How Can One Calculate the Gravitational Constant G by Combining Two Formulas?

[Siebevp]

I wonderd how if could calculate G the gravitational constant by combining the two formulas;

Fg=G x (m1 x m2 / r^2)
Fz=m1 x g

you can only make these two fromulas equal to each other when you stand on the surface of the earth. or when you recalculate your g (accelaration towards the mass).

m1= your mass
m2= mass of the earth
R= the radius of the earth
G= the gravitational constant
g= the gravitational accelarationon the surface of the Earth in 9.81 m/s^2

m1xg = G x (m1 x m2 / r^2)
g = G x (m2 / r^2)

so from here out you can say that

G = (g x R^2)/m2

this is correct.

but now when I take g what is an accelaration in s/t^2

s= traveled distance
t= time

and put it in this formula you wil get..

G = (2s x R^2)/(t^2 x m2)

supposedly you take s=1

then you can calculate your time with the accelaration formula s=0.5 x g x t^2

and so you can calculate the correct G constant,

but assuming you're standing on the surface of the Earth your s (travelled distance) is actually 0 so your time is also 0 and that makes G also 0. Why because you do not move in space towards the mass of the earth, you actually do not travel any distance, if you would jump up a metre then you would have a distance to travel towards earth. giving you a time and so being able to calculate G.

can someone explain me what I did wrong because sofar nobody could answer me the question

 

[Doc Al]

I wonderd how if could calculate G the gravitational constant by combining the two formulas;

Fg=G x (m1 x m2 / r^2)
Fz=m1 x g

you can only make these two fromulas equal to each other when you stand on the surface of the earth. or when you recalculate your g (accelaration towards the mass).

m1= your mass
m2= mass of the earth
R= the radius of the earth
G= the gravitational constant
g= the gravitational accelarationon the surface of the Earth in 9.81 m/s^2

m1xg = G x (m1 x m2 / r^2)
g = G x (m2 / r^2)

so from here out you can say that

G = (g x R^2)/m2

this is correct.

That's fine. But to make use of it, you need to know or measure g.

but now when I take g what is an accelaration in s/t^2

s= traveled distance
t= time

and put it in this formula you wil get..

G = (2s x R^2)/(t^2 x m2)

supposedly you take s=1

then you can calculate your time with the accelaration formula s=0.5 x g x t^2

and so you can calculate the correct G constant,

If you want to measure g by dropping something a distance s, then by reversing the formula you just used you'd get: g = 2s/t^2 (not s/t^2).

but assuming you're standing on the surface of the Earth your s (travelled distance) is actually 0 so your time is also 0 and that makes G also 0. Why because you do not move in space towards the mass of the earth, you actually do not travel any distance, if you would jump up a metre then you would have a distance to travel towards earth. giving you a time and so being able to calculate G.

When calculating g, s is the distance traveled by some falling object in some time. If you don't drop an object, how will you measure g? When you're standing on the surface of the earth, there are several forces acting on you (not just gravity). You are not a falling object.

 

[Siebevp]

I do understand that it is 2s/t^2 it is so because; 2s/t^2 is the primitive of v/t, that is what I was told.

And I do know that if you dropped something of a certain hight or s; a distance, it would have an accelaration towards the mass (earth) because of gravitational forces,

but as you know an accelaration is a movement of an object,

And on Earth you always feel the accelaration, there for you can create the formula to calculate the gravitational force that is working on your mass; by F = m x a

but suppose you're standing on the earth
and you're not falling towards the earth

we still say that you can calculate the force that is working on your mass with:
F = m x a

and a is ofcourse the pre calculated accaleration.

but when you're are standing on the Earth you don't have a movement in comparison of the Earth the distance between the objects does not change, so therefor your traveled distance towards the Earth is zero, and if I would put that into the formula s=0.5 x a x t^2

then my time and accaleration would be zero, and if I put that into my formula to calculate G

with a certain distance and time, which has been shown that if you fall from a certain height and have a movement towards the other mass (earth) then you would be able to calculate G, but that does not happen when you stand on the earth, because you do not move towards the centre of the earth.

G= (2s x R^2)/(t^2 x m2)

G= (0x R^2)/(0 x m2)

G= (0)/(0)

G=0

do you now know what I mean

 

[Dickfore]

0/0 is not defined.

 

[Doc Al]

but as you know an accelaration is a movement of an object,

No, acceleration is the rate of change of velocity.

And on Earth you always feel the accelaration, there for you can create the formula to calculate the gravitational force that is working on your mass; by F = m x a

but suppose you're standing on the earth
and you're not falling towards the earth

we still say that you can calculate the force that is working on your mass with:
F = m x a

and a is ofcourse the pre calculated accaleration.

No. If you're standing on the earth, your acceleration is zero. (Let's ignore the Earth's rotation for now.) The net force on you is zero. Gravity is not the only force acting on you.

Your formula relating acceleration with G only applies to objects in free fall. An object resting on the Earth's surface is not in free fall.

 

[Doc Al]

Just to add to my previous comments...

but when you're are standing on the Earth you don't have a movement in comparison of the Earth the distance between the objects does not change, so therefor your traveled distance towards the Earth is zero, and if I would put that into the formula s=0.5 x a x t^2

then my time and accaleration would be zero, and if I put that into my formula to calculate G

If you measure the distance traveled by anything in zero time you'll get zero no matter how it moves or doesn't move. So that will tell you nothing. Plugging those numbers into your acceleration formula just gives you 0/0, which as dickfore points out is undefined. It does not give you zero.

Of course, you can measure your acceleration using a non-zero time, then you will get zero. (Obviously, since you're just standing there.) But as I pointed out before, that is irrelevant. What you need to measure is g, the acceleration of something in free fall, not the acceleration of something just sitting on the Earth's surface.

 

[Siebevp]

you say that gravitational formula only accounts for falling objects equal and opposite to each other, right..

how do you know that f=m x a because you can
also known in its other form as fz=m x g

you can get that with putting these formulas as shwon before equal

m x g = G x (m1 x m2)/R^2
g = G x (m2)/R^2
a= G x m2 / R^2

G= a x R^2 / m2

G= (2s x R^2)/(t^2 x m2)

so what is G in then...

G is in (Fs^2)/m^2

so

(F x s^2)/(m^2) = (2s x R^2)/(t^2 x m2)

(F x s^2 x t^2 x m) = (2s x R^2 x m^2)

(F x s^2 x t^2) = (2s x R^2 x m)

(F x t^2) = (2s x m)

F = (2s x m) / t^2
F= m x a,,,

and as you have been tought in high school that we constantly experience gravity, otherwise we would be floating in air and space, whatever you want...

and so we have also been told that how to calculate that force in Newtons, you just use this simple formula (that I just showed to you how to come up with that simple form) eaven when you are standing still on the earth

F = m x g

to calculate the force of gravity,

you cannot say that gravity is not there because; what you are saying, that these formulas are based on an object that is falling. not experiencing a movement which refers to the g in this formula the falling accelaration, and so you are telling me and my high school teacher that I cannot use this formula to calculate how much gravitaty I am experiencing right now sitting in my chair, and so not changing my position, and my radial distance towards the centre of matter we call earth. and so you're saying that right now I'm not eaven experiencing that force.

and unfortuneatly for you it is there. and we can eaven make an experiment to test it, it's called "making use of a scale." so do you now understand my issue.?.?.?

 

[Siebevp]

and btw time continues so that is not going to be zero, if we measure my "falling accelaration"

the only thing that is going to be zero is my traveled distance...

 

[Doc Al]

you say that gravitational formula only accounts for falling objects equal and opposite to each other, right..

No, the 'gravitational formula' only accounts for the force of gravity.

how do you know that f=m x a because you can
also known in its other form as fz=m x g

The first equation, better written as ΣF = ma, is Newton's 2nd law and is generally true.
The second, F = mg, is a special case. g is the acceleration of a free falling object.

you can get that with putting these formulas as shwon before equal

m x g = G x (m1 x m2)/R^2
g = G x (m2)/R^2
a= G x m2 / R^2

G= a x R^2 / m2

G= (2s x R^2)/(t^2 x m2)

so what is G in then...

G is in (Fs^2)/m^2

so

(F x s^2)/(m^2) = (2s x R^2)/(t^2 x m2)

(F x s^2 x t^2 x m) = (2s x R^2 x m^2)

(F x s^2 x t^2) = (2s x R^2 x m)

(F x t^2) = (2s x m)

F = (2s x m) / t^2
F= m x a,,,

As long as you use s and t to measure the acceleration of a freely falling object you'll be fine.

and as you have been tought in high school that we constantly experience gravity, otherwise we would be floating in air and space, whatever you want...

So? You should have also learned that when you're standing on the earth, the Earth exerts a force on you to hold you up. Otherwise you'd be in free fall.

and so we have also been told that how to calculate that force in Newtons, you just use this simple formula (that I just showed to you how to come up with that simple form) eaven when you are standing still on the earth

F = m x g

to calculate the force of gravity,

Yes, you can use that formula to calculate the force of gravity.

you cannot say that gravity is not there because; what you are saying, that these formulas are based on an object that is falling. not experiencing a movement which refers to the g in this formula the falling accelaration, and so you are telling me and my high school teacher that I cannot use this formula to calculate how much gravitaty I am experiencing right now sitting in my chair, and so not changing my position, and my radial distance towards the centre of matter we call earth. and so you're saying that right now I'm not eaven experiencing that force.

and unfortuneatly for you it is there. and we can eaven make an experiment to test it, it's called "making use of a scale." so do you now understand my issue.?.?.?

Sorry, but you're still confusing your acceleration while you are sitting in your chair with the acceleration of a freely falling object (which is g).

Obviously, your acceleration while you are in the chair is zero (ignoring the Earth's rotation). But that just means, per Newton's 2nd law, that ΣF = 0. The net force on you is zero. Gravity pulls you down while the chair pushes you up.

and btw time continues so that is not going to be zero, if we measure my "falling accelaration"

the only thing that is going to be zero is my traveled distance...

Your acceleration while sitting in a chair will be zero. But that doesn't mean that the gravitational force is zero. You cannot equate your acceleration in a chair with g.

 

[Siebevp]

i understand that ofcourse the netforce is zero, meaning that the gravitaional force is pulling me down with a certain amount in n (Newtons), and that there is another force that is pushing me up with the same amount of n (Newtons), so you have two different vectors, two different forces with a other direction the complete other direction of the other. so you have two DIFFERENT forces

ΣF = 0

suppose we take a mass of 100 kilos then the force of gravity would be 981 something Newtons, and the other force would be the same -981 Newtons ( because of the direction)

but g (in f = m x g) is not a constant not every where on the surface of Earth is it 9,81 m/s^2

capitol G is a constant

and you can calculate the force of gravity that is working on your mass with a pre calculated g "an accelaration" which you are not experiencing

and what you said about the netforce being zero is corect but f=mg does not give me the netforce, f=ma would, but seeming that you have to experience an accelaration which you are not it doesn't add up, because in these formulas you have your own mass, and an accelaration.

In f=mg in order to get g you need to know the mass of the approaching object the distance between you and the object, and the gravitational constant G

and then again coming back to my point f=mg is not the same as f=ma, f=ma does not have a pre calculated accelaration, for a certain distance and mass.

So that is the gravitational force working on your mass or my mass.

it's Fg witch is the gravitational force in 981 Newtons let's just assume that
the other force working against the gravitational forcein the opposite direction is also 981 Newtons

and there fore -Fg+F=0-981+981=0

the minus ofcourse to indecate the opposite direction
aka
ΣF = 0

is it because the Earth is also accelerating towards me?

 

[Siebevp]

but then the force would be equal and opposite to each other, still showing me that there is no accelarations of the objects toward each other, let's say i weigh in about a 100 kilos, that the gravitational force is 981 Newtons, this force accounts for both of the objects. both of the objects feel that force in 2 directions,

then still your g in f=mg is zero, saying that both of the objects feel no gravitational forces, which also says that your netforce is zero but it also says that there is no force, no gravity do you get me now?

 

[Siebevp]

i understand that ofcourse the netforce is zero, meaning that the gravitaional force is pulling me down with a certain amount in n (Newtons), and that there is another force that is pushing me up with the same amount of n (Newtons), so you have two different vectors, two different forces with a other direction the complete other direction of the other. so you have two DIFFERENT forces

ΣF = 0

suppose we take a mass of 100 kilos then the force of gravity would be 981 something Newtons, and the other force would be the same -981 Newtons ( because of the direction)

but g (in f = m x g) is not a constant not every where on the surface of Earth is it 9,81 m/s^2

capitol G is a constant

and you can calculate the force of gravity that is working on your mass with a pre calculated g "an accelaration" which you are not experiencing

and what you said about the netforce being zero is corect but f=mg does not give me the netforce, f=ma would, but seeming that you have to experience an accelaration which you are not it doesn't add up, because in these formulas you have your own mass, and an accelaration.

In f=mg in order to get g you need to know the mass of the approaching object the distance between you and the object, and the gravitational constant G

and then again coming back to my point f=mg is not the same as f=ma, f=ma does not have a pre calculated accelaration, for a certain distance and mass.

So that is the gravitational force working on your mass or my mass.

it's Fg witch is the gravitational force in 981 Newtons let's just assume that
the other force working against the gravitational forcein the opposite direction is also 981 Newtons

and there fore -Fg+F=0


-981+981=0

the minus ofcourse to indecate the opposite direction
aka
ΣF = 0

is it because the Earth is also accelerating towards me?

but then the force would be equal and opposite to each other, still showing me that there is no accelarations of the objects toward each other, let's say i weigh in about a 100 kilos, that the gravitational force is 981 Newtons, this force accounts for both of the objects. both of the objects feel that force in 2 directions,

then still your g in f=mg is zero, saying that both of the objects feel no gravitational forces, which also says that your netforce is zero but it also says that there is no force, no gravity do you get me now?

 

[Doc Al]

but then the force would be equal and opposite to each other, still showing me that there is no accelarations of the objects toward each other, let's say i weigh in about a 100 kilos, that the gravitational force is 981 Newtons, this force accounts for both of the objects. both of the objects feel that force in 2 directions,

If you have a mass of 100 kg and you are sitting in a chair then two forces act on you that are equal and opposite. Your weight, which equals mg, acts down. And an equal upward force from the chair acts up.

then still your g in f=mg is zero, saying that both of the objects feel no gravitational forces, which also says that your netforce is zero but it also says that there is no force, no gravity do you get me now?

No, g is not zero. g is approximately 9.81 m/s^2. Your weight isn't zero just because you sit in a chair (I wish that were so!), but the net force on you is zero.

One more time: The force of gravity--your weight--is given by mg, not by ma! Your acceleration may go to zero (if the net force on you happens to be zero), but the force of gravity will not.

 

[Siebevp]

thats how you calculate the amount of force with an accelaration you just said it now yourselve, but you do not have an accelaration, nor does the Earth have one towards you. because you don not move ( the distance from your core of your mass, to the core of the mass of earth) so you do not have a movement, because a accelaration is a movement you do not have a g or it would be equal to 0 which it obviously is not; because then we wouldn't eaven excist

 

[Doc Al]

thats how you calculate the amount of force with an accelaration you just said it now yourselve, but you do not have an accelaration, nor does the Earth have one towards you. because you don not move ( the distance from your core of your mass, to the core of the mass of earth) so you do not have a movement, because a accelaration is a movement you do not have a g or it would be equal to 0 which it obviously is not; because then we wouldn't eaven excist

Come on now... To use weight = mg does not require you to have an acceleration. Do you seriously think that the force of gravity depends on how you move? But if you wanted to measure g for yourself by measuring the acceleration due to gravity directly, then you'd need to have a freely falling body. (In order to have weight = ma, you need to have gravity being the only force acting.)

 

[Siebevp]

the point is... to calculate the force of gravity on a object (which is standing on the surface of the earth) we say it will have a movement towards the Earth (the accelaration of 9.81 m/s^2), and a certain mass, multiplie these two givins and you have calculated the force of gravity that is working on the object. in the direction of the Earth's core, the Earth also exerts the same force in the direction of your core of mass. so yeah then the netforce is zero.

for insteance again a mass of 100 kilo, the force from me pointing to the core of the Earth 981 Newtons, the Earth also exerts the same force but in the directions of the core of the 100 kilo object, so then again you would add up the two forces to calculate the netforce

-981+ 981= 0
-Fg1 + Fg2 = Fnet=0
and again the minus sign because Fg1 has the other direction than Fg2

meaning that these forces are equal and opposite of each other. These Forces are in Newtons, it's measurable, and calculatable

calculatable

with this simple formula f=mg

and we say that g is 9.81 m/s^2 saying that a mass like you and me feel that same accelarations towards the core of the Earth but we are not moving towards the core of the earth, so we do not have a movement or a accelaration. what you see and measure, but you also measure a force. so g is not eaven there, but we say it is there with this formula f=mg, and partially f=ma is the same form and then again it is not

 

[Siebevp]

you don't need accelaration, but you will get one because of gravity, otherwise you wouldn't be able to calculate the gravitational force in Newtons

 

[Doc Al]

the point is... to calculate the force of gravity on a object (which is standing on the surface of the earth) we say it will have a movement towards the earth (the accelaration of 9.81 m/s^2),

No, we don't say that. (Well, you do, but you're wrong.)

and a certain mass, multiplie these two givins and you have calculated the force of gravity that is working on the object.

You can certainly find the weight of an object by multiplying mass*g. (Not mass*acceleration, unless the mass happens to be in free fall.)

and we say that g is 9.81 m/s^2 saying that a mass like you and me feel that same accelarations towards the core of the earth

No, we don't say that. 9.81 m/s^2 would be your acceleration if you fell out of a window (ignoring air resistance, etc.). If you're sitting in a chair, you feel the chair holding you up but you won't be accelerating.

but we are not moving towards the core of the earth, so we do not have a movement or a accelaration. what you see and measure, but you also measure a force. so g is not eaven there, but we say it is there with this formula f=mg, and partially f=ma is the same form and then again it is not

Again, you confuse g, which is the acceleration of a freely falling object, with the acceleration you happen to have when you sit in a chair. Still wrong.

 

[Siebevp]

never mind...

 

[Siebevp]

would you mind if i asked you...
what degree do you have?

 

[Doc Al]

would you mind if i asked you...
what degree do you have?

Take a guess! (Or look at my profile.)

 

[hologramANDY]

When someone is sitting at rest, the force of gravity on that person is equal to g. However, there is also an equal and opposite force which cancels out the acceleration due to gravity.

In other words, you are accelerated towards the Earth by gravity, but you are also accelerated by whatever object you are resting on away from the Earth, so your net acceleration is zero. You cannot ignore the opposing reaction or your math will break down.

 

[Siebevp]

well you still did not answer the question for me... because you probably do not understand what i ment,, srry to have bothered you and your precious time

and I hope you will get a nobel price for what you did to get a phd title

 

[hologramANDY]

Its just science, there is no need to get snippy. Maybe if you better articulated what you don't understand you could learn somthing.

 

[Siebevp]

Doctor,
I'm from holland and I'm 18 years old, I will be going to study physics in holland, or where ever. And do not start about articulating because english is not my first language, and for that I will never be as good in articulating *** someone who was born with this language and learned physics in that language and mathematics in that language.

I am trying to understand it in two different languages. And I'm not saying that you're wrong, and maybe I'm not good at articulating and maybe, I shouldn't doubt your "intelligence" Which I do not, and then maybe I do because you can put on the internet whatever you can. Therefor turning you into a bad source of explenation.

I would like to thank you for your help, your logic is correct and I understand it but I know for certain that I mean something else by my question. and I keep asking the same but over and over again in a slitly different way. So we keep getting in a visual cirkel, and I thank you for your help, and I will try to ask this to a dutch person with a PhD or whatever, and maybe he will say the same as you have and maybe then I will understand, and then again maybe not

so thank you anyways

 

[Dale]

Siebevp, you are misunderstanding how to calculate weight. Weight is the force due to gravity, and it is given by
f=GMm/r^2

In the case where r does not change much you can simplify this to
f=mg where g=GM/r^2

This is just a simplified expression for the full expression of the force due to gravity.

Although the form is similar this equation is not the same as f=ma because g=GM/r^2 not g=a. GM/r^2 has units of acceleration, but it is not the acceleration of the object.

In order to determine the acceleration of the object you have to sum up all of the forces to obtain the net force. In the special case where the object is in free fall the only force acting on it is gravity and in that special case a=g. In the special case of an object at rest on the ground the same downwards gravitational force is balanced by the upwards normal force so that the net force is 0 and a=0 even though g still equals GM/r^2.

 

[Siebevp]

i know that it's not similar,

to calculate g you ofcourse need the equation g=GM/r^2,

but every time on one side of the equation there is just one variable of time.

so we keep saying that to calculate g which is in a distance devided by time square, is equal to a constant called G what is in a force multiplied by a distance square and devided by a mass square, and then again multiplie this constant by the mass that is acting out the force on you (the opposite mass) devided by the radial distance between the two masses.

put it in a equation you will get

g=(F x s^2)/(m^2) x (m)/(r^2)

g= (F x s^2 x m)/(m^2 x r^2)

g= (F)/(m)

g= m/s^2

s/t^2=F/m

F x s =m x t^2

am I doing it wrong no because we take values that are being used together, so it should be the same

F x s =m x t^2
9.81 = 9.81

g = m/s^2
9.81 = 9.81

 

[Doc Al]

put it in a equation you will get

g=(F x s^2)/(m^2) x (m)/(r^2)

g= (F x s^2 x m)/(m^2 x r^2)

Looks like you're just playing around with units/dimensions. Sure, G has units of Force*Distance^2/Mass^2. Other than that, I see no meaning in this equation. (You haven't defined what force, distance, or mass you're talking about.)

g= (F)/(m)

OK, still dimensionally correct at least.

g= m/s^2

?? These units don't match.

s/t^2=F/m

OK, at least the units match.

F x s =m x t^2

?? These units don't match.

am I doing it wrong no because we take values that are being used together, so it should be the same

F x s =m x t^2
9.81 = 9.81

g = m/s^2
9.81 = 9.81

Again, I have no idea what you're trying to say. But keep trying.

 

[Siebevp]

something doesn't add up for me, and I will have to think how to formulate the question, which I'm puzzleing about

 

[sophiecentaur]

If you want to measure g then why not use a pendulum?
(T =2π√(l\g)
You can refine it and refine it, experimentally, by using jewelled pivots etc and doing it in a vacuum.
I reckon that method would beat the Stopwatch and falling ball, any day.

 

[Siebevp]

it is not about measuring g

 

[sophiecentaur]

The first words of the thread are about calculating G with a formula containing g. How can you calculate it from g if you don't measure g at some stage? (I now see this has been mentioned at least once before).

Calculating G from g (and other things) is surely just a rearrangement of some formulae. Most of the arguments here seem to involve pulling yourself up by your own bootstraps. You can't do anything without measuring some mass, some forces, some times and some distances, in one form or another. You takes your pick as to which ones and how.

 

[Dale]

to calculate g you ofcourse need the equation g=GM/r^2,

but every time on one side of the equation there is just one variable of time.

I don't know where you are getting that idea. The only variables are g, G, M, and r. There is no variable of time which would usually be denoted by the variable t.

The rest of your post seems to confuse a dimensional analysis of an equation with the variables of the equation.