Can someone please help me with this?

[fireballxl5]

Also, another thing that gets me about the whole moon scenario, is that gravity is relative to mass hence why there is less gravity on the moon. Smaller the planet, less gravity (So we're told) But then we're told that gravity comes from within the Earth, so by that formula wouldn't the gravitational pull be much stronger on the surface of the moon as you would be closer to the gravitational centre?

Is it to do with the comparative density and speed of rotation?

[Retrograde]

May 8, 2008 7:10:19 GMT -4 fireballxl5 said:

Also, another thing that gets me about the whole moon scenario, is that gravity is relative to mass hence why there is less gravity on the moon. Smaller the planet, less gravity (So we're told) But then we're told that gravity comes from within the Earth, so by that formula wouldn't the gravitational pull be much stronger on the surface of the moon as you would be closer to the gravitational centre?

Is it to do with the comparative density and speed of rotation?

The force of gravity is inversely proportional to the square of the distance to the center of mass (for a spherical object, we can treat all mass as if it were located at the center - for other objects, that won't be the case). So if the radius of the moon were 1/4 of the radius of the earth, then the force of gravity on the moon would be 16 times stronger, if it had the same mass as the earth.

However, the moon doesn't have the same mass as the earth. If it had the same density as the earth (and it doesn't, but the difference isn't huge), then the mass would be proportional to the cube of the radius. For example, double the size of the moon (keeping its density the same), and its mass would be multiplied by eight.

So there are two effects. One causes the force of gravity to increase proportionately to the cube of the radius; the other causes the force to decrease proportionately to the square of the radius. The combined effect is (holding density fixed), the force of gravity on the surface of a planet or moon is proportional to its radius. Earth is approximately four times the size of the moon, so its gravity ought to be four times stronger on the surface; the difference is larger because the earth is denser than the moon.

[fireballxl5]

Oh man now ive got to get my head round that lol.

Thanks mate i understand, just about.

[BertL]

We have some nice equations for that. From Newton:

F = (G * M * m) / r²

In which:
F = the force of gravity
G = gravitational constant
M = mass of object 1 (in our case, Earth or the Moon)
m = mass of object 2 (a person, or a block or something; this will be divided away soon)
r = distance between center of mass of object 1 and object 2 (usually the radius of Earth or the Moon)

As you can see, the force of gravity (and, if you combine it with F = m * a, the acceleration) of an object on a certain planet not only depends on the radius, but also on the mass. For example, if you had a planet with a mass four times as small as Earth's, but with a radius only two times as small, the F (or a) would be pretty much the same. (Because your M is .25x as large, but your r² is also .25 times as large; it "divides each other away", more or less. Of course, this is only if you make some big assumptions.

I hope I don't sound too incoherent. I had to manually translate what I learn at school.

[fireballxl5]

Well i quoted via an E mail what i was told and heres what i got back

The force of gravity is inversely proportional to the square of the distance to the center of mass (for a spherical object, we can treat all mass as if it were located at the center for other objects that won't be the case). So if the radius of the moon were 1/4 of the radius of the earth, then the force of gravity on the moon would be 16 times stronger if it had the same mass as the earth. However the moon doesn't have the same mass as the earth. If it had the same density as the earth (and it doesn't but the difference isn't huge), then the mass would be proportional to the cube of the radius. For example double the size of the moon (keeping its density the same), and its mass would be multiplied by eight.

So there are two effects. One causes the force of gravity to increase proportionately to the cube of the radius; the other causes the force to decrease proportionately to the square of the radius. The combined effect is (density fixed) the force of gravity on the surface of a planet or moon is proportional to its radius. Earth is approximately four times the size of the moon, so its gravity ought to be four times stronger on the surface; the difference is larger because the earth is denser than the moon.

That's just a hypothesis, with a bit of guesswork.

If it had the same density as the earth (and it doesn't but the difference isn't huge)

Earth is approximately four times the size of the moon, so its gravity ought to be four times stronger on the surface; the difference is larger because the earth is denser than the moon.

The gravitational pull on the Earth is said to be six times that of the Moon, so where they squeezed in the other two times is anyones guess, especially after the difference isn't huge.

People generally believe that the moon has one sixth the gravity of the earth because they are told that the density of the moon is 3.3 gr/cm3 and that the earth's gravity is 5.5 gr/cm3. And with those densities and the radius of the moon and the earth, the calculation is made:

Fg=G x m1m2/r2

The assumption that the density of the moon is 3.3 gr/cm3 is based on:

1. Earth density is 5.5 gr/cm3 and the gravity of the moon is one sixth that of earth
2. Observations of asteroids and the Earth's polar axis
3. Angular momentum
4. Soil samples from the moon
5. Keplers Third Law
6. Shockwave measurement

In fact, we have no idea what the density of the moon is and we are just assuming that the density is 3.3 gr/cm3 because that is the only explanation apparent to us at this time. It seems to make sense.

[fireballxl5]

Looks like the person is quoting this

lucianarchy.proboards21.com/index.cgi?board=johnlear&action=display&thread=1633&page=1

BTW Khrushchev's Other Shoe, i did give him a link to here so he knew it was your fine work.

Bertls you came over just fine.

[Ginnie]

I saw a "Zero Gravity Lounge Chair" at a local department store.
Does that mean it has no mass? ;D

I suspect that if most of the moon was sold iron, it's gravitational force would be much greater than on earth? Let's say its core started just 10 km from its surface - because of the increase in density, it's mass would be huge and you'd be much closer to its core than on earth.

Remember, I am a simple man.

[Retrograde]

May 8, 2008 8:35:29 GMT -4 fireballxl5 said:

Looks like the person is quoting this

lucianarchy.proboards21.com/index.cgi?board=johnlear&action=display&thread=1633&page=1

BTW Khrushchev's Other Shoe, i did give him a link to here so he knew it was your fine work.

OK, thanks, but I would just as soon not be brought to the attention of someone who says a smaller planet should have higher surface gravity, and then says that Newton's law of gravitation is "just a hypothesis, with a bit of guesswork."

[Retrograde]

May 8, 2008 19:37:39 GMT -4 Ginnie said:

I saw a "Zero Gravity Lounge Chair" at a local department store.
Does that mean it has no mass? ;D

I suspect that if most of the moon was sold iron, it's gravitational force would be much greater than on earth? Let's say its core started just 10 km from its surface - because of the increase in density, it's mass would be huge and you'd be much closer to its core than on earth.

Remember, I am a simple man.

Not bothering to look up exact numbers, but if the earth's radius were four times the moon's radius, then the density of the moon would have to be four times as large as the density of the earth for them to have the same surface gravity. A uniform spherical object can be treated as having all the mass at a point at the center. So with one fourth of earth's radius, the moon would have 1/64th of the earth's volume, and with four times the density, it would have 1/16th of the mass. Then standing on the surface of the moon, you would be four times closer to the center of mass than you would be on the surface of the earth, so the gravity would be 16 times stronger, per unit of mass, than it would be on the earth. So 16 times as much gravity per unit of mass, times 1/16th as much mass, means the same surface gravity.

But remember, "That's just a hypothesis, with a bit of guesswork."

I'm not sure how dense earth is, but for the moon to have four times the density of earth, I'm not sure iron would do it. Maybe uranium

[Ginnie]

May 8, 2008 8:35:29 GMT -4 fireballxl5 said:

Looks like the person is quoting this

lucianarchy.proboards21.com/index.cgi?board=johnlear&action=display&thread=1633&page=1

BTW Khrushchev's Other Shoe, i did give him a link to here so he knew it was your fine work.

Bertls you came over just fine.

There is another theory that is not generally accepted but its occurrence is far more likely and that is that our moon was placed in orbit around our earth by a civilization very far advanced from our own.
Uh, I've been trying to read as many books as I can on Spaceflight, Satellites etc. but haven't encountered this theory in any of them. Can someone direct me to a manual, book, essay, lecture or journal where the above theory is mentioned?

[Tanalia]

There is no "guesswork" involved.

From Newton's law of universal gravitation F = G m₁ m₂ / r², combined with his Second law of motion F = ma, we get a relatively simple formula for acceleration due to gravity a = G m / r². Since the surface acceleration a and radius r of Earth are both known (nothing more than high-school math and science), and G is a constant well defined by measurements from know masses, the mass of Earth is easily calculated.

From Kepler's third law of planetary motion (P / 2p)² = a³ / (G (M +m)) , given the period P and the semi-major axis a of the Moon's orbit, plus the Earth mass from before, we can directly compute the mass of the Moon.

Once the mass for each body is know, simply divide by the volume (trivial for a sphere or even an oblate spheroid) to get the average density.

[Bob B.]

May 8, 2008 19:59:16 GMT -4 Retrograde said:

Not bothering to look up exact numbers...

Suppose that we do look up the numbers.

Earth's mean radius is 6,371 km and the Moon's radius is 1,738 km. If we assume the surface gravity of the Earth and Moon are equal, than we can use the equation Bert posted earlier to find the ratio of Earth mass to Moon mass,

F = (G * M * m) / r²

(G * Me * m) / 6371² = (G * Mm * m) / 1738²

Me / Mm = 6371² / 1738² = 13.44

Earth's mass is 5.974 x 10²⁴ kg. Thus, to have the same surface gravity as Earth, the Moon's mass must be:

Mm = 5.974 x 10²⁴ / 13.44 = 4.445 x 10²³ kg

The volume of the Moon is

Vm = 4/3 * Pi * 1738000³ = 2.199 x 10¹⁹ m³

Therefore, the required lunar density is

Density = 4.445 x 10²³ kg / 2.199 x 10¹⁹ m³ = 20,200 kg/m³

which is a little denser than Plutonium.


If the Moon where solid iron at a density of 7,870 kg/m³, it would have a surface gravity about 0.39 times Earth.

[Bob B.]

May 8, 2008 22:04:28 GMT -4 Tanalia said:

From Kepler's third law of planetary motion (P / 2p)² = a³ / (G (M +m)) , given the period P and the semi-major axis a of the Moon's orbit, plus the Earth mass from before, we can directly compute the mass of the Moon.

You have to be a little bit careful here. If you calculate the mass of the Moon from its observed orbital period, you'll under estimate the Moon's mass by about 22%. This is because the Sun perturbs the Moon's orbit and lengthens its period by about 53 minutes. This may not seem like much but it has a significant effect on the calculations.