"Science vs. Religion: What Scientists Really Think"

humble brother wrote:

<quoted text>
So let me get this straight. You believe that adding 5.37498 * 10^25 kg of mass to Earth (so multiplying Earth by 10) will produce the same exact change in the gravitational force between the 1000 kg object as would adding 9000 kg to that tiny object?
Are you sure about that?

Yes.

You think adding 5.37498 * 10^25 kg to Earth will do the same as to adding 9000 kg to the thing in orbit?
Really?

Having the same force does not mean having the same effects. Remember that F=ma, so if the force stays the same and the mass increases by a factor of 10, the acceleration *decreases* by a factor of 10.

Did you fail to notice that I have the sum of both masses squared there also?

Yes, I did.*laughing uncontrollably*.

And as I said, I'm only in the initial stage with that one. Haven't really run much simulations yet.

Hilarious.

Hey, HB. Care to answer the question from post #10350?

polymath257 wrote:

For moving masses, a better approximation is E=mc^2 +1/2 mv^2 if the velocity is small compared to that of light. Since v<<c for everyday objects, the dominant term is mc^2. of course, it is simply an approximation and the *exact* value is given by the longer equation which is correct even for relativistic velocities.

Was there first actually only E=mc^2 presented?

polymath257 wrote:

it is the standard energy equation in relativity.

Yes, the glue an tape one.

polymath257 wrote:

No, that is not at all what I am saying. For things with mass, the m in the equation is non-zero. For things that are moving, the p (momentum) in the equation is non-zero.

So, what are the things that are not moving?

polymath257 wrote:

Again, you are attacking your lack of understanding, not the equation itself.

You can believe that if you wish :)
I don't mind.

humble brother wrote:

<quoted text>
Do you represent mass somehow differently than you represent weight?

Yes, of course. Technically, weight is a force and is given in Newtons or pounds. Mass is, well, a mass, and is given in kilograms or slugs.

However, it is common to stadardize things and have a pound of mass be the mass that has a weight of one pound at sea level on the earth on the equator.

Can you not for example express the mass of Jupiter in kilograms?

Yes, but not the weight.

Do you not calculate the masses of planets from their paths in the gravitational fields?

Actually, we calculate the mass of a planet by looking at the motion of *other* things in its gravitational field. For example, we can determine the mass of Jupiter by looking at the motions of its moons and using Kepler's laws. it is much more difficult to determine the mass of a planet that has no moons for exactly this reason. For such cases, we have to measure the much smaller effect the planet has on other planets. That is why, for exmaple, the mass of Pluto was very uncertain until we discovered Chiron.

humble brother wrote:

<quoted text>
Was there first actually only E=mc^2 presented?

I believe Einstein's 1905 paper had the full equation. I could look it up.

So, what are the things that are not moving?

You can regard something as being at rest in its own reference frame (assuming it is un-accelerated). The euqation holds in all reference frames.

You can believe that if you wish :)
I don't mind.

Well, it is pretty obvious from your comments that you don't understand the equations you are playing with.

polymath257 wrote:

Hey, HB. Care to answer the question from post #10350?

In the scenarios both the masses will end up orbiting but their orbits will be different.

Care to answer the question in msg #10353?

polymath257 wrote:

<quoted text>Yes.
<quoted text>
Having the same force does not mean having the same effects. Remember that F=ma, so if the force stays the same and the mass increases by a factor of 10, the acceleration *decreases* by a factor of 10.
<quoted text>
Yes, I did.*laughing uncontrollably*.
<quoted text>
Hilarious.

Are you sure you don't want to rethink what you have written here?:D :D :D

Seriously, you don't see the problem there?

humble brother wrote:

<quoted text>
Was there first actually only E=mc^2 presented?

I did a quick search. Einstein's first 1905 paper actually has neither equation. Instead, it has an equation for the kinetic energy of an electron that is equivalent to the 'tape and glue' equation you seem to dislike.

http://www.fourmilab.ch/etexts/einstein/specr...
(section 10 has the equation for kinetic energy, which he denotes by W). Using E=W+mc^2 and a little bit of algebra gives E=sqrt(m^2c^4 +p^2 c^2).

The second paper, however, he determines how much the relativistic mass changes upon emission of light. he determines the *change* in mass corresponds to the energy of the light emitted divided by c^2.

http://www.fourmilab.ch/etexts/einstein/E_mc2...

humble brother wrote:

<quoted text>
Lets say that the two objects (2 tons and 100 tons in calculated weight) have deviated from each other 100 kilometers after traveling a lightyear and initially one was following the other some ~50 meters behind. Would you still say that they're on the same path?
Let's assume they don't encounter any other objects that would separate them even more. After 100 light years they would be 10000 kilometers apart. Would you still say that they're on the same path?

Depends on the accuaracy I need in my situation. For both, I would consider them the same path over the course of the first few million miles.

polymath257 wrote:

<quoted text>
I did a quick search. Einstein's first 1905 paper actually has neither equation. Instead, it has an equation for the kinetic energy of an electron that is equivalent to the 'tape and glue' equation you seem to dislike.
http://www.fourmilab.ch/etexts/einstein/specr...
(section 10 has the equation for kinetic energy, which he denotes by W). Using E=W+mc^2 and a little bit of algebra gives E=sqrt(m^2c^4 +p^2 c^2).
The second paper, however, he determines how much the relativistic mass changes upon emission of light. he determines the *change* in mass corresponds to the energy of the light emitted divided by c^2.
http://www.fourmilab.ch/etexts/einstein/E_mc2...

have you read the wonderful book, Einstein's Dreams? I think you might like it. it is fine literature, and interesting speculation.

humble brother wrote:

<quoted text>
In the scenarios both the masses will end up orbiting but their orbits will be different.

Will both scenarios give the same results?

Please be more specific, in scenario 1, which answer is closest to being true? In scenario 2, which answer is closest to being true? If the difference between an inch and a mile is troubling you, it is fine to replace them by a foot and 10 miles.

humble brother wrote:

<quoted text>
Are you sure you don't want to rethink what you have written here?:D :D :D
Seriously, you don't see the problem there?

Oh, please elaborate. I need the laugh.

polymath257 wrote:

Yes, of course. Technically, weight is a force and is given in Newtons or pounds. Mass is, well, a mass, and is given in kilograms or slugs.

However, it is common to stadardize things and have a pound of mass be the mass that has a weight of one pound at sea level on the earth on the equator.

You've tried to evade the question. The point is that you can represent the weight of any planet in kilograms. This is because it is all relative to the same thing which is gravity and all is rooted in weighing things on Earth. That has been extended to cover things outside Earth. They're still relative to the same thing, which is just called gravity.

polymath257 wrote:

Yes, but not the weight.

Of course. Jupiter is not on Earth.

polymath257 wrote:

Actually, we calculate the mass of a planet by looking at the motion of *other* things in its gravitational field. For example, we can determine the mass of Jupiter by looking at the motions of its moons and using Kepler's laws. it is much more difficult to determine the mass of a planet that has no moons for exactly this reason. For such cases, we have to measure the much smaller effect the planet has on other planets. That is why, for exmaple, the mass of Pluto was very uncertain until we discovered Chiron.

So if you don't know the mass of Jupiter or any of its moons, how can you calculate the mass of Jupiter from its moons' motion?:)

humble brother wrote:

<quoted text>
Lets say that the two objects (2 tons and 100 tons in calculated weight) have deviated from each other 100 kilometers after traveling a lightyear and initially one was following the other some ~50 meters behind. Would you still say that they're on the same path?
Let's assume they don't encounter any other objects that would separate them even more. After 100 light years they would be 10000 kilometers apart. Would you still say that they're on the same path?

If I am computing distances on the order of light years, I would consider a difference of 50 meters to be irrelevant. Same for a distance of 10000 milometers over a distance of 100 light years. So, yes, to the desired accuracy, I would consider them to be the same path.

polymath257 wrote:

Will both scenarios give the same results?
Please be more specific, in scenario 1, which answer is closest to being true? In scenario 2, which answer is closest to being true? If the difference between an inch and a mile is troubling you, it is fine to replace them by a foot and 10 miles.

Both will end up in orbits in both scenarios and the orbits will be different in both scenarios. I'm not doing any calculations now. I will later create an simulation environment for this kind of things so I can easily create graphical illustrations.

humble brother wrote:

<quoted text>
You've tried to evade the question. The point is that you can represent the weight of any planet in kilograms.

No. You can represent the *mass* of a planet in kilograms. Tecnically, the planets are weightless since they are in freefall (although tidal effects are relevant for all planets).

This is because it is all relative to the same thing which is gravity and all is rooted in weighing things on Earth. That has been extended to cover things outside Earth. They're still relative to the same thing, which is just called gravity.

The weight of a hammer on the earth is different than the weight of that same hammer on the moon. The masses are the same. For that matter, the weight of the hammer will be different if it at the top of a mountain of at sea level. The mass will be the same in both cases.

So if you don't know the mass of Jupiter or any of its moons, how can you calculate the mass of Jupiter from its moons' motion?:)

Like I said, Kepler's laws. It really isn't a difficult thing. G*MJ*T^2=4*pi^2*R^3. Here, MJ is the mass of Jupiter, T is the time it takes for a moon to complete an orbit and R is the radius of that orbit.

humble brother wrote:

<quoted text>
Both will end up in orbits in both scenarios and the orbits will be different in both scenarios. I'm not doing any calculations now. I will later create an simulation environment for this kind of things so I can easily create graphical illustrations.

Well, the point is that you are wrong and dramatically wrong. In the scenario where the two have the same momentum (so the smaller has higher velocity), the small object will have enough velocity to leave earth's orbit. The small one will NOT be in orbit.

In the scenario where the two have the same velocity but different momenta, the difference will be small than an inch over the course of 100 orbits.

polymath257 wrote:

Oh, please elaborate. I need the laugh.

You're talking about changing the mass of one of the objects in gravitational pull, and you say it does not affect the acceleration. You already have the evidence that it does.

Take the hammer. Drop it on Earth and trop it on the moon. You have effectively simulated changing the mass of one of the objects while keeping the other the same.

Changing of one of the masses results in different acceleration. The amount of change relative to the sum of the masses dictates how much the gravitational acceleration changes.

polymath257 wrote:

You can regard something as being at rest in its own reference frame (assuming it is un-accelerated). The euqation holds in all reference frames.

Are photons not at rest in their own reference frames?
Is there something in the universe that is un-accelerated?:D

polymath257 wrote:

Well, it is pretty obvious from your comments that you don't understand the equations you are playing with.

We'll see about that soon enough.

Polymath, it is obvious to anyone who cares to look that you are arguing with someone without even the most basic understanding of elementary physics. It certainly is highly amusing to read HB's posts, they are beyond ridiculous.

Has it not occured to him that if his ideas (if I may stretch the definition of that word almost to breaking point) of how gravity and basic orbital mechanics work then NASA would have had a hard job sending all their spacecraft to all the places they have done so?

Don't you feel like you're arguing with someone with an understanding of physics like that of a truculent three year old?