- #1

- 665

- 0

[tex]g=\frac{GM}{r^{2}}[/tex]

...or is that just observed?

Also, if someone knows how to show this, could you post it here or at least a hint?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter amcavoy
- Start date

- #1

- 665

- 0

[tex]g=\frac{GM}{r^{2}}[/tex]

...or is that just observed?

Also, if someone knows how to show this, could you post it here or at least a hint?

- #2

- 37

- 0

apmcavoy said:

[tex]g=\frac{GM}{r^{2}}[/tex]

...or is that just observed?

Also, if someone knows how to show this, could you post it here or at least a hint?

Combining Newton's 2nd law of motion with Newton's law of gravitational attraction gives your result:

[tex]F=\frac{GMm}{r^{2}}[/tex]

[tex]F=ma[/tex]

The acceleration, a, is labelled g.

- #3

- 665

- 0

Theoretician said:Combining Newton's 2nd law of motion with Newton's law of gravitational attraction gives your result:

[tex]F=\frac{GMm}{r^{2}}[/tex]

[tex]F=ma[/tex]

The acceleration, a, is labelled g.

I guess what I'm asking then is how do you prove Newton's Law of Gravitation?

Thank you.

Last edited:

- #4

- 37

- 0

apmcavoy said:I guess what I'm asking then is how do you prove Newton's Law of Gravitation?

I thought that it was too simple a reply. As far as I know, Newton came up with that by finding that the force should be proportional to the mass of each body and inversely proportional to the distance between them squared. I believe that this was a prediction rather than a result of observation. The constant G is found through experiment (its value, of course, depending on the units used for the distance and the masses).

- #5

- 665

- 0

Theoretician said:I thought that it was too simple a reply. As far as I know, Newton came up with that by finding that the force should be proportional to the mass of each body and inversely proportional to the distance between them squared. I believe that this was a prediction rather than a result of observation. The constant G is found through experiment (its value, of course, depending on the units used for the distance and the masses).

Alright, I was just wondering. I have heard that someone proved it with ellipses, but I'm not sure if I'm thinking of the right thing.

Thanks for your help.

- #6

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,549

- 8

apmcavoy said:I guess what I'm asking then is how do you prove Newton's Law of Gravitation?

You can't give a purely mathematical proof of any equation of physics that has empirical content. But it

If we define the "gravitational flux" [itex]\Phi_g[/itex] as the surface integral [itex]\Phi_g\equiv\int_S\vec{g}\cdot d\vec{S}[/itex], then Gauss' law for gravitation states the following:

[itex]\oint_S\vec{g}\cdot d\vec{S}=-4\pi Gm_{enclosed}[/itex]

The closed surface is the so-called "Gaussian surface". Now if the mass is a point particle of mass [itex]M[/itex] situated at the center of a Gaussian sphere, then we have the following for the LHS of Gauss's law:

[itex]\oint_S\vec{g}\cdot d\vec{S}=\oint_SgdS[/itex]

[itex]\oint_S\vec{g}\cdot d\vec{S}=g\oint_SdS[/itex]

[itex]\oint_S\vec{g}\cdot d\vec{S}=g(r)\cdot 4\pi r^2[/itex].

Plugging that into the LHS of Gauss' law gives:

[itex]g(r)=-\frac{GM}{r^2}[/itex].

So, the force [itex]F[/itex] exerted by our point mass on a second particle of mass [itex]m[/itex] is:

[itex]F=mg(r)=-\frac{GMm}{r^2}[/itex].

- #7

- 665

- 0

Tom Mattson said:You can't give a purely mathematical proof of any equation of physics that has empirical content. But itcanbe shown that Newton's law of gravitation can be derived from a more general priniciple, "Gauss' law" for gravitation.

If we define the "gravitational flux" [itex]\Phi_g[/itex] as the surface integral [itex]\Phi_g\equiv\int_S\vec{g}\cdot d\vec{S}[/itex], then Gauss' law for gravitation states the following:

[itex]\oint_S\vec{g}\cdot d\vec{S}=-4\pi Gm_{enclosed}[/itex]

The closed surface is the so-called "Gaussian surface". Now if the mass is a point particle of mass [itex]M[/itex] situated at the center of a Gaussian sphere, then we have the following for the LHS of Gauss's law:

[itex]\oint_S\vec{g}\cdot d\vec{S}=\oint_SgdS[/itex]

[itex]\oint_S\vec{g}\cdot d\vec{S}=g\oint_SdS[/itex]

[itex]\oint_S\vec{g}\cdot d\vec{S}=g(r)\cdot 4\pi r^2[/itex].

Plugging that into the LHS of Gauss' law gives:

[itex]g(r)=-\frac{GM}{r^2}[/itex].

So, the force [itex]F[/itex] exerted by our point mass on a second particle of mass [itex]m[/itex] is:

[itex]F=mg(r)=-\frac{GMm}{r^2}[/itex].

I understand that proof (and it's what I was looking for) but it seems like each proof I find depends on a different law. The rest, about the surface integrals is clear and makes perfect sense, but the original law is giving me some trouble. Where did this come from? Sorry, I hate to keep bothering you with these questions.

[tex]\oint_S\vec{g}\cdot d\vec{S}=-4\pi Gm_{\text{enclosed}}[/tex]

??

Thanks a lot

- #8

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,549

- 8

apmcavoy said:Where did this come from?

[tex]\oint_S\vec{g}\cdot d\vec{S}=-4\pi Gm_{\text{enclosed}}[/tex]

I took it as fundamental. You could alternatively take either of the following as fundamental:

[itex]\vec{\nabla}\cdot\vec{g}=-4\pi \rho[/itex] (Gauss' law in differential form)

[itex]\vec{g}=\int_Vd\vec{g}=-G\int_V\frac{dM}{r'^2}\hat{r}dV'[/itex] (Newton's law of gravitation in differential form, integrated over a body)

You can derive any of these from the others, but none of them comes from a purely mathematical formalism. That is because they have empirical content.

- #9

- 665

- 0

Just what I wanted to know! Thanks a lot Tom Mattson.

- #10

- 782

- 1

apmcavoy said:Alright, I was just wondering. I have heard that someone proved it with ellipses, but I'm not sure if I'm thinking of the right thing.

Thanks for your help.

Maybe you're thinking of Feynmen's Lost Lecture?

- #11

- 665

- 0

I don't know... where can I find it (if it isn't really "lost" )?

- #12

- 10

- 0

Does that answer the original question?

- #13

- 3,042

- 16

Was that the same force which pulled all falling objects downward?

Supposedly, the above question occured to Newton when he saw an apple falling from a tree. John Conduitt, Newton's assistant at the royal mint and husband of Newton's niece, had this to say about the event when he wrote about Newton's life:

..................................

If it was the same force, then a connection would exist between the way objects fell and the motion of the Moon around Earth, that is, its distance and orbital period. The orbital period we know--it is the lunar month, corrected for the motion of the Earth around the Sun, which also affects the length of time between one "new moon" and the next. The distance was first estimated in ancient Greece--see here and here.

To calculate the force of gravity on the Moon, one must also know how much weaker it was at the Moon's distance. Newton showed that if gravity at a distance R was proportional to 1/R2 (varied like the "inverse square of the distance"), then indeed the acceleration g measured at the Earth's surface would correctly predict the orbital period T of the Moon.

.........................

"Interestingly, however, the proof which Newton published did not use calculus, but relied on intricate properties of ellipses and other conic sections. "

You can read the full article here, smart guy that newton, too bad he was not a nice person.

http://www-istp.gsfc.nasa.gov/stargaze/Sgravity.htm

- #14

reilly

Science Advisor

- 1,077

- 1

Regards,

Reilly Atkinson

Share: