
#1
Jun1104, 09:15 PM

P: 12

Hello,
Does anyone can tell me how to derivate the forumla for the acceleration due to gravity, i mean how to find the g of other planets like the moon?... 



#2
Jun1104, 09:36 PM

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Look at http://www.physicsforums.com/showthread.php?t=30396
The last post tells you how to derive g = MG/R^2 M : planet's mass, G : Universal Grav. Const R : planet's radius 



#3
Jun1204, 02:35 AM

P: 1,322

By the way, the term "acceleration due to gravity" is a misnomer, because it assumes g is an acceleration. It is not. The correct term, in my opinion, should be "gravitational field constant."




#4
Jun1204, 04:49 AM

P: 12

Acceleration due to gravity
Sorry, do u mean that using the equation above, the g of the moon or other planets can be solve?!...Actually, I still don't really understand...




#5
Jun1204, 05:08 AM

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P: 40,886

Start with Newton's law of gravity. Find the gravitational force on a mass (m) on the surface of a planet/moon (radius = R) due to the mass of that planet/moon (M). F = GMm/R^2. Does that make sense? Once you have the force due to gravity, the "acceleration" due to gravity is gotten via Newton's 2nd law: a = F/m, so: a = g = GM/R^2 Make sense? 



#6
Jun1204, 06:12 AM

P: 182





#7
Jun1204, 06:44 AM

P: 51

GM/R^2=V/R is gravitational potential (V) over that distance(R). I don't se why it would be accel. By the way, is every body in some gravity field if accelerating? 



#8
Jun1204, 06:50 AM

P: 2,955

Why don't you think that g is an acceleration? The term "field" as applied to both EM and Newtonian gravity refers to what happens to a free particle when a test particle is at a particular location. For gravity and EM its defined as F/m and F/q respectively. 



#9
Jun1204, 07:47 AM

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P: 40,886





#10
Jun1204, 10:26 AM

P: 12





#11
Jun1204, 10:50 AM

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P: 40,886





#12
Jun1204, 11:05 AM

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If you did not actually look at that post, here it is :
[tex]r=R+h [/tex] So, [tex]U =\frac {GMm} {r} = \frac {GMm} {R+h} = \frac {GMm} {R} (1 + h/R)^{1}[/tex] For h<<R, you can expand the last term binomially, and neglect terms of second order and up. So, [tex]U = \frac {GMm} {R} (1  h/R) =\frac {GMm} {R} + \frac {GMmh} {R^2}[/tex] [tex] = Constant + (m)*(\frac {GM} {R^2})*h = Constant + mgh [/tex] Since we are interested only in changes in potential, we can throw away the constant term. 



#13
Jun1204, 11:09 AM

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Thanks
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On the other hand "G", in [itex]F= \frac{GmM}{r^2}[/itex] is the "universal gravitational constant". Gokul43201's original response told how to calculate "g" for other planets. 



#14
Jun1204, 12:44 PM

P: 1,322

RE: "I disagree. The term "acceleration due to gravity" refers to the fact that if you place a point test mass at a particular location then a free particle will accelerate at a rate g."
Not if there are other forces acting on the particle, hence the problem. So defining g as the acceleration due to gravity causes confusion unless you stick the qualifier onto the definition, which is awkward. If I had a dime for every student who had an object accelerating at g when it wasn't supposed to, I would be a rich man. And students make this mistake because we use confusing terminology. Defining g "the acceleration due to gravity" is not really wrong, but unfortunate in my opinion. But I think some definitions are more practical than others. The constants g and E are directly analagous. One indicates the force that would act on a mass if the mass was placed at that point in space. The other indicates the force on a charge if placed at that point in space. One we call the electric field. The other then is obviously the gravitational field. So it makes sense to call g "the gravitational field constant." When an object is placed at a point in space where g is defined, it may or may not accelerate at a value coinciding with g depending on the value of the other forces acting on the object. 



#15
Jun1204, 03:03 PM

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If there are other forces involved then there will of course be accelerations due to them and the net acceleration would be the result of all the accelerations combined. 



#16
Jun1204, 06:30 PM

P: 1,322

RE: "The qualifier is already there in "due to gravity. That's what it means; that it is that part of any acceleration the object undergoes (if any), that is attributed to gravity."
A ball dropped in molasses does not accelerate at g. To say that it accelerates downward at g, and that this motion is partially cancelled by an acceleration upward, makes no intuitive sense. Sure it would work mathematically, but that doesn't make it sensible. However, the Earth apply a gravitational force down on the ball and the molasses can apply a separate force upwards. But there is only one resulting motion. 



#17
Jun2504, 01:13 AM

P: 18

If you have trouble getting the planet's weight then you could still use another notsopractical method to measure "g". You can get on a rocket on that planet or celestial body you want to measure "g" at. Start the engines and accelerate until your velocity (Vel) is such that you manage to scape from the gravitational field of that planet or celestial body. You might have to do this several times until you get it right... =(
Then: g = (Vel^2)/(2*R) 



#18
Jun2504, 10:39 AM

P: 698

calling it 'acceleration due to gravity' is a correct interpretation, if you've ever studied general relativity, there is no difference between a gravitational field force and acceleration. An observer cannot distinguish between the two. Therefore making gravity and acceleration the same thing.



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