'What is difference between free fall acceleration g and gravitational acceleration a?***a is with subscript g.In my textbook it is written that "free fall acceleration = gravitational acceleration - centripetal acceleration." so i'm confused if there is any difference between these two?
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$\begingroup$ Your original question attracted correct answers. Rather than continuing to edit this question that already attracted correct answers to what you wrote: I would suggest reverting this question back to the original version. Then ask a whole new question where you put centripetal acceleration into the title and possibly in the tags as well and specifically state (in the new question) that you want to know about about "how the rotation of the earth and the acceleration due to gravity combine to make an effective inertial force" and feel free to use that exact phrase in the new question. $\endgroup$– TimaeusJun 9, 2015 at 21:15
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$\begingroup$ Oh, and even after you revert this one back to the original, the software should allow you to find this version from the edit log and get the source and then you can copy that to your new question. Your new question is probably going to be classified as homework, so read the policy on asking homework like questions. $\endgroup$– TimaeusJun 9, 2015 at 21:19
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3 Answers
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In the second law of Newton appears the acceleration $a$. It refers to a generic acceleration due to any phenomenon. $g$ has the same role of $a$, but it refers specifically to the acceleration of gravity (free fall particular case) on the Earth. Usually we approximate $g$ to be constant $\left(9.81\, \mathrm{m}/\mathrm{s}^2\right)$, but in the real case the value of $g$ change at different altitudes (as mentioned by Edward Newgate).
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$\begingroup$ Your textbook is considering a particular case: an object that is orbiting around the Earth. You have to decompose the motion in 2 singular motions: 1) the falling motion due to the gravitational force and 2) the circular motion. 1) is forced by the gravitational force and 2) is forced by the centripetal force. So the resulting motion is the sum of 2 motions. The total force is the sum of these 2 forces: F_total = F_gravitational - F_centripetal (these 2 forces lie on the same axe, but they are opposite). Forces are linear in acceleration, so the same relation is true for accelerations. $\endgroup$– mm5Jun 9, 2015 at 16:12
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$\begingroup$ Here the total acceleration is called free fall acceleration, but this is only your textbook convention. $\endgroup$– mm5Jun 9, 2015 at 16:16
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Newton devised a very good law of gravity (until Einstein came along) where the force between the two bodies is scaled by a very small number usually written as a capital G. It's a general law that applies to any two bodies. But if you plug in the mass of the earth, the mass of a test ball, and the distance between the center of earth and the test ball, then plug in the experimentally determined G you will deduce the force on that test object. Next if you plug the calculated force on the test object along with its mass into another of Newton's laws F=ma and solve for a you will get the acceleration that the ball will experience at earths surface due to gravity. Since it is known that the calculated acceleration is independent of mass (it pretty much comes out the same for all test balls unless you do something ridiculous like plug in the moon but then you've got a radius issue) they give it a special symbol little g and give you the approximation 9.8 some odd meters per second per second. Edit: In the text book images above a seems to be the usual g and g is shown to include other accelerations. Another way you could complicate the equation is to consider the acceleration imposed by various other planets on the test ball. Also give it a charge and imagine some other charges pulling it in various other directions. Then you'd get an equation like mg=ma+fifty other things... But for almost all practical purposes g=a so you want to look at F=mg only. Here's an example of a situation that breaks that rule: in the equation above the extra term subtracts from ma resulting in smaller mg (weight) when you're lifting things it's easier to lift smaller dumb bells than larger ones right? Well in the space rocket industry easier equates to less fuel. which means less costs. So to maximize that second term they tend to build space centers nearer to the equator.
answered Jun 9, 2015 at 4:48
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I think you are confused between gravitational CONSTANT g, and gravitational acceleration a which can be thought of as a VARIABLE.
Gravitation acceleration g is around 9.81m/s^2 near sea level. But as you go higher the gravitational acceleration is no longer g, but another number, let's say a. a is a more generic gravitational acceleration that is not necessarily g.
Also, a is not only gravitational acceleration, but can be acceleration due to other phenomena (ex. Tension force).
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1$\begingroup$ No, $g$ changes depending on the planet or the reference frame. It is called gravitational field strength or gravitational acceleration. It is NOT limited to the surface of the Earth. $\endgroup$– Bill NJun 9, 2015 at 1:55