# Acceleration due to gravity

by cutesoqq
Tags: acceleration, gravity
 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?...
 Emeritus Sci Advisor PF Gold P: 11,155 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
 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."
 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...
Mentor
P: 41,323
 Quote by cutesoqq 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...

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?
P: 182
 Quote by Doc Al so: a = g = GM/R^2 Make sense?
Yup,if the planet is a perfect sphere with radially symmetric mass distribution (density function in spherical coordinate system)
P: 51
 Quote by Doc Al a = g = GM/R^2 Make sense?
Not quite.
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?
P: 2,954
 Quote by JohnDubYa 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."
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.

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.
Mentor
P: 41,323
 Quote by dedaNoe Not quite. GM/R^2=V/R is gravitational potential (V) over that distance(R).
What does that statement mean?
 I don't se why it would be accel. By the way, is every body in some gravity field if accelerating?
The term "acceleration due to gravity" means the acceleration a test mass would experience just due to the gravitational force. (See pmb_phy's post above.) It is a measure of the strength of the field.
P: 12
 Quote by Doc Al Short answer: yes. 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?
Do u mean that the equation is available for any planets?! Then, I must put G be the universal gravitational constant, M be mass of the moon as well as R be the radius of the moon, right?!
Mentor
P: 41,323
 Quote by cutesoqq Do u mean that the equation is available for any planets?! Then, I must put G be the universal gravitational constant, M be mass of the moon as well as R be the radius of the moon, right?!
Yes. This of course assumes a simplified model of the moon (or planet) of being spherically symmetric. (As TeV explained.) But close enough!
 Emeritus Sci Advisor PF Gold P: 11,155 If you did not actually look at that post, here it is : $$r=R+h$$ So, $$U =\frac {-GMm} {r} = \frac {-GMm} {R+h} = \frac {-GMm} {R} (1 + h/R)^{-1}$$ For h<
Math
Emeritus
Thanks
PF Gold
P: 39,353
 Quote by JohnDubYa 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."
Are you talking about "g" or "G"? The "g" in f= mg is definitely an acceleration: it is the acceleration of any object on the surface of the earth: approximately 9.81 m/s2.

On the other hand "G", in $F= \frac{GmM}{r^2}$ is the "universal gravitational constant".

Gokul43201's original response told how to calculate "g" for other planets.
 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.
Emeritus