Law of gravitation

shounakbhatta

I was going through a site which tells:

F=G.m1.m2r^2

Also

F=-G.m.m/r^2.

Can it be written?

Thanks.

Doc Al

The first version makes no sense.

See: Newton's law of universal gravitation

shounakbhatta

Sorry,

F=G.m1.m2/r^2

Can it be written:
F=-G.m.m/r^2.

DaleSpam

Yes, if and only if m1=m2=m.

EDIT: actually I just noticed the minus sign. You have to be careful that gravity is always attractive, never repulsive.

Doc Al

Sometimes when writing the force as a vector expression a minus sign is used to indicate that it's an attractive force:

[tex]\vec{F} = - \frac{G m_1 m_2}{r^2} \hat{r}[/tex]
Where [itex]\vec{r}[/itex] is the position vector of m_2 with respect to m_1 and [itex]\vec{F}[/itex] is the gravitational force on m_2 due to m_1.

See: Vector form

shounakbhatta

Ok, so writing force as a vector we use minus.

As gravity is always attractive, so can we always use minus sign and don't use minus sign when any force is repulsive, like electric?

BruceW

yep. As Doc Al said, the important part is if on the left-hand side, we have the force on 1 due to 2, and if on the right hand side we have the position of 1 with respect to 2, then we need the minus sign for an attractive force. (And this is the most common way it is written). And yes, for a repulsive force, it will be positive. (this is automatically taken into account by multiplying the two charges together in the case of electric force).

shounakbhatta

Ok. Thank you very much.

kweba

Yes, as what with the others have said, we use a minus sign to indicate (an attractive) force as a vector.

So the equation, with R² proportional to Force, should and would be:

F = G M₁M₂ r^-2

So that R^-2 would mean to be inversely proportional to the Force (F), equal to the second equation in your original post.

The equation is derived from the Inverse Square Law: The greater the (square) distance between objects/masses, the lesser the force; and vice versa.