vector form of the law of gravitation

[ahmeeeeeeeeeed]

Why is the vector form of the law of gravitation :

 

F = -GMm/r^2

 

 

Why is there a negative sign ? Aren't the force direction and the displacement directiona the same ?

 

And also in derivating the law of gravitaional potential energy

 

The work done by the external force to move the body from point r1 to infinity equals ecatly the potential energy

 

the force of the extenal force is the same direction as the displacement so Work is the INTEGRATION of ( the magnitude of F * the magnitude of the displacement * cos 0 )

 

and this turns that work from r1 to infinity = -GMm/infinity - (-GMm/r1) = GMm/r1

 

(No negative sign)

 

What is the Error in these calculations, please ?

Edited November 4, 2011 by ahmeeeeeeeeeed

[swansont]

Why is the vector form of the law of gravitation :

 

F = -GMm/r^2

 

 

Why is there a negative sign ? Aren't the force direction and the displacement directiona the same ?

 

And also in derivating the law of gravitaional potential energy

 

The work done by the external force to move the body from point r1 to infinity equals ecatly the potential energy

 

the force of the extenal force is the same direction as the displacement so Work is the INTEGRATION of ( the magnitude of F * the magnitude of the displacement * cos 0 )

 

and this turns that work from r1 to infinity = -GMm/infinity - (-GMm/r1) = GMm/r1

 

(No negative sign)

 

What is the Error in these calculations, please ?

 

The negative sign is the convention for an attractive force. (You can see that in the electrostatic force equation, since you can get differing signs)

 

The work done is not equal to the PE, it is equal to the change in PE. You go from -GMm/r to zero. You have to add GMm/r to the system to get that.

[DrRocket]

Why is the vector form of the law of gravitation :

 

F = -GMm/r^2

 

 

Why is there a negative sign ? Aren't the force direction and the displacement directiona the same ?

 

 

Because that is NOT a vector equation. In fact the right hand side includes nothing that could be interpreted as a vector. The vector form would be

 

[math] \vec F = G \frac {Mm \vec r}{||\vec r||^3}[/math] where [math] \vec F[/math] is the force exerted on [math]m[/math] by [math]M[/math] and [math]\vec r[/math] is the vector from [math]m[/math] to [math]M[/math].

[imatfaal]

Because that is NOT a vector equation. In fact the right hand side includes nothing that could be interpreted as a vector. The vector form would be

 

[math] \vec F = G \frac {Mm \vec r}{||\vec r||^3}[/math] where [math] \vec F[/math] is the force exerted on [math]m[/math] by [math]M[/math] and [math]\vec r[/math] is the vector from [math]m[/math] to [math]M[/math].

 

I realise it is kinda equivalent and you have made it quite clear what your vector F represents - but I learnt that the more traditional approach is the force should be that on [math]M[/math] by [math]m[/math] when the vector is [math]m[/math] to [math]M[/math] so that if using subscripts we maintain order ie the associated force with [math]r_{12}[/math] is [math]F_{12}[/math]. This would require a minus sign in the above equation. Now it is obvious which is the direction in terms of gravity - but it is important to be consistent so that dealing with others forces and vectors doesn't become confusing .