Universal Gravitation help

Beholder · Jul 17, 2005

Hello I need some help understanding Newtons universal gravitation. I read on a site that an objects Force on earth is its mass x gravity(on earth) hence MG (like F=MA). Then they say using newtons inverse square law that the same object at a distance of the moon would have the force(g) MG/distance to moon² and so it would be that much weaker.

Now with the universal gravitation between two masses we know F=GxM1xM2/R²
my questions are how did newton come to this conclusion?, what is the inverse square law?, and why are the two masses multiplied not added? I'm a little confused about this the only answer I can come up with is:

say we have the first equation they talked about F=MG (g for gravity on earth) if we substitute G with its components M x Gravitational constant we get two seperate masses mass 1 for the object and mass 2 for the earth and we also get G (grav. const.) so thats our GxM1xM2, now divide that by the square of the distance between them and we get the answer. Is that even close? can anyone answer where this equation comes from?

abercrombiems02 · Jul 17, 2005

First an inverse squared law is a big generalization. Basically all the inverse square law really means is that some quantity is proportional to 1/x^2 where x is some variable of interest. so the equation y = c/x^2 (c is some constant) is a general example of an inverse square law.

The masses are multipled because if we added them we would not have the correct dimensions. 1kg + 3kg = 4kg where as 1kg x 3kg = 3kg^2

As for the actual derivation I dont really know how he came up with it. I would have started assuming g is variable, thus...

F = mg
F/m = g
g = GMe/r^2

At this point I assume the mass and radius of the Earth had rough values. I think it could be reasoned that the mass should be on top because more mass leads to a greater weight force and that the radius should be on the bottom as increasing distance from a body yields a weaker gravitational force. Thus, g can be defined as follows

g = c*Me/r^n
where c is the constant of proportionality and n is the order of the radius

how these values were determined.....I have no idea, but thats a start

HallsofIvy · Jul 18, 2005

Essentially, Newton showed that this was the only force law that resulted in Kepler's laws for planetary motion.

James R · Jul 18, 2005

Newton's constant G was first determined accurately, I think, by Cavendish, using a torsion balance.

tony873004 · Jul 18, 2005

Beholder said:

Hello I need some help understanding Newtons universal gravitation. I read on a site that an objects Force on earth is its mass x gravity(on earth) hence MG (like F=MA). Then they say using newtons inverse square law that the same object at a distance of the moon would have the force(g) MG/distance to moon² and so it would be that much weaker.

Now with the universal gravitation between two masses we know F=GxM1xM2/R²
my questions are how did newton come to this conclusion?, what is the inverse square law?, and why are the two masses multiplied not added? I'm a little confused about this the only answer I can come up with is:

say we have the first equation they talked about F=MG (g for gravity on earth) if we substitute G with its components M x Gravitational constant we get two seperate masses mass 1 for the object and mass 2 for the earth and we also get G (grav. const.) so thats our GxM1xM2, now divide that by the square of the distance between them and we get the answer. Is that even close? can anyone answer where this equation comes from?

Don't confuse G with g. Big G is the gravitational constant (6.673e-11 N m^2 / kg^2), and little g is acceleration due to gravity at the surface of the Earth (9.8 m/s^2).

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

This solves for the gravitational force between them. Why do we multiply instead of add? Think of what would happen if we added. If a dust grain with a mass so close to 0 that you might as well call it 0 were on a bathroom scale, and we ADDED, the numerator of the equation would be enormous. But if we multiply, the numerator goes to 0, which is exactly what the bathroom scale would read.

Since
[tex]F = \frac{GMm}{d^2}[/tex]
and
[tex]F=ma[/tex]

you can conclude that

[tex]\frac{GMm}{d^2} = ma[/tex]

You've got a little m on each side of the equation, so they cancel:

[tex]a = \frac{GM}{d^2}[/tex]

giving you the acceleration formula.

Beholder · Jul 23, 2005

tony873004 said:

Don't confuse G with g. Big G is the gravitational constant (6.673e-11 N m^2 / kg^2), and little g is acceleration due to gravity at the surface of the Earth (9.8 m/s^2).

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

This solves for the gravitational force between them. Why do we multiply instead of add? Think of what would happen if we added. If a dust grain with a mass so close to 0 that you might as well call it 0 were on a bathroom scale, and we ADDED, the numerator of the equation would be enormous. But if we multiply, the numerator goes to 0, which is exactly what the bathroom scale would read.

Since
[tex]F = \frac{GMm}{d^2}[/tex]
and
[tex]F=ma[/tex]

you can conclude that

[tex]\frac{GMm}{d^2} = ma[/tex]

You've got a little m on each side of the equation, so they cancel:

[tex]a = \frac{GM}{d^2}[/tex]

giving you the acceleration formula.

I'm still a little confused here, I'm just begining physics and only know the basics of motion and things, its not the math that i can't work out its just the conceptual part, I mean I thought (so far) that acceleration was change in velocity divided by change in time. Although I can picture that given a distance and a velocity you can derive the acceleration but I'm not sure, any additional help would be much appreciated.

:shy:

James R · Jul 24, 2005

You're approximately right that acceleration is change of velocity divided by time. Technically, that is average acceleration, but never mind for now.

When we say the acceleration due to gravity is, say, 9.8 m/s^2 near the surface of the Earth, we mean that, if you drop an object (e.g. a brick), its speed will change by 9.8 metres per second every second. So, if you drop it from rest, then after 1 second it will be travelling at 9.8 m/s. After 2 seconds, it will be travelling at 19.6 m/s, and so on.