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We know thanks to Newton that:

$$F=G\frac{m_1\cdot m_2}{r^2}$$

Where $G$ is the Gravitational Constant that is about $6.673\cdot10^{-11}$

$m_1$ and $m_2$ are the masses of two different objects

$r$ is the distance between them.

We also know that $F=ma$ so can we write this? : $$ma=G\frac{m_1\cdot m_2}{r^2}$$

If so what is $m$ in the first term of the equation? $m_1$ or $m_2$? Thanks a lot for help!!!

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    $\begingroup$ You should first decide upon what system to observe. The expression for $F$ gives the force on $m_1$ due to $m_2$ "OR" the force on $m_2$ due to $m_1$. The mass $m$ would correspond to that of the system under observation, because the force "ON" the system is the cause of acceleration "OF" the system. $\endgroup$
    – Shubham
    Commented Apr 14, 2014 at 13:57

4 Answers 4

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If $a$ is the acceleration of object 1 (should write as $a_1$), then $m$ should be $m_1$. Vice versa.

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The equation for gravitational force F=Gm1m2/r^2 gives the force of attraction b/n any 2 bodies with point mass m1 and m2 and separated by a distance 'r'..it means both the objects are attracted towards each other by a force F=Gm1m2/r^2..It is also coherent with newtons 3rd law i.e action and reaction forces are equal.. The expression F=ma or a=F/m represents the acceleration of any object under observation of mass m if subjected to a force F.. So if you wan't to equate there two equation,from 2nd equation ie newtons 2nd law you wish to find the acceleration of body under observation.. First decide which body do you want to observe..if it is m1 you want put m1 in 2nd equation i.e F=m1a then put the 1st equation in place of F ie force on m1(also on m2)and solve for 'a'..you will get Gm2/r^2.. Similarly he you want the acc of m2 proceed as above and you 'll get Gm1/r^2

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Yes, this is all correct so far.

What you need to remember here is that Force is a vector quantity. That is, it has a direction associated with it.

A force pushing you into the ground is the not same as one pushing you up into the sky, like the seat of a flying airplane.

So here you need to lable your force $\vec{F}_1$, say, and this would be the force acting on particle $m_1$.

Then if you draw a nice force diagram, of the Sun and Earth or anything in particular, you'll be able to assign the directions correctly since you know that Newtonian gravity is an attractive force.

Does this clear it up?

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  • $\begingroup$ Thanks a lot! Also another little question: if the Forces $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ are parallels but have opposite directions shouldn't them cancel out? $\endgroup$
    – PunkZebra
    Commented Apr 14, 2014 at 14:15
  • $\begingroup$ @Peterix Yes exactly, this is a statement of Newton's 3rd Law. Remember that when you're working out the acceleration of a particle, the acceleration $\vec{a}$ is also vector-valued. So you need to make a force equation for each body seperately. $\endgroup$
    – Flint72
    Commented Apr 14, 2014 at 14:24
  • $\begingroup$ Note that this does not apply to situations in which $\overrightarrow {F_1}$ and $\overrightarrow {F_2}$ act on 'different' objects(i.e. not belonging to the same system) $\endgroup$
    – Shubham
    Commented Apr 14, 2014 at 16:32
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$\begingroup$ 
1.F = G(M1x m2)/r2
2.F = G(M1 x m2)/r2 = F = G(M2 x m1)/r2
3.F = G(M1 x m2)/r2 = G(M2 x m1)/r2
4.F = G(M1 x m2)/r2 x r2/G(m2 X M1)
5.f = 1

Objects exactly identical do not fall toward each other. They still attract. Just not each other. Because neither can overcome the others Inertia

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    $\begingroup$ You may want to rethink line 4 $\endgroup$
    – QCD_IS_GOOD
    Commented Mar 12, 2015 at 3:48

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