Understanding Gravitational Force: A Simple Explanation

In summary, Professor gave out a hand out with formulas to calculate the gravitational force. The student was having trouble understanding how to calculate the force due to not knowing the difference between vectors and scalars. The student was able to understand the difference and then was able to plug in the information into the correct equation to get the final equation.
  • #1
Shaad
13
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:
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  • #2
Starting from the beginning, what is the first point where you become stuck, and why?
 
  • #3
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.
 
  • #4
If I understand your question, that is a defintion, you don't have to prove it.
 
  • #5
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:
  • #6
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?
 
  • #7
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.
 
  • #8
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.
 
  • #9
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)
 

Related to Understanding Gravitational Force: A Simple Explanation

1. What is gravitational force?

Gravitational force is a natural phenomenon by which objects with mass attract one another. It is one of the four fundamental forces of nature and is responsible for keeping planets in orbit around the sun, and objects on Earth from floating away into space.

2. How does gravitational force work?

Gravitational force is caused by the curvature of space and time created by massive objects. The more massive an object is, the stronger its gravitational force. Objects with larger masses have a greater ability to bend the fabric of space and time, causing other objects to be pulled towards them.

3. What is the formula for calculating gravitational force?

The formula for calculating gravitational force is F = G * (m1 * m2) / d^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and d is the distance between them. This formula was first discovered by Sir Isaac Newton.

4. How does distance affect gravitational force?

The force of gravity follows an inverse square law, meaning that as the distance between two objects increases, the force of gravity decreases exponentially. This means that the closer two objects are to each other, the stronger the gravitational force between them.

5. What are some real-life examples of gravitational force?

Some examples of gravitational force in everyday life include the Earth's gravitational pull on objects, the moon's gravitational pull on Earth's tides, and the gravitational pull between the sun and the planets in our solar system. Additionally, gravitational force is responsible for keeping our feet on the ground and for the formation of stars and galaxies.

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