Newton's gravitational constant.

[dstebbins]

My high school physics teacher says that, when Newton invented the laws of universal gravitation, he devised a constant to be implemented when considering the gravitational relationship between two spacial bodies. According to my teacher, it was using this constant that was used to "find" the supermassive black hole in the center of the galaxy (by studying the behavior of the stars around it). He says that this constant's unit is kg^2/m^2 (kilgrams squared per meters squared); however, that's as far as his knowledge goes. He remembers virtually nothing about this constant, not even the numerical value of it, so I'm turning to you guys.

 

1. What is the numerical value of this constant? Is it in decimal form or fraction form?

 

2. How did Newton come up with this constant? What was his mathematical reasoning?

 

3. How is this constant utilized in things like the discovery of a supermassive black hole?

 

Thanks ahead of time.

[Klaynos]

I shall provide a link because it goes into far more depth than I can possibly do:

 

http://en.wikipedia.org/wiki/Gravitational_constant

 

 

1. 6.67*10^-11m^3kg-1s-2 (Nm^2kg^-2)

 

2. I can't remember the derivation offhand, any good low level university text book should have one in it... If I remember tomorrow I shall have a look.

 

3. The GR theory of gravity is used more than the newtonian one, which also uses the constant. But the way it works is you put the experimental evidence into some equations and the results of these says that there should be a very large mass at the centre of the galaxy, on investigation and other evidence it is surmissed that this mass must be in the form of a blackhole.

[dstebbins]

I shall provide a link because it goes into far more depth than I can possibly do:

 

http://en.wikipedia.org/wiki/Gravitational_constant

Thanks. I'll check it out when I have more than five minutes.

 

 

1. 6.67*10^-11m^3kg-1s-2 (Nm^2kg^-2)

Can you put that in a number and unit combo that a high school student like myself can understand?

 

2. I can't remember the derivation offhand, any good low level university text book should have one in it... If I remember tomorrow I shall have a look.

Thanks for that. Science like this drives me nuts.

 

3. The GR theory of gravity is used more than the newtonian one, which also uses the constant. But the way it works is you put the experimental evidence into some equations and the results of these says that there should be a very large mass at the centre of the galaxy, on investigation and other evidence it is surmissed that this mass must be in the form of a blackhole.

That's not exactly what I'm talking about. What's their mathematical reasoning? Suppose I was a science professor grading these scientists' exams. If they were to show all their work, what would it look like?

[timo]

Can you put that [G] in a number and unit combo that a high school student like myself can understand?

I don´t know many high school students such as yourself (I don´t even know what "high school" is) but there you go: [math] G= 0.00000000001667 \frac{N m^2}{kg^2} [/math] (value copied from Klaynos' post without verification).

[dstebbins]

I don´t know many high school students such as yourself (I don´t even know what "high school" is) but there you go: [math] G= 0.00000000001667 \frac{N m^2}{kg^2} [/math] (value copied from Klaynos' post without verification).

 

High school means grades 9-12. Where are you from? Africa?

 

So what you're saying is that the constant is (5/3)*10^-11 J^2/kg^2. Is that it?

 

EDIT: Hold the phone. That link that Klaynos gave me suggests that the constant isn't what you claim. Wtf?

[ydoaPs]

High school means grades 9-12. Where are you from? Africa?

I'm going to take a stab in the dark and say Darmstadt/Germany.

[Sisyphus]

The gravitational constant has much less significance than any of this.

 

How gravitational force varies is proportional to the product of the two masses divided by the square of the distance between them. In order to write it as an equation instead of just a proportion, the two side have to be in terms of the same units. The gravitational constant is just a made up value such that if you multiply it by a mass^2/distance, you get a force. That's ALL there is to it.

[swansont]

Since F = GMm/r^2, you can find the units of G by rearranging the equation

 

G = F r^2/Mm, or units of (force x distance^2)/mass^2

 

In MKS/SI units, force is in Newtons, length in meters , mass in kg

 

A Newton is a kg-m/s^2 (since F=ma) You can verify that this results in the same units Klaynos gave

 

G was originally determined by Cavendish, using a torsion balance. I suggest Googling that if you want more detail.

[swansont]

High school means grades 9-12. Where are you from? Africa?

 

"High school" is pretty much a US term, and the world-wide web is, well, world-wide.

[timo]

I looked it up: The value on Wikipedia is pretty much correct; the 1.667 is wrong (I misread Klaynos' post and took the "1." as part of the value). Since you´re obviously not a stupid bimbo, you should be able to figure out all the rest yourself.

[Klaynos]

*really geeky suggestion* put together a little talk about it and ask your teacher if you can give it in class....

[dstebbins]

I looked it up: The value on Wikipedia is pretty much correct; the 1.667 is wrong (I misread Klaynos' post and took the "1." as part of the value). Since you´re obviously not a stupid bimbo, you should be able to figure out all the rest yourself.

 

I swear, I used to have respect for you, but if that last comment is meant sarcastically, you can kiss my hairy caucasion ass.

[Klaynos]

I swear, I used to have respect for you, but if that last comment is meant sarcastically, you can kiss my hairy caucasion ass.

 

Just to say (yes I know this is going WILDLY off topic) that I've spoken to Athiest quite a lot, and I would not read it sarcastically at all.

[dstebbins]

Just to say (yes I know this is going WILDLY off topic) that I've spoken to Athiest quite a lot, and I would not read it sarcastically at all.

 

In that case, I apologize. I just hate it to the bone when someone thinks I'm stupid just because I don't know how to change a u-joint on a car or suchlike. It's not a fact that I'm stupid; it's the fact that I've never been taught.

[Spyman]

That's not exactly what I'm talking about. What's their mathematical reasoning? Suppose I was a science professor grading these scientists' exams. If they were to show all their work, what would it look like?

A link to read for question 3: http://en.wikipedia.org/wiki/Orbit

[dstebbins]

Okay, let's put this theory of universal gravitation to the test. My hypothesis (since we are, technically, scientists) is that the mass of the earth will be an unimaginatively large number of kilograms.

 

A 100.0kg person standing 1.000m above sea level on Earth weighs approximately 980.0N. By that logic, the mass of the earth should be equal to M in M=(d^2F)/(Gm) where d=distance between the objects, F equals force applied by one object on the other object, G is the gravitational constant, and m is the mass of the person. Therefore, plug amounts into the proper places, and you get (1.000^2 x 980.0)/(6.674*10^-11 x 100.0), or M= (980.0)/(6.674 x 10^-9). This means that the mass of the earth equals approximately 1.468 x 10^11. So my hypothesis was kind of right and kind of wrong, since 1.468 x 10^11 is not a very small number, but isn't a very large number either.

 

Does that sound right to you guys?

 

Edit: Don't answer that. I looked it up on ask.com, and I found here that the mass of the earth is 5.97 * 10^24 kilograms. What gives with this? Where did I go wrong?

[fredrik]

I awaiting edit#2 where you exclaim we don't have to answer that

 

> d=distance between the objects

 

Look closer at this.

 

/Fredrik

[swansont]

Is ALL of the mass 1 m away?

 

An application of Gauss's law shows that for a uniform mass distribution, you can treat all the mass as if it were at the center-of-mass location, for r > R (i.e. all mass "below" you behaves as if it were at the center of the sphere)

[dstebbins]

Is ALL of the mass 1 m away?

 

An application of Gauss's law shows that for a uniform mass distribution, you can treat all the mass as if it were at the center-of-mass location, for r > R (i.e. all mass "below" you behaves as if it were at the center of the sphere)

 

I'm aware of that. Don't you know that 100% of Newtonian physics is based on the assumption that the conditions are ideal? For example, when we say that the free-fall acceleration due to gravity is 9.8m/s/s, that's assuming (emphasis on that last word) that A) The object is EXACTLY at sea level, B) the portion of the earth directly beneath you is the same size and density as the average size and density of earth, and C) the object is accelerating in a vacuum, so air resistance can be ignored. Therefore, it is safe to assume that the person is the same density throughout, the earth is the same density throughout, and so on. Yes, 1m is not right at sea level, but the difference that one meter makes is negligible in the equation.

 

EDIT: Nevermind. I now see what you're saying. Let me rework my calculations.

 

According to ask.com, the radius of the earth is about 4000mi. That's approximately 6667km, or 6.667*10^6m. Square that, you get 4.445*10^13 m^2. Since the weight of a 100.0kg object is 980N, that means that means that the numerator in this equation is equal to 4.356*10^16. The denominator stays the same at 6.674*10^-9, so compute that, and I get 6.527*10^24, and it STILL comes out wrong! WTF?

[Spyman]

d = distance (Earth mean radius) = 6 372 797 m

m1 = Earth mass = 5 973 600 000 000 000 000 000 000 kg

G = Gravitational constant = 0.00 000 000 006 674 2 Nm²kg^-2

m2 = Body mass = 100 kg

 

Calculation of force at mean radius:

 

[MATH]F = G* \frac{m1*m2}{d^2} = 0.00 000 000 006 674 2 \frac{ 5 973 600 000 000 000 000 000 000 * 100 }{ 6 372 797 ^ 2} = 981.7 N[/MATH]

 

 

Calculation of Earth mass at 980 N:

 

[MATH]m1 = \frac{F*d^2}{G*m2} = \frac{980*6 372 797 ^ 2}{0.00 000 000 006 674 2*100} = 5 963 305 080 930 271 792 874 052 kg[/MATH]

[swansont]

the radius of the earth is about 4000mi

...

I get 6.527*10^24, and it STILL comes out wrong! WTF?

 

You got it right to 10%, which is OK considering that you approximated the radius, which is then squared, compounding any error of approximation.

[Klaynos]

According to ask.com, the radius of the earth is about 4000mi. That's approximately 6667km, or 6.667*10^6m. Square that, you get 4.445*10^13 m^2. Since the weight of a 100.0kg object is 980N, that means that means that the numerator in this equation is equal to 4.356*10^16. The denominator stays the same at 6.674*10^-9, so compute that, and I get 6.527*10^24, and it STILL comes out wrong! WTF?

 

Your problem is that you say round things "radius of the earth is about", this will result in difference in the final answer related to the amount of rounding you did (it is possible to calculate this using error analysis), but you're the same order of magnitude as what you expect to be the answer so it's pretty much correct.

 

Edit: and I've just noticed swansont's reply, this is the problem with browsing without images

[dstebbins]

Now I see. The radius of the earth was just an estimate.

 

One last question (for the time being, at least): On that Wikipedia page, it says that the mass of the sun, aka the "solar mass," is 1.9891 x 10^30 kg, but how can they be so sure? The sun forces tons and tons of Hydrogen to undergo nuclear fusion every second, meaning it's mass is constantly decreasing. How can they give such a precise measurement for a star?

[swansont]

Now I see. The radius of the earth was just an estimate.

 

One last question (for the time being, at least): On that Wikipedia page, it says that the mass of the sun, aka the "solar mass," is 1.9891 x 10^30 kg, but how can they be so sure? The sun forces tons and tons of Hydrogen to undergo nuclear fusion every second, meaning it's mass is constantly decreasing. How can they give such a precise measurement for a star?

 

Express "tons and tons" (actually ~4e9 kg) as a fraction of 10^30 kg and you'll see.

[insane_alien]

all things in consideration +/- 5 *10^26 kg isn't all that accurate. thats a hundred earths either way.