<p>Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, ..., the period of a satellite of negligible mass (even the International Space Station qualifies as a "satellite of negligible mass") is $T=2\pi\sqrt{\frac {a^3}{\mu_\mathrm{Earth}}}$. Neither Newton's gravitational constant nor the mass of the Earth are involved in this expression. This means that, ignoring those details, calculating $\mu_\mathrm{Earth}$ is merely a matter of calculating a satellite's rotational period and its semimajor axis.</p>

<p>Humanity has lots and lots of artificial satellites in orbit, and the people who model the orbits of those satellites don't ignore those details. A few of those satellites were specially designed to enable the determination of the Earth's non-spherical gravitational field (e.g., GRACE and GOCE), and a few were specially designed to enable extremely precise orbit determination (e.g., LAGEOS). Even with all of those details, the Earth's gravitational parameter is a directly inferable quantity (i.e., knowledge of G is not required). Moreover, the value is known to a very high degree of precision.</p>

<p>The Earth's mass? Not so much. The most precise way to "weigh the Earth" is to divide the high precision Earth's gravitational parameter by the low precision universal gravitational constant G. There's a problem here, which is the notoriously low precision of the gravitational constant when expressed in SI units.</p>

CC BY-SA 3.0
2016-02-06T23:56:56.280Z
Thanks for the explanation. When you say "notoriously low precision...when expressed in SI units" are you implying that there are more accurate measures of the gravitational constant in non-SI units?
doug5085
CC BY-SA 3.0
2016-02-07T00:03:03.763Z
@doug5085 -- Yes. For example, consider a system where the unit of mass is one solar mass, the unit of distance is the astronomical unit, and the unit of time is a mean solar day. Here, G is known to about 11 places of accuracy. This is because we known $\mu_\mathrm{Sun}$ to about 11 places of accuracy. This system of units effectively wipes out the distinction between $GM$ and $\mu$, at least with regard to the Sun and the planets.
David Hammen
CC BY-SA 3.0
2016-02-07T00:13:31.277Z
This is the system Gauss introduced 200+ years ago (1809, to be precise). Solar system astronomers use this system of units to this day. Gauss initially defined the AU to be the length of the semimajor axis of the Earth's orbit about the Sun. By the late 19th century, it had become obvious that the ten places of accuracy Gauss used for his gravitational constant (essentially the square root of G) was inconsistent with observation. The solution was simple: Make the astronomical unit be something consistent with observation. The Earth's major axis length is about 1.000001 AU.
David Hammen
CC BY-SA 3.0
2016-02-06T23:53:18.207Z
2016-02-06T23:57:06.403Z
2016-02-06T23:57:06.403Z
David Hammen
<p>$\mu$ is a quantity that can be easily observed (semi-) directly. It can be easily derived from orbital period of orbiting bodies or acceleration of falling bodies, even if their mass is significantly smaller than the mass of the body you're measuring $\mu$ for.</p>

<p>Now the only way to really obtain G is to divide $\mu$ by mass of your celestial body. And measuring this mass without knowing G is opening a can of worms. You may try to get geophysical studies of density distribution, or try to see how this body reacts to other masses (of comparable magnitude), or a number of increasingly complex methods of dubious accuracy.</p>

<p>Measuring mass of a simple body of range of milligrams to tons without use of gravity as a reference is not a big problem. You accelerate it with a known force, measure the acceleration, and get the mass. (possibly, the force is known because it was used to accelerate a known mass - or a model mass, that defines, say, 1kg.)</p>

<p>But how do you go about accelerating a planet - or any body heavy enough that it exhibits gravitational force strong enough to be measurable with any reasonable precision - with a known force, to measure its mass? About the only forces that are capable of doing so in any observable manner, are the forces of gravity - and they are all proportional to the unknown G.</p>

CC BY-SA 3.0
2016-02-06T23:36:53.140Z
2016-02-07T00:02:07.180Z
2016-02-07T00:02:07.180Z
SF.
<p>When a body spins around another body due to gravity and maintains a consistent orbit, we can know clearly two things about the body: 
1. The speed of the orbit
2. The average radius of the orbit</p>

<p>Think about it. The Moon spins around the Earth due to gravity of the Earth. Right? And it's orbit always stays the same(when averaged). For it to stay in this balance where it isn't flying away from the Earth, requires that it be traveling at a speed and distance that equal out the effect of gravity on the object. So by knowing this speed(orbital speed) and distance(orbital radius), we can determine something very useful about the gravitational effect of the Earth on the Moon. From the gravitational effect, we can then determine the mass of the Earth.</p>

<p>Finding the mass of the Earth otherwise is quite difficult because we don't know exactly what is inside, and there are too many factors involved. We can't just put the Earth on a big scale. Also, determining it from gravity is flawed because there are too many factors that cause high uncertainty(Earth is not a perfect sphere, the moon exhibits gravity too, atmospheric pressure, centrifugal force, etc.). BUT we DO know the distance of the Moon from the Earth very accurately using Trigonometry. We also know the time the Moon takes to go around the Earth as we can easily measure this(lunar cycle). So if you take the orbital speed squared and multiply it by the orbital radius of the Moon's spin around the Earth, you obtain the Geocentric Gravitational Constant of the Earth. </p>

<p>Using that constant, we can determine the mass of the Earth. But we also need to know the Gravitational Constant(G) to make this determination. However, G has been difficult to determine with a high level of certainty. This is because the experiments to determine G have so far given inconsistent results. Either G is changing OR the factors in the experiments have not all been ruled out, likely the latter. You can read more on Geocentric Gravitational Constant here: <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow">https://en.wikipedia.org/wiki/Standard_gravitational_parameter</a></p>

CC BY-SA 3.0
2016-06-23T16:45:26.913Z
2016-06-23T16:45:26.913Z
user2299067
<p>I am looking for an answer as to why the geocentric gravitational constant, μ, defined as the product of the gravitational constant, G, and the mass of a body (earth in this case), M, can be calculated to a higher degree of precision than either G or M alone.</p>

<p>From wikipedia</p>

<blockquote>
  <p>...for celestial bodies such as Earth and the Sun, the value of the product GM is known much more accurately than each factor independently. Indeed, the  limited accuracy available for G limits the accuracy of scientific 
   determination of such masses in the first place.</p>
</blockquote>

<p>Can someone explain this? I can't find a good answer anywhere I look. </p>

<gravity><newtonian-gravity>
Precision of geocentric gravitational constant
doug5085
# Precision of geocentric gravitational constant

(Asked by: username_0 on 2016-02-06)

I am looking for an answer as to why the geocentric gravitational constant, μ, defined as the product of the gravitational constant, G, and the mass of a body (earth in this case), M, can be calculated to a higher degree of precision than either G or M alone.

From wikipedia

> ...for celestial bodies such as Earth and the Sun, the value of the product GM is known much more accurately than each factor independently. Indeed, the limited accuracy available for G limits the accuracy of scientific
> determination of such masses in the first place.

Can someone explain this? I can't find a good answer anywhere I look.

**Tags:** gravity, newtonian-gravity

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## 3 answers

### Answer by username_1 on 2016-02-06. Score: 3 | Accepted answer

Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, ..., the period of a satellite of negligible mass (even the International Space Station qualifies as a "satellite of negligible mass") is $T=2\pi\sqrt{\frac {a^3}{\mu\_\mathrm{Earth}}}$. Neither Newton's gravitational constant nor the mass of the Earth are involved in this expression. This means that, ignoring those details, calculating $\mu\_\mathrm{Earth}$ is merely a matter of calculating a satellite's rotational period and its semimajor axis.

Humanity has lots and lots of artificial satellites in orbit, and the people who model the orbits of those satellites don't ignore those details. A few of those satellites were specially designed to enable the determination of the Earth's non-spherical gravitational field (e.g., GRACE and GOCE), and a few were specially designed to enable extremely precise orbit determination (e.g., LAGEOS). Even with all of those details, the Earth's gravitational parameter is a directly inferable quantity (i.e., knowledge of G is not required). Moreover, the value is known to a very high degree of precision.

The Earth's mass? Not so much. The most precise way to "weigh the Earth" is to divide the high precision Earth's gravitational parameter by the low precision universal gravitational constant G. There's a problem here, which is the notoriously low precision of the gravitational constant when expressed in SI units.

> **Comments:**
> **username_0**: Thanks for the explanation. When you say "notoriously low precision...when expressed in SI units" are you implying that there are more accurate measures of the gravitational constant in non-SI units?  
> **username_1**: @doug5085 -- Yes. For example, consider a system where the unit of mass is one solar mass, the unit of distance is the astronomical unit, and the unit of time is a mean solar day. Here, G is known to about 11 places of accuracy. This is because we known $\mu_\mathrm{Sun}$ to about 11 places of accuracy. This system of units effectively wipes out the distinction between $GM$ and $\mu$, at least with regard to the Sun and the planets.  
> **username_1**: This is the system Gauss introduced 200+ years ago (1809, to be precise). Solar system astronomers use this system of units to this day. Gauss initially defined the AU to be the length of the semimajor axis of the Earth's orbit about the Sun. By the late 19th century, it had become obvious that the ten places of accuracy Gauss used for his gravitational constant (essentially the square root of G) was inconsistent with observation. The solution was simple: Make the astronomical unit be something consistent with observation. The Earth's major axis length is about 1.000001 AU.  

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### Answer by username_2 on 2016-02-06. Score: 2

$\mu$ is a quantity that can be easily observed (semi-) directly. It can be easily derived from orbital period of orbiting bodies or acceleration of falling bodies, even if their mass is significantly smaller than the mass of the body you're measuring $\mu$ for.

Now the only way to really obtain G is to divide $\mu$ by mass of your celestial body. And measuring this mass without knowing G is opening a can of worms. You may try to get geophysical studies of density distribution, or try to see how this body reacts to other masses (of comparable magnitude), or a number of increasingly complex methods of dubious accuracy.

Measuring mass of a simple body of range of milligrams to tons without use of gravity as a reference is not a big problem. You accelerate it with a known force, measure the acceleration, and get the mass. (possibly, the force is known because it was used to accelerate a known mass - or a model mass, that defines, say, 1kg.)

But how do you go about accelerating a planet - or any body heavy enough that it exhibits gravitational force strong enough to be measurable with any reasonable precision - with a known force, to measure its mass? About the only forces that are capable of doing so in any observable manner, are the forces of gravity - and they are all proportional to the unknown G.


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### Answer by username_3 on 2016-06-23. Score: 2

When a body spins around another body due to gravity and maintains a consistent orbit, we can know clearly two things about the body:
1. The speed of the orbit
2. The average radius of the orbit

Think about it. The Moon spins around the Earth due to gravity of the Earth. Right? And it's orbit always stays the same(when averaged). For it to stay in this balance where it isn't flying away from the Earth, requires that it be traveling at a speed and distance that equal out the effect of gravity on the object. So by knowing this speed(orbital speed) and distance(orbital radius), we can determine something very useful about the gravitational effect of the Earth on the Moon. From the gravitational effect, we can then determine the mass of the Earth.

Finding the mass of the Earth otherwise is quite difficult because we don't know exactly what is inside, and there are too many factors involved. We can't just put the Earth on a big scale. Also, determining it from gravity is flawed because there are too many factors that cause high uncertainty(Earth is not a perfect sphere, the moon exhibits gravity too, atmospheric pressure, centrifugal force, etc.). BUT we DO know the distance of the Moon from the Earth very accurately using Trigonometry. We also know the time the Moon takes to go around the Earth as we can easily measure this(lunar cycle). So if you take the orbital speed squared and multiply it by the orbital radius of the Moon's spin around the Earth, you obtain the Geocentric Gravitational Constant of the Earth.

Using that constant, we can determine the mass of the Earth. But we also need to know the Gravitational Constant(G) to make this determination. However, G has been difficult to determine with a high level of certainty. This is because the experiments to determine G have so far given inconsistent results. Either G is changing OR the factors in the experiments have not all been ruled out, likely the latter. You can read more on Geocentric Gravitational Constant here: <https://en.wikipedia.org/wiki/Standard_gravitational_parameter>


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