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Gravitational Force Formula

by uranium_235
Tags: force, formula, gravitational
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uranium_235
#1
Jun30-04, 12:03 AM
P: 36
I am sure most here are familiar with it
Fg = G*m1*m2/d^2
With this formula, if one were to venture to the centre of the earth, the distance between he and the centre of the earth == 0.
Therefore we get the bottom equal to 0^2 or 0. If we then continue we get an undefined answer. Certainly the gravitational force between the earth and a body and the centre of the earth is not infinite or else everything would collapse inwards towards the centre. So does this mean the gravitational force between the earth and a body at its centre is Null and void, or in other words equal to 0?

Please excuse my ignorance; I am very new to this.
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chroot
#2
Jun30-04, 12:10 AM
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When you're at the center of the earth, gravity is balanced in every direction, so there is zero field there. In other words, the same amount of matter is above your head as below your feet, so their attractions cancel.

On your way down to the center, you should note that only the mass interior to you matters. In other words, if you're 1 km from the center, only the mass within a 1 km radius of the center has any effect on you. The exterior mass all cancels out. When you get to the center, there is no longer any mass interior to you, and thus no force.

- Warren
uranium_235
#3
Jun30-04, 12:13 AM
P: 36
My question is not so much concerning the centre of the earth, but rather when d=0.

gerben
#4
Jun30-04, 01:02 AM
P: 540
Gravitational Force Formula

The distance (d) will never be 0, because the matter occupies some volume (it is not at one single position). In case of the earth, if you move to its center some mass will be in front of you and, if you are under ground level, some behind you. All the mass of the earth will pull on you but not all of it will pull in the same direction, some will pull you forwards (the mass that is in front of you) and some will pull you backwards (the mass that is behind you). When you are in the center of the mass(es) you will be pulled with equal force in all directions.

If two pieces of matter get very close together other kinds of forces (other than gravity) will come into play that are much larger than the force of gravity. You can mathematically make d go to zero so that Fg goes to infinity but physically this is not possible, at very small d you will get into molecules, into atoms etc. then things get really complicated physically and Fg will be of no importance compared to the other forces.

If you think about masses getting very close together, you will need to imagine very small objects and you will get into the part of physics that has not so much to do with gravity but with atomical and subatomical forces, these will cancel out if you compare your body (an enormous complex of atoms) with the world (an even larger complex of atoms).
pmb_phy
#5
Jun30-04, 08:22 AM
P: 2,954
Quote Quote by uranium_235
I am sure most here are familiar with it
Fg = G*m1*m2/d^2
With this formula, if one were to venture to the centre of the earth, the distance between he and the centre of the earth == 0.
Therefore we get the bottom equal to 0^2 or 0. If we then continue we get an undefined answer. Certainly the gravitational force between the earth and a body and the centre of the earth is not infinite or else everything would collapse inwards towards the centre. So does this mean the gravitational force between the earth and a body at its centre is Null and void, or in other words equal to 0?

Please excuse my ignorance; I am very new to this.
The expression you gave is not the correct one. That holds only for a point body or outside a body with a spherically symmetric mass density. Inside a body with a uniform mass density the force is linearly proportional to the mass density and to the distance to the center of the body. Think of it like this - Inside a spherically symetrical shell of mass the gravitational field is zero. Now consider the field inside the body. You can think of this as being inside a sphericall shell and outside a body of radius r with a uniform mass density. The closer you get to the center the less mass there is inside this imaginary shell which you're standing on.

You can use Gauss's law to calculate this. Gauss's law is the surface integral of the gravitational field vector g over the sphere containing a quantity of mass M. The surface of this sphere is called a "Gaussian surface". The value of that integral is -4*pi*GM where M is the mass inside the Gaussian surface. Solving for g gives the magnitude of

g = (4/3)*pi*G*rho*r

where rho is the mass density of the body. The direction is toward the center of the sphere. As expected g -> 0 as r - > 0. To get the force on the body multiply g by the mass of the body. You can express the above relation in terms of the total mass of the body by replacing rho with rho = M/V where M is the total mass and V is the volume of the entire body

Pete
FZ+
#6
Jun30-04, 03:19 PM
FZ+'s Avatar
P: 1,954
Quote Quote by uranium_235
I am sure most here are familiar with it
Fg = G*m1*m2/d^2
With this formula, if one were to venture to the centre of the earth, the distance between he and the centre of the earth == 0.
Therefore we get the bottom equal to 0^2 or 0. If we then continue we get an undefined answer. Certainly the gravitational force between the earth and a body and the centre of the earth is not infinite or else everything would collapse inwards towards the centre. So does this mean the gravitational force between the earth and a body at its centre is Null and void, or in other words equal to 0?

Please excuse my ignorance; I am very new to this.
This is wrong, because you are using the approximation of the Earth to a sizeless particle, in a situation where it cannot be correct. Instead, you must mathematically break up your massive object into a large number of small components, work out the gravitational attraction from each component, and then sum them.
uranium_235
#7
Jul1-04, 12:22 AM
P: 36
Thank you for all your responses. I have much to learn.
Michael F. Dmitriyev
#8
Jul4-04, 04:21 AM
P: 348
Quote Quote by pmb_phy
The expression you gave is not the correct one. That holds only for a point body or outside a body with a spherically symmetric mass density. Inside a body with a uniform mass density the force is linearly proportional to the mass density and to the distance to the center of the body. Think of it like this - Inside a spherically symetrical shell of mass the gravitational field is zero. Now consider the field inside the body. You can think of this as being inside a sphericall shell and outside a body of radius r with a uniform mass density. The closer you get to the center the less mass there is inside this imaginary shell which you're standing on.

You can use Gauss's law to calculate this. Gauss's law is the surface integral of the gravitational field vector g over the sphere containing a quantity of mass M. The surface of this sphere is called a "Gaussian surface". The value of that integral is -4*pi*GM where M is the mass inside the Gaussian surface. Solving for g gives the magnitude of

g = (4/3)*pi*G*rho*r

where rho is the mass density of the body. The direction is toward the center of the sphere. As expected g -> 0 as r - > 0. To get the force on the body multiply g by the mass of the body. You can express the above relation in terms of the total mass of the body by replacing rho with rho = M/V where M is the total mass and V is the volume of the entire body

Pete
Gravity force must be equal to
G (4pi) M1*M2/(4pi)r^2
I.e. a gravity force must be inversely to area of sphere with radius r. Then G is not correct and should be multiplied on 4pi.
On my glance it is consequence of 4D action.

Michael
pmb_phy
#9
Jul4-04, 07:31 AM
P: 2,954
Quote Quote by Michael F. Dmitriyev
Gravity force must be equal to
G (4pi) M1*M2/(4pi)r^2
I.e. a gravity force must be inversely to area of sphere with radius r. Then G is not correct and should be multiplied on 4pi.
On my glance it is consequence of 4D action.

Michael
The gravitational force must equal GM1*M2/(4pi)r^2 only for the force of one particle (or outside a spherically symmetric mass distribution) on another. It does not have that value in all cases and especially not in the present case.

Pmb
Michael F. Dmitriyev
#10
Jul4-04, 09:30 AM
P: 348
Quote Quote by pmb_phy
The gravitational force must equal GM1*M2/(4pi)r^2 only for the force of one particle (or outside a spherically symmetric mass distribution) on another. It does not have that value in all cases and especially not in the present case.

Pmb
Agree, for conformity to experimental data G should be multiplied on 4pi.
There should be also a logic explanation "visualizing" this formula. Why a sphere for mass and a circle for charge or magnetic force?

Michael


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