The mass of Earth.

**mgb_phys** · Nov27-09, 10:04 PM

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

**D H** · Nov27-09, 11:28 PM

Originally Posted by mgb_phys

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~~ · Nov28-09, 06:48 PM

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** · Nov28-09, 09:29 PM

Originally Posted by D H

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** · Nov28-09, 09:41 PM

Originally Posted by mikelepore

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** · Nov28-09, 09:46 PM

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

~~JANm~~ · Nov29-09, 06:20 AM

Originally Posted by D H

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** · Nov30-09, 01:25 PM

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

~~JANm~~ · Nov30-09, 02:22 PM

Originally Posted by Thermodave

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** · Nov30-09, 02:56 PM

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~~ · Dec27-09, 06:21 PM

Originally Posted by 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.

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** · Dec27-09, 07:15 PM

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

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

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

**milford30** · Dec28-09, 05:08 AM

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~~ · Dec28-09, 05:30 PM

Originally Posted by D H

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