Gravitational force between two masses

[Oliver321]

Hello everyone.
Probably this question is trivial, but nevertheless I am confused about Newtons law of motion:
$$F=G\frac{m_1m_2}{r^2}$$

Now, some sources say, that F is the force between the two masses m1 and m2. Other sources say, that F is the force that m1 exhibits on m2. But isn’t this a contradiction? Because of Newtosns third law, if F is the force between the two masses, m1 acts on m2 by a a force of F/2 and vice versa.
On the other hand, if F is the force of m1 on m2, the whole force between both particles is 2F.

Also it is a bit confusing for me: I can rewrite F as
$$F=m_2a$$
if F is the force of m1 on m2.
But if F is the force between two masses rather than the force of one mass on the other, it should be stated that
$$F=\mu a$$
with μ the reduced mass. In this form it is used to solve the Kepler-problem. But on the other hand it is often stated, that the acceleration is only dependent on the mass which produces the field (the acceleration of a falling stone is independent of the mass of the stone), suggesting the first form F=m2*a.
Where is now my mistake? Or are both views compadible with each other (and why)?

I hope you understand why I am confuse and in this sense I would appreciate every help. Thanks!

 

[mitochan]

The forces on m_1 and m_2 are same amount of $G\frac{m_1m_2}{r^2}$ and opposite direction, attracting each other.

 

[Oliver321]

The forces on m_1 and m_2 are same amount of $G\frac{m_1m_2}{r^2}$ and opposite direction, attracting each other.

Thank you! But would this mean, that the following is true:
So let’s say, that we have two masses m1, m2. Mass m1 is accelerated toward m2 with a magnitude of
$$a1=G\frac{m2}{r^2} $$
And then also m2 would be accelerated by
$$a2=G\frac{m1}{r^2} $$
So the acceleration of the approaching of both body’s is
$$a1+a2=G\frac{m1+m2}{r^2}$$
(same could be arrived by using relative coordinates and the reduced mass as mentioned above).
And therefore a heavy object in Earth's gravitational field would really fall faster than a light one, even if the difference is tiny. Also the force needed to hold the two body’s m1 and m2 apart of each other (so that they don’t approach) would be 2F.

 

[256bits]

Also the force needed to hold the two body’s m1 and m2 apart of each other (so that they don’t approach) would be 2F.

That doesn't follow.
If there is a force F on mass 1 and a force F on mass 2 , that does not become a 2F force.

 

[Ibix]

And therefore a heavy object in Earth's gravitational field would really fall faster than a light one, even if the difference is tiny.

Yes and no. You are correct that the distance between two masses depends on $a_1+a_2$. However, as you note, one of those accelerations is utterly negligible. Furthermore, if you work in an inertial frame then the accelerations of each object genuinely are independent of their mass. It's only if you adopt the (accelerating) rest frame of one object that the other's acceleration depends on its mass.

So, in short, there are (usually unwritten) caveats to the "gravitational acceleration is independent of mass" claim. It's not wrong, but it has some assumptions about how you are doing the measuring of acceleration.

 

[Oliver321]

That doesn't follow.
If there is a force F on mass 1 and a force F on mass 2 , that does not become a 2F force.

Thanks for the awnser.
But why not? Isn’t the force on m1 the same as on m2 because of Newton’s third law?

Yes and no. You are correct that the distance between two masses depends on $a_1+a_2$. However, as you note, one of those accelerations is utterly negligible. Furthermore, if you work in an inertial frame then the accelerations of each object genuinely are independent of their mass. It's only if you adopt the (accelerating) rest frame of one object that the other's acceleration depends on its mass.

So, in short, there are (usually unwritten) caveats to the "gravitational acceleration is independent of mass" claim. It's not wrong, but it has some assumptions about how you are doing the measuring of acceleration.

Thank you!

 

[256bits]

Thanks for the awnser.
But why not? Isn’t the force on m1 the same as on m2 because of Newton’s third law?

Correct.

Take this case, something more familiar maybe.
Two people are pulling a rope - one to the left and one to the right so that the rope is in tension.
Each exerts a force F on their end of the rope to have no acceleration.
The 'total' force does not become 2F.
The force on the left person from the rope is F.
The force on the rope by the left person is F, but of the opposite sign, or direction.
The force on the right person from the rope and on the rope by the right person follow the same statics.
Nowhere does a force of 2F come into play.

 

[vanhees71]

First of all forces are vectors. The gravitational interaction between two forces is given by the force on particle 1,
$$\vec{F}_{12}=-\frac{G m_1 m_2}{|\vec{r}_1-\vec{r}_2|^2} \frac{\vec{r}_1-\vec{r}_2}{|\vec{r}_1-\vec{r}_2|} = - G m_1 m_2 \frac{\vec{r}_1-\vec{r}_2}{|\vec{r}_1-\vec{r}_2|^3}.$$
The force on particle 2 is
$$\vec{F}_{21}=-\vec{F}_{12},$$
in accordance with Newton's 3rd Law.

The equations of motion are
$$m_1 \ddot{\vec{r}}_1=\vec{F}_{12}, \quad m_2 \ddot{\vec{r}}_2=\vec{F}_{21}.$$
Now, because $\vec{F}_{12}+\vec{F}_{21}=0$, it's a good idea to introduce the center-of-mass coordinates
$$\vec{R}=\frac{1}{m_1+m_2} (m_1 \vec{r}_1+m_2 \vec{r}_2).$$
Using the equations of motion leads to
$$\ddot{\vec{R}}=0.$$
The center of mass is thus moving with constant velocity.

To describe the motion further it's obviously a good idea to introduce the relative position vector $\vec{r}=\vec{r}_2-\vec{r}_1$. Using $\vec{R}$ and $\vec{r}$ you can easily express $\vec{r}_1$ and $\vec{r}_2$ via these new coordinates, and the equation of motion for $\vec{r}$ turns out to be
$$\mu \ddot{\vec{r}}=-G m_1 m_2 \frac{\vec{r}}{|\vec{r}|^3}=\vec{F}(\vec{r}), \qquad (*)$$
where the "reduced mass" is defined by
$$\mu=\frac{m_1 m_2}{m_1+m_2}.$$
Thus the further solution of the equations of motion is reduced to the motion of one "quasi-particle" with the reduced mass $\vec{\mu}$ with the force given on the right-hand side of (*).

It describes, e.g., the motion of a planet and the Sun around their common center of mass. The solution of this remaining equation makes use of energy conservation, because the force can be derived from the potential
$$V(\vec{r})=V(r)=-\frac{G m_1 m_2}{r}\; \Rightarrow \; \vec{F}(\vec{r})=-\vec{\nabla} V.$$
Further it's a radial force, i.e., $\vec{F} \propto \vec{r}$, from which you get by taking the cross product of (*) with $\vec{r}$
$$\mu \vec{r} \times \ddot{\vec{r}}=\frac{\mathrm{d}}{\mathrm{d} t} (\mu \vec{r} \times \dot{\vec{r}}) = \dot{\vec{L}}=0.$$
This means that the angular momentum of the relative motion,
$$\vec{L}=\mu \vec{r} \times \dot{\vec{r}}$$
is conserved, which means (a) that the motion is in a fixed plane perpendicular to $\vec{L}=\text{const}$ and (b) it implies Kepler's 2nd Law: The radius vector $\vec{r}$ swipes out the same area per unit time. It's obviously valid for any central force.

The rest is a bit more complicated. Introducing polar coordinates for the motion you can derive Kepler's 1st and 3rd Law: The orbits of the planet and the Sun are ellipses with the center of mass in one of its foci. Since in our solar system the mass of the Sun is much larger than the mass of any of the planets, it's a good approximation to say that the planet goes around the Sun in an elliptical orbit with the Sun in one of its foci (Kepler's 1st law) and that $T^2/a^3=\text{const}$ for all planets in our solar system, i.e., the square of the orbital periods of all planets scales with the cubes of the semi-major axes of the elliptic orbits.

 

[PeroK]

Hello everyone.
Probably this question is trivial, but nevertheless I am confused about Newtons law of motion:
$$F=G\frac{m_1m_2}{r^2}$$

Now, some sources say, that F is the force between the two masses m1 and m2. Other sources say, that F is the force that m1 exhibits on m2. But isn’t this a contradiction? Because of Newton's third law, if F is the force between the two masses, m1 acts on m2 by a a force of F/2 and vice versa.
On the other hand, if F is the force of m1 on m2, the whole force between both particles is 2F.

There are two forces here that form a Newton's third-law pair. First, there is the gravitational force on $m_1$ exercted by $m_2$. This force is: $$F_1 = +\frac{Gm_1 m_2}{r^2}$$ where we will take the direction from $m_1$ to $m_2$ as positive. There is also the gravitational force on $m_2$ exerted by $m_1$, which is in the opposite direction (from $m_2$ to $m_1$): $$F_2 = - \frac{Gm_1 m_2}{r^2}$$ And, as expected, the forces are equal and opposite.

Note that if we consider the sum of these two "internal" forces on the two-body system of $m_1, m_2$, we find that: $$F_{net} = F_1 + F_2 = 0$$ and this is an example of the sum of all internal forces being zero, as implied by Newton's third law.

 

[Dale]

Thanks for the awnser.
But why not? Isn’t the force on m1 the same as on m2 because of Newton’s third law?

Because you never add the forces on two different systems. You only add the forces on a single system.

Also, you are forgetting that vectors add vectorially. If you consider the two objects to comprise a single system then the forces would add vectorially to 0. In other words, you are forgetting the “opposite” part of “equal and opposite” in Newton’s 3rd law.

There is just no sensible way to get a total force of 2F.

 

[Oliver321]

Thank you all, I think I do understand!

 

[etotheipi]

well I don't know if I understood the question but should be clear that for $(\alpha, \beta) \in \{ (1,2), (2,1) \}$ you have$$m_{\alpha} \ddot{\boldsymbol{x}}_{\alpha} \cdot \dot{\boldsymbol{x}}_{\alpha} = \frac{\gamma}{r_{\alpha \beta}^3}(\boldsymbol{x}_{\beta} \cdot \dot{\boldsymbol{x}}_{\alpha} - \boldsymbol{x}_{\alpha} \cdot \dot{\boldsymbol{x}}_{\alpha})$$with $r_{\alpha \beta} := |\boldsymbol{x}_{\alpha} - \boldsymbol{x}_{\beta}|$. add these two equations,$$\begin{align*}\frac{d}{dt} \left( \frac{1}{2} \sum_{\lambda = 1}^2 m_{\lambda} \dot{\boldsymbol{x}}_{\lambda}^2\right) &= \frac{\gamma}{r_{12}^3} \left( \boldsymbol{x}_2 \cdot \dot{\boldsymbol{x}}_1 - \boldsymbol{x}_1 \cdot \dot{\boldsymbol{x}}_1 - \boldsymbol{x}_2 \cdot \dot{\boldsymbol{x}}_2 + \boldsymbol{x}_1 \cdot \dot{\boldsymbol{x}}_2 \right) \\

&= \frac{-\gamma}{r_{12}^2} \frac{d}{dt} \left( \sqrt{(\boldsymbol{x}_2 - \boldsymbol{x}_1)^2} \right) \\

&= \frac{d}{dt} \left( \frac{\gamma}{r_{12}} \right)

\end{align*}$$this means the quantity$$\mathscr{E} := \frac{-\gamma}{r_{12}} + \frac{1}{2} \sum_{\lambda = 1}^2 m_{\lambda} \dot{\boldsymbol{x}}_{\lambda}^2$$is an integral of hte motion