Role of gravitational force F=GMm/d^2 b/t two objects?

  • 91 Replies
Re: Role of gravitational force F=GMm/d^2 b/t two objects?
« Reply #90 on: September 03, 2020, 02:15:31 PM »
Iím still wanting to make a real post but until then, is it just me or is it supposed to be r2 not d2?
It can be either, as long as you know what it stands for, and I have seen it presented in many ways.
Some people use r for radius as that works nicely for an orbit and is consistent with F=m omega^2 r=m v^2/r.
Others use d for distance, sometimes to not confuse it with the radius of the object.

What is important is the meaning, not the variables used to represent them.

And if you want to get into the more technical level of how to represent them, as they represent physical quantities they should be in italic font, and then because they are both masses, both should be represented by a lowercase, italic m, with superscripts to distinguish between them.
e.g. F = G m1 m2 / r2

Re: Role of gravitational force F=GMm/d^2 b/t two objects?
« Reply #91 on: September 05, 2020, 10:31:29 AM »
In many ways, this is quite analogous to electrostatic interactions.
One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss. 

G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2,

Considering “g” separately from force F as g = Gm1/d^2, means G is considered separately from force F which is impossible. Therefore I would say “G” is wrongly placed in g=GM/d^2.

Two objects are must for the presence of gravitational force “F” [F=GMm/d^2, where g=GM/d^2] and gravitational constant “G”. Gravitational force “F“ should generate gravitational acceleration “g” but “g” which is equal to GM/d^2 independent of second mass (falling mass) and gravitational force “F”. 

As said, gravitational constant “G” is wrongly placed in g=GM/d^2 as explained above.

If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.

Still stuck on this I see.

Seems the only hope is for you to start plugging some numbers into the equations and see what happens.

I suggest this.

1.  Open up a spreadsheet as it’s the easiest way to see a bunch of different values side by side.  (Download Open Office if you don’t have one).

2.  First Row:  Mass m.  Write a bunch of  different masses for the falling object.  Perhaps  1kg, 10kg, 100kg, etc but it doesn’t matter.

3. Second row:  Force F.  Use F=GMm/d^2 with M as mass of Earth, d as radius of Earth (for gravity at sea level) and m taken from above row.

4.  Third row:  Acceleration a.  Use a = F/m using the numbers from  rows 1 and 2.

5.  Report back.  What do you see?
« Last Edit: September 05, 2020, 10:45:15 AM by Unconvinced »