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Law of gravitationby shounakbhatta
Tags: gravitation 
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#1
Jan713, 12:59 PM

P: 272

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. 


#2
Jan713, 01:04 PM

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#3
Jan713, 01:11 PM

P: 272

Sorry,
F=G.m1.m2/r^2 Can it be written: F=G.m.m/r^2. 


#4
Jan713, 01:35 PM

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Law of gravitation
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. 


#5
Jan713, 01:36 PM

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[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 


#6
Jan713, 02:57 PM

P: 272

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? 


#7
Jan813, 02:10 PM

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P: 3,446

yep. As Doc Al said, the important part is if on the lefthand 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).



#8
Jan913, 01:27 AM

P: 272

Ok. Thank you very much.



#9
Jan913, 07:13 AM

P: 43

So the equation, with R^{2} proportional to Force, should and would be: F = G M_{1}M_{2} 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. 


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