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Concept of Free Fall in question! 
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#1
Jun1114, 07:59 PM

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OK so let me apologies if I don't seem to fully understand the concept of free fall and am making a far fetched connection.
I've been recently reviewing some of my old course material such as vector calculations, Newtonian mechanics,etc, and while reading up on basic vector calculation the book made a reference to Newton's second law while setting an example for scaling a vector by a constant. Saying that you can change the length of a vector without necessarily changing its direction. That is B = cD. Now having read that, when I reached the topic of free fall, it occurred to me that although the mass is irrelevant in increasing or decreasing the objects acceleration, does it not even have an effect on the gravitational force? So to sum it up, do objects of different masses also feel the same gravitational force while in free fall? And if they don't feel the same gravitational force, how come they all accelerate at the same rate. ( I know that when an object acts as a projectile the gravitational force is in the opposite direction as acceleration until it reaches its maximum height, and has the same direction afterwards, but what about the magnitude of the gravitational force, doesn't it change with varies masses?) Once again my apologies if something is missing in my understanding of these concepts, will appreciate it if someone can clear things up for me! Thank you! 


#2
Jun1114, 08:23 PM

Sci Advisor
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There's FAQ at the top of this forum: http://www.physicsforums.com/showthread.php?t=511172 


#3
Jun1114, 11:27 PM

P: 27

In other words you are asking if F is different how can g be same? You know here F = mg or g = F/m. So you see, as m changes F also changes certainly, but their ratio will always be the same. Thus, objects of different masses feel different gravitational forces but they accelerate (free fall) at same rate. EDIT: Here I am considering the free fall of a small object (of small mass, compared to the mass of earth) under Earth's gravitational force. The above link gives a more general discussion. 


#4
Jun1214, 04:13 AM

P: 3,085

Concept of Free Fall in question!
What Nugatory said.
The force F = Gm_{1}m_{2}/r^{2} If the small mass is m_{2} then the acceleration of m_{2} is... a = F/m_{2} (from Newtons f=m_{2}a) So a = Gm_{1}m_{2}/r^{2}m_{2} m_{2} cancels so a is independent of m_{2} a = Gm_{1}/r^{2} 


#5
Jun1414, 07:52 AM

P: 32

The mass is not irrelevant, but the difference in results in this situation are negligable and so are ignored, simplifying the maths.
Consider the equation for (total ) acceleration (a) between two bodies : a = ( G * ( m1 + m2 ) ) / d ² To keep the maths easy, and exaggerate the difference, let : G = 1 m1 = 100 d = 10 For experiment 1, let m2 = 1 So, a = 1.01 For experiment 2, let m2 = 2 So, a = 1.02 So, altering the mass of m2 does alter the total acceleration rate. If you apply this to the earth and variable masses the values for a are all but identical, but do exist. Comments ? deanbarry365@yahoo.com 


#6
Jun1414, 08:07 AM

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P: 3,412




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