Question

[Johnny5]

Suppose I want to measure the mass of the moon in the following way.

 

I want to do a handstand on the moon.

 

Suppose that the moon is a perfect sphere, and that it has a uniform density.

 

Will it matter where on the moon I do the handstand?

 

Let us suppose that I have an excellent "muscle memory" and that I remember how hard it is to do a handstand on earth. Define the earth to have one inertial mass unit. I go to the moon, do my handstand, and go wow, that was 1/3 easier therefore

 

[math] M_{moon} = \frac{1}{3} M_{earth} [/math]

 

So that the moon has a mass of 1/3 imu

 

The question really has to do with the vacuum, not the moon, nor my method of measurment. Is inertial mass a direction dependent quantity? That's really what I'm asking.

 

 

Suppose the following diagram illustrates where the center of mass of the universe is in relation to the moon.

 

CMU . _______________________________( L-moon-R )

 

I can either choose to do the handstand on the left side of the moon, which is closer to the CMU, or the right side of the moon, which is further away from the CMU. Will which side I'm on affect my measurement?

 

And if the location of the center of mass of the universe doesn't affect the measurement, then let me ask this instead...

 

Does the vacuum have any effect on the measurement? And if it does, at what level does this effect become appreciable?

 

Another way to ask this kind of question is this...

 

Is there friction in the vacuum? if yes, how much does this vary, as a function of position?

 

To a good approximation, the vacuum does not impede the foward motion of objects. But this cannot be literally true, since there are hydrogen atoms scattered throughout the universe, and also the vacuum is filled with EM radiation. Does the EM radiation impede our foward motion at all?

 

Thank you

[swansont]

Gravity depends on R as well as M, so your initial analysis is flawed.

 

There is no center-of-mass for the universe; as has been stated several times, this is contrary to the big bang. It's tough to analyze any physics based on an invalid premise.

 

Yes, the vacuum will have an effect, because there is a bouyancy force of the atmosphere, though it is small. It is the mass of the atmosphere the body displaces.

 

Anisotropic EM radiation would also affect the measurement, since there is radiation pressure. This is very small.

 

When you discuss a vacuum, you have to define which definition you are using - relatively low pressure or absolutely zero content.

[Johnny5]

Gravity depends on R as well as M, so your initial analysis is flawed.

 

Yes of course, R matters as well as M. If the moon was squeezed into a volume the size of say a football stadium , the inertial mass M would have remained constant, but the density of the moon would have increased dramatically. Then when I went to do my handstand, the force upon me would be much greater, because of gravity.

 

The gravitational force of an object A on an object B is inversely proportional to the distance object B is from the center of mass of object A.

 

So by squeezing the moon into the volume of a football stadium, I can get much closer to the center of gravity, (R decreases a lot, so F increases a lot), and M stayed the same (i think).

 

There is no center-of-mass for the universe; as has been stated several times' date=' this is contrary to the big bang. It's tough to analyze any physics based on an invalid premise.[/quote']

 

The number of bodies is finite, and the inertial mass per body is finite, so the total mass of the universe is finite, so there is a center of mass somewhere in space. But my question didn't really center on whether there is one or not.

 

Yes' date=' the vacuum will have an effect, because there is a bouyancy force of the atmosphere, though it is small. It is the mass of the atmosphere the body displaces.

 

[/quote']

 

The moon has an atmosphere? Explain this more to me please.

 

 

Anisotropic EM radiation would also affect the measurement' date=' since there is radiation pressure. This is very small.

[/quote']

 

Yes radiation pressure, what is the formula for it again?

[swansont]

The moon has an atmosphere? Explain this more to me please.

 

Yes radiation pressure' date=' what is the formula for it again?[/quote']

 

The moon has little atmosphere, but the earth does. This affects the comparison.

 

Radiation pressure is from the momentum of the photons. p = E/c. From that you can get a force, if the photons are absorbed at some rate: F = dp/dt = dE/cdt, but dE/dt is power. So F=P/c. Of course if it's isotropic, it all cancels. This all gets mofified if the object absorbing/reflecting the photons isn't uniform. (reflecting gives you an extra factor of 2 in the force)

[Johnny5]

The moon has little atmosphere, but the earth does. This affects the comparison.

 

I guess it would. Are you saying that when I do my handstand on earth, that I am also pushing slightly on our atmosphere (on the other side of the world?

 

 

Radiation pressure is from the momentum of the photons. p = E/c.

 

Ok suppose that formula is correct.

 

[math] \vec F = \frac{d\vec P}{dt} = 1/c \frac{dE}{dt} [/math]

 

Pressure is force per unit area.

 

So when I do my handstand on the moon, I push the moon away slightly from me. So the area in question is half of the surface area of the moon?

[Martin]

Johnny you say

 

[math] M_{moon} = \frac{1}{3} M_{earth} [/math]

 

 

but my astronomy handbook says

 

[math] M_{moon} = \frac{1}{81} M_{earth} [/math]

 

how did you get the 1/3?

 

it sounds to me like you have mixed up the surface gravity of the moon, which is 1/6 of earth's, with the mass of the moon

 

you would weigh only 1/6 as much on the moon, that's clear.

so it would be that much easier to do handstands

if you weigh 180 pound on earth the force on moon would be 1/6

(like supporting 30 pound) so it would be easier for your arms to support the weight.

 

But moon mass is not 1/6 of earth mass.

Moon mass is 1/81 of earth mass.

[Johnny5]

Johnny you say

 

 

but my astronomy handbook says

 

[math] M_{moon} = \frac{1}{81} M_{earth} [/math]

 

how did you get the 1/3?

 

 

 

I just needed a number' date=' I pulled it out of thin air. I'm not really talking about the real moon, and the real earth, but it helps to visualize them though.

 

As was correctly pointed out, my weight on the moon depends upon the mass of the moon, and the density (through R). So that really I should have written this:

 

[math'] W_{moon} = \frac{1}{3} W_{earth} [/math]

 

That is the fact that I would discover when I did my handstand. If the acceleration of gravity at the surface of the moon, were identical to the acceleration of gravity at the surface of the earth, then what I wrote would have been correct. But the acceleration of the moon's gravity is 1/6 that of the earth.

 

At any rate, I wasn't talking about the real moon, and real earth, the question centers more around... what am I pushing against, when I do the handstand. What I was really interested in was, does the vacuum offer resistance or not.

[Martin]

I think it would be great to talk about the real moon.

 

for example (in case anyone else is reading the thread)

 

Would it make any difference, on the real moon, if you were at the moon's north pole versus at its equator?

 

How do you figure out if it would be noticeably easier or harder to support your weight?

[Martin]

Suppose I want to measure the mass of the moon in the following way.

 

I want to do a handstand on the moon.

 

Suppose that the moon is a perfect sphere' date=' and that it has a uniform density.

 

Will it matter where on the moon I do the handstand?

 

...[/quote']

 

I think this is a great way to start a thread.

 

How, in fact, would a person determine the mass of the moon this way?

 

Suppose the moon is a perfect sphere with uniform density, like J5 says.

 

Suppose a person is allowed to measure the force of his weight, standing anywhere on the moon he chooses.

 

A. does it make any difference where he stands?

 

B. suppose we build a box for him on the moon containing earth-like air so that he can breathe and do handstands comfortably without a spacesuit, and so the buoyancy effect will not be important either. weighing oneself always has the small buoyancy effect but we can neglect it because now its the same air being displaced.

suppose he finds his weight on earth is 180 poundforce and on moon is 30 poundforce.

How can he figure out the mass of the moon from this?

[Johnny5]

I think this is a great way to start a thread.

 

How' date=' in fact, would a person determine the mass of the moon this way?

[/quote']

 

I don't think you can determine the mass this way, unless you also measure the acceleration due to gravity, unless I am forgetting something.

 

The force that you feel, is your weight. Your weight is equal to your mass M, times the acceleration of gravity of the moon which I will denote as g_m.

 

In other words, your weight on the moon is given by the following:

 

W_m = M g_m

 

And your weight on earth is given by:

 

W_e = M g_e

 

Your weight on earth is not the same as your weight upon the moon, you are heavier on earth. So W_e is greater than W_m.

 

Suppose you do the handstand, and discover that

 

W_e = 3W_m

 

Using the definitions above, it follows that:

 

3W_m = M g_e

 

And since W_m = M g_m, it follows that:

 

 

3(M g_m) = M g_e

 

Your mass in non-zero, so we can divide both sides of the equation above by your mass M to obtain:

 

3 g_m = g_e

 

To go any further, would require the use of a formula for the force due to gravity, for example we could use Newton's formula.

 

Let the force of earths gravity on us be given by the following formula:

 

[math] F_e = \frac{G M_e M}{R_e^2} [/math]

 

Where M_e is the mass of the earth, and R_e is the radius of the earth.

 

Let the force of the moon's gravity on us be given by the following formula:

 

[math] F_m = \frac{G M_m M}{R_m^2} [/math]

 

Where M_m is the mass of the moon (which is what we want to know,) and R_m is the radius of the moon.

 

The force of gravity of earth on us, is equal to our weight on earth, in other words:

 

[math] F_e = W_e = Mg_e = \frac{G M_e M}{R_e^2} [/math]

 

From the equation above, we can infer that:

 

[math] g_e = \frac{G M_e }{R_e^2} [/math]

 

Doing the same thing for our weight on the moon reveals that:

 

[math] g_m = \frac{G M_m }{R_m^2} [/math]

 

In a previous line of work we concluded that 3g_m = g_e

 

Therefore:

 

[math] g_e = 3g_m = \frac{3G M_m }{R_m^2} [/math]

 

From which it follows that:

 

[math] \frac{G M_e }{R_e^2} = \frac{3G M_m }{R_m^2} [/math]

 

G is the Newtonian gravitational constant, which is non-zero, so we can divide both sides by G to obtain:

 

[math] \frac{M_e }{R_e^2} = \frac{3M_m }{R_m^2} [/math]

 

Mathematically adjusting things, the following statement is true:

 

[math] \frac{M_e }{M_m} = 3 (\frac{R_e}{R_m})^2 [/math]

 

So you can't figure out the proper mass ratio from just your weight ratio. You would also need to know the radii of each body.

[Martin]

So you can't figure out the proper mass ratio from just your weight ratio. You would also need to know the radii of each body.

 

Right!

 

now suppose we find that your weight force standing on moon is 1/6 of what it is on earth and that

the moon's radius is 27 percent of the earth's

from this, what do you say the moon's mass is compared with earth?

[J.C.MacSwell]

If the moon is a perfect sphere it's probably not spinning. If it's spinning anyway there should be an effective weight difference.

 

If it's orbiting the earth there should be tidal effects making you lighter on the point closest to Earth and also lighter (by almost as much) on the point farthest from Earth.

[swansont]

If the moon is a perfect sphere it's probably not spinning. If it's spinning anyway there should be an effective weight difference.

 

The moon is spinning, albeit very slowly.

[swansont]

I guess it would. Are you saying that when I do my handstand on earth' date=' that I am also pushing slightly on our atmosphere (on the other side of the world?

 

 

 

Ok suppose that formula is correct.

 

[math'] \vec F = \frac{d\vec P}{dt} = 1/c \frac{dE}{dt} [/math]

 

Pressure is force per unit area.

 

So when I do my handstand on the moon, I push the moon away slightly from me. So the area in question is half of the surface area of the moon?

 

When you do the handstand on earth, you are lifted by the weight of atmosphere you displace (Archimedes principle)

 

For photon pressure, use power/area (aka power density) instead of power and you get pressure. The area depends on what the force is exerted on, and what the source is. In your example the area is your area, since you are the one doing the handstand.

[J.C.MacSwell]

The moon is spinning, albeit very slowly.

 

I was referring to a hypothetical moon. Spin (or tidal effects) would tend to make it non-spherical.

[Johnny5]

Right!

 

now suppose we find that your weight force standing on moon is 1/6 of what it is on earth and that

the moon's radius is 27 percent of the earth's

from this' date=' what do you say the moon's mass is compared with earth?[/quote']

 

 

Well using the analysis I previously gave, it would be the case that:

 

[math] \frac{M_e }{M_m} = 6 (\frac{R_e}{R_m})^2 [/math]

 

Because the measurment would have revealed...

 

[math] W_m = \frac{1}{6} W_e [/math]

 

In the case of our real moon, 1/6 is more accurate than 1/3, so let us use 1/6 instead of 1/3.

 

Well, disregarding how we obtained the knowledge that the radius of the moon is 27% of the radius of earth, let us suppose we do know it. So if we divide the radius of the earth into 100 segments of equal length, then the radius of the moon is equal to 27 of them.

 

Suppose additionally that we know the following is true:

 

[math] R_m = \frac{27}{100} R_e [/math]

 

We can now go a littler further in the mathematical analysis of the problem:

 

 

[math] \frac{M_e }{M_m} = 6 (\frac{R_e}{R_m})^2 = 6 (\frac{R_e}{\frac{27}{100} R_e })^2 [/math]

 

From which it follows that:

 

[math] \frac{M_e }{M_m} = 6 (\frac{100}{27})^2 [/math]

 

100² = 10000

 

27² = 729

 

10000/729 = 13.71

 

(13.71)(6) = 82.26

 

Therefore:

 

[math] \frac{M_e }{M_m} = 82.26 [/math]

 

Therefore:

 

[math] M_{moon} = \frac{1}{82.26} M_{earth}[/math]

 

 

Johnny you say

 

 

[math] M_{moon} = \frac{1}{3} M_{earth} [/math]

 

but my astronomy handbook says

 

[math] M_{moon} = \frac{1}{81} M_{earth} [/math]

 

 

 

It appears that your astronomy book is correct, if indeed the radius of the moon is really 27% the radius of the earth. The earth is an oblate spheroid actually, and therefore doesn't have a single 'radius' to speak of. The real earth buldges at the equator, because it is spinning (as JC and swansont both pointed out).

[Martin]

...

 

We can now go a littler further in the mathematical analysis of the problem:

 

 

[math] \frac{M_e }{M_m} = 6 (\frac{R_e}{R_m})^2 = 6 (\frac{R_e}{\frac{27}{100} R_e })^2 [/math]

 

From which it follows that:

 

[math] \frac{M_e }{M_m} = 6 (\frac{100}{27})^2 [/math]

 

100² = 10000

 

27² = 729

 

10000/729 = 13.71

 

(13.71)(6) = 82.26

 

 

Cheers! it's great to meet someone who thinks with both sides of head----use imagination and explore out-of-ordinary questions (on the one hand) but on the other hand ready and able to calculate!

 

if we would only make the 1/6 number slightly more accurate and use 27.2 percent instead of plain 27 percent then we could get that 82 to be a more accurate 81 something. but I am very happy you can get that close with simple data.

 

I guess earth and moon are revolving around a common point which is not exactly the center of the earth. how far is that point (in miles or kilometers or fractions of the earths radius or whatever) from the center of the earth.

 

could it be for example in Pittsburgh Pennsylvania or more precisely could Pittsburgh at this point be passing thru this point?

 

hmmmm I dont want to make a pest of myself, dont worry about it unless you are curious.

[J.C.MacSwell]

 

I guess earth and moon are revolving around a common point which is not exactly the center of the earth. how far is that point (in miles or kilometers or fractions of the earths radius or whatever) from the center of the earth.

 

.

 

Should be very close to the centre of gravity of the earth/moon system.

[Johnny5]

I guess earth and moon are revolving around a common point which is not exactly the center of the earth. how far is that point (in miles or kilometers or fractions of the earths radius or whatever) from the center of the earth.

 

could it be for example in Pittsburgh Pennsylvania or more precisely could Pittsburgh at this point be passing thru this point?

 

hmmmm I dont want to make a pest of myself' date=' dont worry about it unless you are curious.

 

[/quote']

 

I am curious, but what you say ignores local gravitational field effects of all other bodies in the universe, which matters according to Mach, they contribute to the local gravitational potential, which displaces the CM of the earth-moon system from where it would be if there were no other material in the universe, yes? At any rate...

 

You say that the earth and moon are orbiting their common center of mass. I am not going to ask you to justify that, but you ask, "How far is that CM point from the center of the earth?"

 

Supposing what you say is true...

 

If the earth had an infinite mass, then the center of mass of the two body system would be located at the center of the earth (in fact the center of mass of the universe would be located at the center of the earth). But the earth does not have an infinite mass. The ratio of the distances of the centers of inertia of the earth and moon from the CM of the earth/moon system is equivalent to the ratio of their inertial masses. That is:

 

[math] \frac{M_{earth}}{M_{moon}} = \frac{R_{mc}}{R_{ec}} [/math]

 

Where R_mc is the distance from the center of inertia of the moon to the center of inertia of the earth-moon system, and R_ec is the distance from the center of inertia of the earth to the center of the earth-moon system.

 

We previously found that:

 

[math] \frac{M_{earth}}{M_{moon}} = 81 [/math]

 

Where I am using the values found in your astronomy book. So if the formula above is true, that means that the following is true:

 

[math] 81 = \frac{R_{mc}}{R_{ec}} [/math]

 

Which means that the following is true:

 

[math] R_{mc} = 81 R_{ec} [/math]

 

But in order to answer your question, I need the two distances above, in terms of earth radial units. All I know at this point, is that the center of inertia of earth is 81 times closer to the center of mass of the earth-moon system than the center of inertia of the moon is. I do not know if the CM is located at pittsburg or inside the surface of the earth, or outside the surface of the earth. In order to answer that, I would need to know how far apart the CM of earth and the CM of the moon are, say in meters, or in earth radial units or whatever.

[Johnny5]

Should be very close to the centre of gravity of the earth/moon system.

 

That location is the center of gravity of the earth-moon system. I think he wants to know if that location is inside the earth, at the surface of the earth (possibly Pittsburg PA), or outside the surface of the earth. He wants to know where that location is, in relation to the CM of the earth, and the CM of the moon.

[Martin]

...Which means that the following is true:

 

[math] R_{mc} = 81 R_{ec} [/math]

 

But in order to answer your question' date=' I need the two distances above, in terms of earth radial units. All I know at this point, is that the earth is 81 times closer to the center of mass of the earth-moon system than the moon is. I do not know if the CM is located at pittsburg or inside the surface of the earth, or outside the surface of the earth. In order to answer that, I would need to know how far apart the CM of earth and the CM of the moon are, say in meters, or in earth radial units or whatever.[/quote']

 

you are great. people so rarely calculate.

 

around 130 BC a greek named Hipparchus measured the distance to the moon in earth radii as 60.

 

amazingly accurate, he did it by drawing pictures and timing the duration of eclipses probably with some kindof klypshydra (spelling?) or waterclock. what a guy. had to be very imaginative to figure out how to measure distance to moon with such primitive instruments as were then available.

there is a satellite observatory named after him. I think the europeans (ESA) put it up.

 

wait awhile, I will get the earthmoon distance (center to center) in meters after i have my coffee.

I know it is about 1 and 1/4 or 1 and 1/3 light second, that is how long it takes signals but I will get it in kilometers

[Johnny5]

 

Would it make any difference' date=' on the real moon, if you were at the moon's north pole versus at its equator?

 

[/quote']

 

Yes, it would because the real earth and real moon are not perfectly spherical, the earth is an oblate spheroid (a sphere which got squashed at the poles), so points on the equator are further away from the center of mass of the earth than the pole points are from the earth's CM.

 

So using Newtonian Gravity formula, the force of earth's gravity on you is less at the equator, than it would be if you were at either of the poles, because the distance from your CM to earth's CM is greater when you are at the equator, then when you are at say the North pole.

 

Since the moon is spinning too, albeit very slowly as was pointed out, the moon also buldges at its equator. Thus, if you did your handstand on the equator of the moon, it would be easier than if you did it at either of the moon's poles.

[Martin]

...

 

Since the moon is spinning too' date=' albeit very slowly as was pointed out, the moon also buldges at its equator. Thus, if you did your handstand on the equator of the moon, it would be easier than if you did it at either of the moon's poles.[/quote']

 

right.

sorry I am late getting back with the average earthmoon distance

my handbook says 384,400 km

the orbit isnt exactly circular and the distance ranges from 356,000 to 407,000 km it says. but I habitually ignore such imperfections in nature.

 

anyway Hipparchus calculated that the distance earth to moon center to center was 30 times earth diameter which we know is about 12,740 km so multiply by 30 to get the old greek estimate: 382,200 km.

So close to the right answer!

 

the way I remember the date for Hipparchus is that it was right around 137 BC (when he found the distance to the moon) and 137 is one of the few numbers I am confident of always remembering.

[Martin]

Since the moon is spinning too' date=' albeit very slowly as was pointed out, the moon also buldges at its equator. Thus, if you did your handstand on the equator of the moon, it would be easier than if you did it at either of the moon's poles.[/quote']

 

yes it rotates (relative to the distant stars) about once every 27.3 days

so there would be a slight centrifugal force lifting you up

which would be some tiny fraction of the weight you would feel without the rotation

 

what about that centrifugal force for a person standing on the earth's equator?

 

my book says the earth's equatorial radius is 6378 km.

 

if you were standing at equator, by what fraction would your weight-force be diminished by the centrifugal business?

 

my book says the net gee at the equator is 9.78 meters per second^2

(this takes into account both the bulge, being farther from center, and also the centrif effect, so 9.78 is the acceleration that the person actually feels at sealevel at equator) what I want is a comparison. how big is the centrif effect compared to that

 

the centrif effect on the moon would be extremely less than that, of course because slower rotation and smaller radius.

 

do you know the V^2/R formula? have no idea of where you are age and schoolwise.

[Johnny5]

right.

sorry I am late getting back with the average earthmoon distance

my handbook says 384' date='400 km

the orbit isnt exactly circular and the distance ranges from 356,000 to 407,000 km it says. but I habitually ignore such imperfections in nature.

 

anyway Hipparchus calculated that the distance earth to moon center to center was 30 times earth diameter which we know is about 12,740 km so multiply by 30 to get the old greek estimate: 382,200 km.

So close to the right answer!

 

the way I remember the date for Hipparchus is that it was right around 137 BC (when he found the distance to the moon) and 137 is one of the few numbers I am confident of always remembering.[/quote']

 

Modern answer for average center to center distance of earth-moon system: 384,400 kilometers

 

Ok.

 

You say that Hipparchus of Rhodes managed to figure out that the distance from the center of the earth to the center of the moon is 30 times the diameter of the earth?? It seems amazing that he could approximate it at all. Is the method he used the one discussed in the Hipparchus link i just gave? I haven't looked at it carefully yet, it looks rather complicated. How do you use that method to conclude that the center to center distance is 60 earth radii? Did they know the distance from Alexandria to Syene back in those days?

 

137.036 BC

 

Regards