Calculate Mass of Earth - Ideas & Formulas

[God's Pen]

Hello
I want to know the ideas and the formulas used to calculate the mass of Earth.
thanks.

 

[mgb_phys]

The speed that an object orbits a much larger object only depends on the mass of the larger object.
So if you have something orbiting the Earth, like the moon, and know its distance and how long it takes to go around - you can find the mass of the earth.

A physicist called Cavendish measured this in about 1800.

 

[Bob S]

Here is a quick way. The volume is (4/3) pi R³. Plug in the radius of the Earth, and the density (~5.515 grams/cc), and you will get the correct answer within a factor of two. Mgb-phys has a much more accurate way, but you need to know the separation distance.
Bob S

 

[mgb_phys]

Knowing the density of the Earth is a bit of a cheat.
Interestingly that's what Cavendish was trying to work out when he measured G - they wanted to know if the Earth was hollow !

 

[D H]

Before I answer: Is this homework?

 

[Bob S]

Knowing the density of the Earth is a bit of a cheat.
!

So is knowing the distance to a satellite, like Sputnik, the ISS, or the Moon.
Bob S

 

[D H]

Knowing the density of the Earth is a bit of a cheat.

So is knowing the distance to a satellite, like Sputnik, the ISS, or the Moon.

The distance to a satellite or the Moon is a directly measurable quantity. The density of the Earth is not. It is inferred from the Earth's size and mass. Using density to derive mass is a circular derivation.

 

[ideasrule]

The distance to a satellite or the Moon is a directly measurable quantity. The density of the Earth is not. It is inferred from the Earth's size and mass. Using density to derive mass is a circular derivation.

The density of Earth might not have been well-known in Cavendish's day, but scientists could at least assume it had the same density as molten rock and get the approximate mass of the Earth from that. Now, in the 21st century, we have a decent understanding of Earth's density profile thanks to scientists' analysis of the refraction of seismic waves as they pass through the Earth.

 

[DaveC426913]

BobS: note that, even in principle, the best you can get for such a heterogenous object as Earth is the average density.

And the only way to derive the average density of the Earth is to first determine the volume ... and the mass.

scientists could at least assume it had the same density as molten rock and get the approximate mass of the Earth from that. Now, in the 21st century, we have a decent understanding of Earth's density profile thanks to scientists' analysis of the refraction of seismic waves as they pass through the Earth.

Sure, if you only need an answer within an order of magnitude or so. The Earth is quite heterogenous, and our tools for measurement are not very accurate. There's a lot of assumption in the method you propose.

 

[ideasrule]

The mass of solid basalt is 3000 kg/m^3, so even if you take that and multiply it by Earth's volume, you'd be within a factor of 2 of the right answer. If you take into account all the latest sesmic data and computer models, you'd probably get within 10% of the actual mass. Of course that's less accurate than watching a satellite orbit and calculating Earth's mass from that, but it refutes DH's comment that the density of Earth is not directly measurable.

 

[DaveC426913]

The mass of solid basalt is 3000 kg/m^3, so even if you take that and multiply it by Earth's volume, you'd be within a factor of 2 of the right answer. If you take into account all the latest sesmic data and computer models, you'd probably get within 10% of the actual mass. Of course that's less accurate than watching a satellite orbit and calculating Earth's mass from that, but it refutes DH's comment that the density of Earth is not directly measurable.

As you said yourself: you're relying on computer models. We make computer models to predict what's probably there when we can't directly measure it.


10% eh? So, you'd only be off by give-or-take one hundred thousand million million million kg or so...

 

[mgb_phys]

The density of Earth might not have been well-known in Cavendish's day, but scientists could at least assume it had the same density as molten rock and get the approximate mass of the Earth from that.

They tried that.
The equations for gravity have a GM term in them where M is the mass of the earth, so you can find M by measuring 'G' in Cavendish's famous experiment or you can find M from the density and use that to get G.
There was a long and careful attempt to measure the volume of a large conical mountain by a lot of surveying - and then measure it's mass by the effect it had on a small pendulum.
It didn't work very well because you don't know what shape the mountain is under the surface.

Cavendish got GM to within about 1%, assuming the Earth has the density as surface rocks gets you about half the correct value.

 

[ideasrule]

As you said yourself: you're relying on computer models. We make computer models to predict what's probably there when we can't directly measure it.


10% eh? So, you'd only be off by give-or-take one hundred thousand million million million kg or so...

And if you calculate the mass of the visible universe, you'd be off by at least a million million million million million million million million million million million million million kilograms with the most precise tools available. I'm not sure why you think the absolute value of the error is more important than the percentage error.

I feel that we're arguing only about semantics. Here's what I'm saying:

1) The density profile of the Earth can be inferred using the refraction of seismic waves.
2) This density profile can be used to arrive at a reasonable figure for Earth's mass.
3) Computer models can improve the accuracy of this result.
4) The most accurate calculations of Earth's mass involve measuring G and g at a known distance r from the Earth.

 

[D H]

Here's what I'm saying:

1) The density profile of the Earth can be inferred using the refraction of seismic waves.

Can it really? Or do siesmologists instead calibrate those inferred density profiles against the already known mass of the Earth?

The current limiting factor in the accuracy in the Earth's mass is the uncertainty in the measured value of G. Suppose seismologists could independently arrive at an estimate of the Earth's mass that is more accurate than the paltry 1 part in 10⁴ to which we know G. That improved estimate of the Earth's mass we would directly lead to a better estimate for G because we know the product G*M_e to an accuracy of 2 parts in 10⁹.

 

[DaveC426913]

I'm not sure why you think the absolute value of the error is more important than the percentage error.

Well played... 10% is 10%.

I feel that we're arguing only about semantics. Here's what I'm saying:

1) The density profile of the Earth can be inferred using the refraction of seismic waves.
2) This density profile can be used to arrive at a reasonable figure for Earth's mass.
3) Computer models can improve the accuracy of this result.
4) The most accurate calculations of Earth's mass involve measuring G and g at a known distance r from the Earth.

It is possible in principle. I'll grant you that. In practice, it is an exceedingly backwards and impractical way of doing it.

 

[D H]

4) The most accurate calculations of Earth's mass involve measuring G and g at a known distance r from the Earth.

This is the freshman physics version. The truth is a lot more complex. The Earth cannot be treated as a point mass for any distance r for which the acceleration due to gravity is observable.

 

[mgb_phys]

Are tidal effects on the moons period significant - compared to the accuracy in 'G' ?

 

[D H]

Are tidal effects on the moons period significant - compared to the accuracy in 'G' ?

To get that precision of 2 parts in 10⁹ in GM_e, yes, tidal effects are important, and in many ways. To assess the GM_e to a precision of one part in 10⁵ (an order of magnitude better than the precision in G), tidal effects are still important. They obviously are of reduced importance, and the subtle effects of tides can probably be ignored.

That two parts in a billion precision was based on five years of LAGEOS-1 data and was published in 2000 (the 2005 results increase the precision to one part in a billion). The kinds of errors that contributed to a few parts in a billion precision include:

• Uncertainties in the location in the Earth's center of mass. An uncertainty of 3 millimeters corresponds to about one part in a billion uncertainty at LAGEOS' altitude.
• Uncertainties in the Earth's and the Moon's tidal Love numbers. The Moon and creates tides in the Earth, which changes the Earth's gravity field, which in turn changes LAGEOS' orbit.
• Uncertainties in the oceanic tidal responses. The ocean tides also affect the orbits of satellites; the effect is about 1/10 that of the solid body tides.
• Uncertainties in the positions of the ground stations that measured the distance to the LAGEOS via laser retroreflector returns. A 2 mm error RMS corresponds to a part per billion error in GM_e.
• Which means properly modeling the Earth tides are important.
• Even modeling the snow load in Siberia is important for this kind of accuracy.

 

[JANm]

A Robinson Crusoe question, nice problem for a mathematician with low memory for numbers and high for principal formulae. We know gravitation at the surface pi^2=GM/r^2.
The deep pit method of measuring circumference of the Earth gives the radius r=31900/5.
So indeed one wishes to know G. In the museum in Munchen I have seen a iron ball of radius 30 cm. This thing is used to measure G...
But I remember there must be an astronomical solution based on sun, Earth and moon. I can't just remember which astronomical facts were known to Robinson Crusoe when he had all the time of the world to calculate its mass,
i'll be back,
greetings Janm

 

[mikelepore]

This is the freshman physics version. The truth is a lot more complex. The Earth cannot be treated as a point mass for any distance r for which the acceleration due to gravity is observable.

Would we be able to treat the Earth as a point mass if, instead of defining distance r to the geometric center, we defined r to the Earth's center of mass?

If so, could we locate the Earth's center of mass by the intersection of several plumb lines?

 

[ideasrule]

Would we be able to treat the Earth as a point mass if, instead of defining distance r to the geometric center, we defined r to the Earth's center of mass?

No. The equation F=GMm/r^2, where r is the distance to the center of mass, only works for perfect spheres and point masses. It does not work for arbitrary solids.

In any case, I wasn't implying that measurements of GM were as simple as I made it seem. There are obviously a lot of complexities. Besides the ones already mentioned, the Earth is not a perfect sphere even if no mountains, oceans, etc. existed, so the equation F=GMm/r^2 can't be used anyways for high-precision calculations.

 

[D H]

If you want high accuracy you cannot treat the Earth as a point mass. The Earth is not spherical.

 

[JANm]

If you want high accuracy you cannot treat the Earth as a point mass. The Earth is not spherical.

Hello D H
Thank you for reminding that to Robinson. Every winter he recalculates
omega_earth=40000/day by head and finds out that r_earth*omega_earth gives V_equator=0,54 km/sec. But the Earth knows that too and adapts its shape to that.
Robinson is pleased with Mercator-projection and makes a drawing of the curve:

omega_earth(NB)=0,54*Cos(NB)

to know the velocity in km/sec of every point on the earth.
greetings Janm

 

[Thermodave]

This thread just made me think of something. What fraction of the Earth's mass is made up by the atmosphere?

 

[JANm]

This thread just made me think of something. What fraction of the Earth's mass is made up by the atmosphere?

Hello Thermodave
Nice question. In meteorology there is a coordinate system which weighs air with coordinate z. Linear in pressure related to pressure at surface the atmosfere is 8 km high. Actually we know it is much higher, but think of it as a fluid with constant density...
One kubic metre of air weighs 1,25 kg. So now you can calculate the answer to your question...
greetings Janm

 

[Vanadium 50]

There is an even easier way to calculate the mass of the atmosphere - atmospheric pressure is due to the weight of the atmosphere. 14.7 pounds per square inch, times the number of square inches, and then suitably converted to your favorite units.

 

[JANm]

There is an even easier way to calculate the mass of the atmosphere - atmospheric pressure is due to the weight of the atmosphere. 14.7 pounds per square inch, times the number of square inches, and then suitably converted to your favorite units.

Ok Vanadium
Perhaps it is simpler in your part of Europe. An inch is 2,54 cm that I know but in all the rest of Europe even the greengroceries are not allowed since a law of about 1990 to use the word pound. Do you mean Nautical Pounds or Australian pounds? I am not sure there and don't see the simplicity you seem to have added.
greetings Janm

 

[D H]

Fine. Do it in metric. Standard atmospheric pressure is 101.325 kPa. The mean radius of the Earth is 6371 km. Standard gravity is 9.80665 m/s². Putting it all together, the mass of the atmosphere is

$$\aligned<br /> m_{\text{atm}} &= 1\,\text{atm}\times 4\pi {R_e}^2/g_0 \\<br /> &= 101.325\,\text{kPa}\,\times4\,\pi\,(6371\,\text{km})^2/(9.80665\,\text{m/s}^2) \\<br /> &\approx 5.27\times10^{18}\,\text{kg}<br /> \endaligned$$

This very simple calculation compares extremely favorably to the estimates here: http://hypertextbook.com/facts/1999/LouiseLiu.shtml.

 

[milford30]

i'm just wondering which approximation would be more accurate using the Sun as center to find the mass of the Earth or use say... the moon?
the Sun center approximation would be affected by the other planet orbitals, but can be treated as point particles due to the large radius...
the moon would have errors in the mass distribution of the Earth being significant?

 

[JANm]

Fine. Do it in metric. The mean radius of the Earth is 6371 km...

Hello D H
Thanks for this calculation. With S_air=1,25 kg/m^3 and reduced height=8 km I get:
m_atm=5,1E18 kg that does not differ so much with you value: 5,27E18 kg.
Now we can answer the original question of Thermodave...
greetings Janm