gm/cc?
https://www.google.com/search?&q=gravitational+map+of+earth

Cheers,
Tom
 
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  • #17
marc873a said:
What other formula calculates the attractive force, that he could have used?
##w=mg##. You know the force exerted on a terrestial body, ##w##, such as the force your own body exerts on a force-measuring scale. You know your own mass ##m##, and you know the value of the free fall acceleration ##g##.

Note: To increase the precision you'd have to know that at most the value of ##g## at sea level is reduced by about 0.51% (depending on latitude). About ##\frac{2}{3}## of that difference is due not to gravity but Earth's spin and the other ##\frac{1}{3}## is due to the equatorial bulge.
 
  • #18
After reading all of these responses, it seems like you are all over-complicating things. We know the acceleration of gravity at the Earth's surface is $$g=\frac{GM}{r^2}$$

The values g and r (the radius of the Earth). were known to pretty good accuracy in Cavendish' time. So GM was known to pretty good accuracy. Cavendish measured G with the famous Cavendish experiment. Once he knew G, since he knew GM, he could calculate M. Then the average density of the Earth is just $$\rho = \frac{3 M}{4 \pi r^3}$$
 
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  • #19
@phyzguy He didn't calculate G as constant explicitly, that's what causing the confusion.

The acceleration is proportional to the ratio of mass divided by distance squared. The mass is proportional to the average density multiplied by the cube of the radius of the object: ##a \propto \frac{M}{r^2} = \frac{\rho R^3}{r^2}##
He found the ratio between the acceleration his lead balls caused and Earth caused, took the different distances and sizes into account, and found that Earth has roughly half the density of lead.
 
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  • #20
@mfb. I don't see why you are saying that he didn't measure G explicitly. He measured the force between his large lead balls and the smaller balls on the end of the tension wire by measuring how much the wire twisted. Since he knew the masses of the balls and the distance between them, he was able to calculate G. Why do you say he didn't measure G?
 
  • #21
phyzguy said:
@mfb. I don't see why you are saying that he didn't measure G explicitly.
If you read carefully, @mfb did not say anything about "measuring" G. He talked about "calculating" G.
mfb said:
He didn't calculate G as constant explicitly
Taken together with the information known at the time, the result of the Cavendish experiment would have allowed for a calculation of the numeric value of ##G## in some chosen set of units. In that sense it amounts to a measurement of ##G##.

@mfb is simply saying that no such calculation was made/published by Cavendish at the time.
 
  • #23
The way physical laws are written and used often changes over time even if the underlying physics doesn't change. Newton expressed the law only as proportionality and Cavendish used the same approach. The modern version F = GmM/r2 with an explicit constant G was introduced in ~1900, about 100 years after the Cavendish experiment. Cavendish didn't calculate a value for G because that concept didn't exist yet.

You can see the same pattern in many places. The Maxwell equations were put into their modern form by Heaviside, who was still a child when Maxwell discovered the relations. The way you learn general relativity today isn't following the approach Einstein took. And so on.
 
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  • #24
phyzguy said:
Since he knew the masses of the balls and the distance between them, he was able to calculate G.
He certainly had all the data he needed. Whether or not he was able is another issue. Equations were not in common use at that time IIRC. They used proportions.

But regardless, he never made that calculation. Others did, though, using his data.
 
  • #25
From the link I provided above:

Newton had published his law of gravitation in 1687, but he hadn’t made any attempt to determine the constant G or the mass of Earth. By the 1700s, astronomers wanted to know the density of Earth, as it would make it possible to determine density of the other planets. ...

In 1772 the Royal Society set up a “Committee of Attraction” to determine the density of Earth. Some people had proposed measuring this by finding a very uniformly shaped mountain and measuring how much it deflected a plumb bob. Since gravity is so weak, this would be a tiny effect, but the committee, including Cavendish, nonetheless tried it, using a large mountain in Scotland. They came up with a value for the density of Earth of about 4.5 times the density of water. But they had made assumptions that Cavendish thought unfounded.

He considered the problem for years, until in 1797, at age 67, he began his own experiments. He started with a torsion balance apparatus given to him by his friend, the geologist Reverend John Michell ...

So Cavendish was trying to determine the density / mass of the earth. That's the question he set out to answer. I'm sure that's why his experiments are written up that way. The algebraic equivalence to determining G simply follows
 

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