Understanding Gravitational Force: A Simple Explanation

Click For Summary

Discussion Overview

The discussion revolves around understanding the gravitational force, particularly the mathematical formulation and the concepts of vectors and scalars involved in Newton's law of gravitation. Participants seek clarification on the derivation of formulas and the distinction between different types of quantities in physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in understanding the gravitational force as presented in a handout and requests a simpler explanation.
  • Another participant asks for clarification on the specific point of confusion regarding the mathematical representation of gravitational force.
  • A participant asserts that the relationship between the masses and distance is a definition rather than something to be proven.
  • There is a discussion about the correct formulation of the gravitational force, with one participant indicating that the formula presented by another is incorrect due to missing elements such as parentheses and vector considerations.
  • One participant attempts to clarify the vector representation of the gravitational force and explains how to derive the force vector from the positions of the masses.
  • Another participant shares their confusion regarding the output of a specific vector equation and seeks further explanation.
  • A participant explains the concept of unit vectors and how they relate to the direction of the gravitational force.
  • There is a clarification on the difference between scalars and vectors, with an emphasis on the representation of quantities in physics.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and disagreement regarding the mathematical formulations and concepts involved. Some participants agree on the definitions and the need for clarity in vector representation, while others challenge the correctness of specific formulations and seek further understanding.

Contextual Notes

Participants express uncertainty about the derivation of certain formulas and the application of vector mathematics in the context of gravitational force. There are unresolved questions about the specific steps needed to arrive at the correct expressions.

Shaad
Messages
13
Reaction score
0
Hi Guys... This topic was discussed in the class. Professor gave out the hand out. I tried to pay attention but couldn't follow it.

Can someone explain this to me in simple English? I'm having hard time understanding this.

I'm attaching a handout that I scanned.

http://img115.imageshack.us/img115/4284/gravitationalforce.jpg
 
Last edited by a moderator:
Physics news on Phys.org
Starting from the beginning, what is the first point where you become stuck, and why?
 
I understand the Newton's law. I just don't know how to prove, that if the mass m1 and m2 are the points located at p1 and p2, the distance of those two points are equals r. I don't know how to prove this.
 
If I understand your question, that is a definition, you don't have to prove it.
 
Right. I understand. This is not a theory which i have to prove (if this then that).

But I have (Gm1m2 / r^2 )

then i have to prove F = (Gm1m2 / p1 - p2) * (p1 - p2)

In the hand out, it says # 3 and shows the above formula. How did they get it? (and rest of the formula) that's what i want to know...
 
Last edited:
Shaad said:
Right. I understand. This is not a theory which i have to prove (if this then that).

But I have (Gm1m2 / r^2 )

then i have to prove F = (Gm1m2 / p1 - p2) * (p1 - p2)

In the hand out, it says # 3 and shows the above formula. How did they get it? (and rest of the formula) that's what i want to know...

That formula you wrote out there is not nearly correct. You are missing parentheses, you are missing a third power, and you are missing the most important factor: vectors!

Answer yourself these questions. Draw the points P1 and P2 on paper (arbitrary positions) and then try to draw it out:

What is
[tex]\vec{P_1} - \vec{P_2}[/tex]
(Hint, it is still a vector!)

Then what is
[tex]\left| \vec{P_1} - \vec{P_2} \right|[/tex]
(Hint, this is not a vector anymore!)

Finally, what is
[tex]\frac{\vec{P_1} - \vec{P_2}}{\left| \vec{P_1} - \vec{P_2} \right|^3}[/tex]

Do you now understand the third power also?
 
Sorry about my typing. I had no idea, how to code this formula and stuff and yes
I do know the difference between vectors and non-vectors (absolute value).

However, I have no idea the output of
[tex] \frac{\vec{P_1} - \vec{P_2}}{\left| \vec{P_1} - \vec{P_2} \right|^3}[/tex]



My question is how to get
[tex] {\vec F}({\vec p}) = - \frac { c } { \left | \vec {P} \right | ^3} } { \vec P }[/tex]

ps: wow it took me 20 min to type it. glad i got it though.
 
If you define the vector r as:
[tex]\vec{r} = \vec{P_1} - \vec{P_2}[/tex]

With length:
[tex]r = \left| \vec{P_1} - \vec{P_2} \right|[/tex]

Then your vector equation for the force becomes:
[tex]\vec{F} = \frac{G m_1 m_2 \, \vec{r}}{r^3}[/tex]

Then finally you have to think about the direction of the force. The direction of the gravitational force is along the vector r. It has the same direction as the vector r. You can also see that in the equation.
Because the direction of the force should have nothing to do with it's magnitude, you have to normalize the vector r:
[tex]\vec{e_r} = \frac{\vec{r}}{|\vec{r}|} = \frac{\vec{r}}{r}[/tex]
This is often called the unit vector in the direction of r. The length of this vector is 1, because we have divided the vector r by it's length (hence the name unit vector).

Then finally, plugging this into the equation, we get:
[tex]\vec{F} = \frac{G m_1 m_2}{r^2} \frac{\vec{r}}{r} = \frac{G m_1 m_2}{r^2} \vec{e_r}[/tex]
Which is simply the magnitude of the force (the first equation on your paper), pointing in the direction of r. To get the very last equation, you simply take P1 to be the origin, which is point (0,0,0) in three dimensions, or (0,0) in two dimensions. In each case, the vector P1 is now the zero vector. So:
[tex]\vec{P_1} - \vec{p} = \vec{O} - \vec{p} = -\vec{p}[/tex]
And because you take the magnitude of the vector, the minus sign does not matter.
 
Shaad said:
yes, I do know the difference between vectors and non-vectors (absolute value).

That's not the difference between vectors and non-vectors. The non-vectors Nick89 was referring to are scalars.

Scalar = a number
Vector = a list of numbers, usually to represent coordinates in some space

A vector can be used to represent many things. It could be a 3D point in that space. Or it could be a direction in that space, if you consider the direction between (0,0,0) and the given point. Or it could be a velocity, if you consider the direction and the length of the vector to be the speed.

ie,in Euclidean space [tex]v = (1,0,0)[/tex] is a vector pointing in the direction of X axis, or it is a point located on the X-axis.

[tex]||v||[/tex] = [tex]\sqrt{ v.x^2 + v.y^2 + v.z^2 }[/tex] = length of v = a number (scalar)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K