# Newton's Law of Universal Gravitation equation(s) ?

1. Jan 31, 2015

### bijou1

1. The problem statement, all variables and given/known data
I am currently reading about Newton's Law of Universal Gravitation and I am so confused as to why there is a negative sign in front of the equation Fg = (G* m1m2)/r^2.

2. Relevant equations
Fg = (G* m1m2)/r^2
There is a vector form of the magnitude of the gravitational force F12 = - ((G*m1m2)/r^2 )r12
F12 = force exerted by object 1 on object 2
G = 6.67 x 10^-11 N⋅m^2 /kg^2
m1 = mass of object 1
m2 = mass of object 2
r = distance between the centers of masses of objects 1 and 2

3. The attempt at a solution
The text states that the negative sign indicates that object 1 is attracted toward object 2 and that through Newton's 3rd law, the force exerted by m2 on m1, F21 is equal in magnitude to F12 but opposite in direction. Therefore, these forces form an action-reaction pair⇒F21=-F12.
So if F12= (G*m1m2)/r^2, then F21 = -(G*m1m2)/r^2
I know that force is a vector quantity, but I am not sure when to use the vector form; and why is the vector form negative? I am so confused...Could anyone please help me understand this mathematically and conceptually?

2. Jan 31, 2015

### Andrew Mason

The force vector is negative because the direction of the force is opposite to the direction of the radius vector. So, for example, the direction of the gravitational force on you is toward the centre of the earth, which is opposite to the direction of the radius vector from the centre of the earth to you.

AM

3. Feb 3, 2015

### bijou1

Hi, since the gravitational force on me is toward the center of the earth, and is pointing in the downward direction and opposite to the direction of the radius vector, if I designate the downward direction as positive, then how is the direction of the force vector negative? Do we always designate the direction of the force vector negative and the radius vector positive? Please correct me if I am not understanding this...Thank you, any help would be greatly appreciated!

4. Feb 3, 2015

### Andrew Mason

It is still opposite to the direction of the unit radius vector. That is all the equation says.

If you want to establish the down direction as positive you would just have to use $\hat d = -\hat{r}$ as the unit vector. i.e.:

$$F_g = \frac{GMm}{r^2}\hat{d} \text{ where } \hat{d} = -\hat{r}$$

It is just not a common way of expressing the universal law of gravitation.

AM

5. Feb 3, 2015

### Staff: Mentor

Your first post practically has what you are looking for. It just isn't quite precise enough. Let $\vec{F}_{12}$ represent the force that body 1 exerts on body 2, and let $\vec{r}_{12}$ represent the position vector drawn from body 1 to body 2. Then:
$$\vec{F}_{12}=-\frac{Gm_1m_2}{r^3_{12}}\vec{r}_{12}$$
where $r_{12}$ is the magnitude of the position vector $\vec{r}_{12}$. So, the force that body 1 exerts on body 2 is opposite in direction to the position vector drawn from body 1 to body 2.

Now, let $\vec{F}_{21}$ represent the force that body 2 exerts on body 1, and let $\vec{r}_{21}$ represent the position vector drawn from body 2 to body 1. Then:
$$\vec{F}_{21}=-\frac{Gm_1m_2}{r^3_{21}}\vec{r}_{21}$$
where $r_{21}$ is the magnitude of the position vector $\vec{r}_{21}$. So, the force that body 2 exerts on body 1 is opposite in direction to the position vector drawn from body 2 to body 1.

But, $\vec{r}_{21}$ = -$\vec{r}_{12}$, and $r_{21}$ = $r_{12}$

Therefore, $\vec{F}_{21}$ = - $\vec{F}_{12}$

Chet

6. Feb 3, 2015

### bijou1

Now, let $\vec{F}_{21}$ represent the force that body 2 exerts on body 1, and let $\vec{r}_{21}$ represent the position vector drawn from body 2 to body 1. Then:
$$\vec{F}_{21}=-\frac{Gm_1m_2}{r^3_{21}}\vec{r}_{21}$$
where $r_{21}$ is the magnitude of the position vector $\vec{r}_{21}$. So, the force that body 2 exerts on body 1 is opposite in direction to the position vector drawn from body 2 to body 1.

But, $\vec{r}_{21}$ = -$\vec{r}_{12}$, and $r_{21}$ = $r_{12}$

Therefore, $\vec{F}_{21}$ = - $\vec{F}_{12}$

Chet[/QUOTE]
Hi, thank you so much for your help, so substituting r21 as -r12 and magnitudes r21 as r12 into equation:
F
21 = -[Gm1m2/(r21)^2] -r21
F21 = -[Gm1m2/(r12)^2]-r12
F21 = [Gm1m2/(r12)^2]r12
F12 = [Gm1m2/(r12)^2]r12
since F21 = -F12
⇒{Gm1m2/(r12)^2]r12 = -([Gm1m2/(r12)^2]r12)
⇒ or F21 = -F12

7. Feb 3, 2015

### Staff: Mentor

Yes, except the it should be $r^3_{12}$ in the denominator. One of the r12's in the denominator can be absorbed into the $\vec{r}_{12}$ to give a unit vector in the direction that the force is acting. This will leave the usual $r^2_{12}$ in the denominator.

Chet

8. Feb 5, 2015

### bijou1

Hi! I think I see where the r12^3 came from:
r12 ⇒unit vector
r12/(r12)
F21 = -[Gm1m2/(r21)^2]r21
⇒-[Gm1m2/(r21)^2] ⇒(-Gm1m2)(r21)/(r21)^3
since r21 =-r12 and r21 = r12
therefore,
F21 = -Gm1m2/(r12)^2 [-r12/(r12)]
⇒Gm1m2 (r12)/ (r12)^3
⇒Gm1m2/(r12)^2
and since Newton's 3rd law states: F12 = -F21
F12 = - [Gm1m2/(r12)^2]

**Thank you once again for helping me understand this concept. This was great!