# How to measure mass of earth, moon, independently of G

1. Apr 9, 2015

### exponent137

In Principle, gravitational constant is measured very precise, but the problem is that it is measured together with mass of sun, earth or moon, G M_s or G M_e; or G M_m. But, is it possible to measure mass of the moon without use of G? Maybe the principle, where the rocket was accelerated when in flies close to Jupiter or Saturn. It seems to me, that this is not only gravitational principle, but it is really a principle of exchanging of momentum?

2. Apr 9, 2015

### Orodruin

Staff Emeritus
One option (and historically perhaps the first) is the Cavendish experiment. It is essentially a measurement of G.

3. Apr 9, 2015

### exponent137

Yes, but all direct measurements of G are not precise enough. Thus, if we could tell what is mass of moon, without inclusion of G, than we can calculate G from the product G M_m, and this G could be very precise.
This is the reason, that I ask for alternative ways of measuring masses of the large objects.

4. Apr 9, 2015

### Orodruin

Staff Emeritus
The big problem here is that you need a method of determining the mass of at least two objects and also determine the gravitational attraction between them. Since you need a very large object for gravitational forces to become large, it boils down to determining the mass of at least one very big object if you want to go down in uncertainty on both the masses themselves and G.

"Not precise enough" for what exactly? Do you need to determine something to one part in $10^4$?

Edit: The fine structure constant is determined to one part in $10^9$ (roughly). How many applications do you know of that actually need to know this very precise value? (Apart from our own curiosity in checking that our physics models make sense.)

5. Apr 9, 2015

### exponent137

One option of measurement would be, to have a rocket with quantum atomic gravimeter, which lands o a asteroid. He put gravimeter on it. Then it gives it a known jerk of momentum and then it measures its relative velocity. Thus it can measure its mass and its gravity. This is in principle enough to measure G more precisely.

But, my question is: when Jupiter's gravity pushes a rocket further into space, (that the rocket flies toward end of solar system) is this only a gravitational impact, or it can give any information about solely mass of Jupiter, this means $m_J$, not only product $G m_J$?

About fine structure constant: it is used for confirmations of QFT theory, I do not know precisely. more precise G can be used for development of quantum gravity theory, as Quinn of BIPM said.

6. Apr 9, 2015

### Orodruin

Staff Emeritus
A priori, you might be able to measure the mass of something large by doing that. However, you really need to escape the gravitational pull without distorting the result, which will not be the case due to tidal forces from the Sun etc. I doubt you can get more precision out of that procedure than the current precision on G. Remember that the fine structure constant is one of the absolutely most well known experimental quantities.

7. Apr 11, 2015

### exponent137

As one your old question: I think that very precise fine structure constant helps that we today believe in quantum electrodynamics, because it is so precisely proved. I hope that G will also help.

I agree with you that "I doubt you can get more precision out of that procedure than the current precision on G.", thus it is possible that such measurement will not improve G precision.

But, anyway it is theoretically interesting. Please, how can you prove me, or clearly explain, that rocket which uses jupiter as acceleration, can measure $m_J$, not $G m_J$.?

8. Apr 11, 2015

### Orodruin

Staff Emeritus
You would use conservation of momentum and you would need to know the mass of the rocket and be able to measure velocity changes of Jupiter due to the rocket passage. It seems quite challenging.

9. Apr 11, 2015

Staff Emeritus
This is turning a problem of measuring G to a part in 10,000 to measuring a very large force to better than a part in 10,000 (hard) and a very small acceleration to better than a part in 10,000 (even harder). I very much doubt this will work well.

10. Apr 11, 2015

### exponent137

At this priniciple it is almost imposible to see any change of velocity of Jupiter.

But I thought that at the same speed of the planet, at the same distance from it, at the same radius of the planet, but at the different mass of a planet, is there any difference which change of rocket momentum will be given to the rocket?

Anyway, in some 50 years, one many tones heavy monocristal of Si will be built. If we will add to it quantum gravimeter, the measured G will be much preciser.

As a detail, one of the bigest successes of measurements of G in the last years was Quinns' or BIPM's one, who in more than 10 years distance measured twice the same value of G.

V50, I agree, I am interested only about principle of measuring $M_j$ without measuring $G M_j$.

Last edited: Apr 11, 2015