Can we equate mg with the gravitational force formula GMm/h²?

[amjad-sh]

The force of gravity is F=mg.We know that that the gravitational force between Earth and object is F=MmG/h2(sqared).Can we say that mg=MmG/h2(sqared)?

 

[Orodruin]

No, first of all F=mgh is not true as you have mixed the gravitational force with the gravitational potential energy. Second, you are mixing two things. E=mgh is an approximation which is valid as long as the gravitational field g is homogeneous. If you are working at the scale where F=MmG/r^2, this is no longer the case. In fact, the gravitational field is then g = MG/r^2.

 

[amjad-sh]

I know,I edited the question.

 

[Orodruin]

Still no. The h means different things in your different equations. The r in F = MmG/r^2 is the distance from the center of a spherical mass distribution, in the homogeneous approximation it is just the vertical distance from some arbitrary reference level. It is not the same h as appears in mgh for the potential energy.

 

[amjad-sh]

Yes, but how the both equations refers to the term "gravitational force" ?

 

[Orodruin]

The gravitational force is given by $\vec F = m\vec g$, where $m$ is the mass of an object and $\vec g$ is the gravitational field. If $\vec g$ is constant, this results in the potential energy being given by $mgh$, where $h$ is the vertical distance from an arbitrary reference level and $g$ the magnitude of $\vec g$. That the gravitational field is constant is a good approximation for relatively small systems, i.e., in your daily life.

When you start looking at things on a larger scale, this is no longer true. Instead, the gravitational field will be given by
$$
\vec g = -\frac{GM}{r^2}\vec e_r
$$
where $G$ is Newton's gravitational constant, $M$ the mass of the gravitating body, $r$ the distance to the body's center, and $\vec e_r$ a unit vector pointing towards the body's center. The potential energy will no longer be given by $E = mgh$.

 

[amjad-sh]

Ok,now let's take the approximated equation for which F=mg.
If we take the system as (earth+stone),the Earth will make a force F=mg on the stone and the stone will react by the same force but in opposite direction on the earth.
Now put the stone on a table(the system is now:earth+table+stone).Does this cancel the reaction force of the stone on earth?

 

[Drakkith]

Now put the stone on a table(the system is now:earth+table+stone).Does this cancel the reaction force of the stone on earth?

Certainly (as long as 'cancel' means 'opposes'). The table exerts a force on the stone, keeping it from accelerating downward, but it also exerts a force on the Earth, keeping it from accelerating upward.

 

[FactChecker]

The force of gravity is F=mg.We know that that the gravitational force between Earth and object is F=MmG/h2(sqared).Can we say that mg=MmG/h2(sqared)?

Yes. They are the same. For G = 6.67384e-011N.(m/kg)², M = mass of Earth = 5.97219e+024 kg, h = radius of spherical Earth = 6,371,000 meters, we get F = 9.8196 * m. So it gives the correct acceleration of gravity (9.8196 m/s²) at the Earth's surface, ignoring centrifugal acceleration. (see https://en.wikipedia.org/wiki/Gravity_of_Earth section "Estimating g from the law of universal gravitation")

 

[amjad-sh]

Certainly (as long as 'cancel' means 'opposes')

.
I meant by "cancel" that the force will disappear temporarily until the table is removed,because if this doesn't happen the summation of the internal forces will not end up to zero and this is impossible according to Newton's third law.

 

[Orodruin]

No, there is still a gravitational force on the Earth from the stone, it is a force pair with the gravitational force from the Earth on the stone, as required by the third law.

 

[jtbell]

if this doesn't happen the summation of the internal forces will not end up to zero

How do you figure that?

 

[amjad-sh]

How do you figure that?

As showed in the figure if you add the internal forces,all will be canceled except the normal force will remain.
[F earth/stone +F stone/earth +F earth/table +F table/earth +normal force of the table (due to the interaction with the stone)]is not equal to zero.

 

[amjad-sh]

No, there is still a gravitational force on the Earth from the stone, it is a force pair with the gravitational force from the Earth on the stone, as required by the third law.

Yes, but this will lead that the summation of internal forces is not zero.

 

[jtbell]

Let's call your "normal force" F_stone/table. By Newton's 3rd law, there is also an F_table/stone which is equal in magnitude and opposite in direction.

 

[amjad-sh]

Let's call your "normal force" F_stone/table. By Newton's 3rd law, there is also an F_table/stone which is equal in magnitude and opposite in direction.

YES, YOU ARE RIGHT!
I don't know how I missed it haha.
thanks all .