Derivate the forumla for the acceleration due to gravity

[cutesoqq]

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

 

[Gokul43201]

Look at https://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

 

[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."

 

[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...

 

[Doc Al]

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...

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?

 

[TeV]

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)

 

[dedaNoe]

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?

 

[pmb_phy]

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.

 

[Doc Al]

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.

 

[cutesoqq]

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?!

 

[Doc Al]

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!

 

[Gokul43201]

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<<R, you can expand the last term binomially, and neglect terms of second order and up. So,

U = \frac {-GMm} {R} (1 - h/R) =\frac {-GMm} {R} + \frac {GMmh} {R^2}
= Constant + (m)*(\frac {GM} {R^2})*h = Constant + mgh

Since we are interested only in changes in potential, we can throw away the constant term.

 

[HallsofIvy]

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.

 

[JohnDubYa]

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.

 

[Janus]

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.

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.

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.

 

[JohnDubYa]

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 canceled 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.

 

[Nenad]

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.

 

[JohnDubYa]

RE:"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."

Sure there is. When I have a box sitting on a desk, it is located inside a gravitational field. But it is not accelerating.

Besides, explanations that resort to general relativity are of little practical use to a person studying purely classical physics.

RE: "If you have trouble getting the planet's weight then you could still use another not-so-practical 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)"

This equation assumes that the spaceship does not develop any thrust after time t = 0, right? A more apt example would be to use a cannon ball instead of a rocket ship.

 

[Nenad]

yes, general relativity is of little use to clessical physics, but it is a lot more accurate.

 

[JohnDubYa]

RE: "yes, general relativity is of little use to clessical physics, but it is a lot more accurate."

If accuracy is what you desire then we should drop Newton's second law altogether and use general relativity to solve classical physics problems.

Using the term "acceleration due to gravity" is misleading and causes students to miss problems they normally wouldn't. Calling g the gravitational field constant does not, in my experience, lead students to assign accelerations of 9.8 m/s/s to bodies that are not in free fall.

 

[zoobyshoe]

Using the term "acceleration due to gravity" is misleading and causes students to miss problems they normally wouldn't. Calling g the gravitational field constant does not, in my experience, lead students to assign accelerations of 9.8 m/s/s to bodies that are not in free fall.

So, if you don't use the acceleration due to gravity to define and describe the force that keeps everything stuck to the planet's surface, what do you use?

 

[JohnDubYa]

I am not sure about your question, but here goes:

We define a gravitational field at every point in space surrounding a planet, such as Earth. This field is a vector and points towards the center of the planet. We denote the magnitude of this field as g. The purpose of this value is to indicate the magnitude of the gravitational force that would exert on any mass placed at that point,

F_g = mg.

If there are no other forces acting on the mass, then

\sum F = F_g = mg = ma,

g = a,

and so in this situation (free fall) the mass accelerates at a value that corresponds numerically to g.

Now there is no confusion. Since you are not calling g an acceleration, students are less likely to assign an acceleration of 9.8 m/s/s to objects that are not in free fall. They consider g merely a constant. Objects in free fall happen to accelerate at a value that agrees with that constant.

I have used this approach and found it very effective. In fact, in those situations where a student assigns g to non-freely objects, I ask them "Where did you get the idea that the object was falling at 9.8 m/s/s?" Their answer is invariably "Because that is the acceleration due to gravity, as it says in the book (or in their last class)."

 

[zoobyshoe]

I guess I'm not clear about the situations where the students misapply 9.8 m/sec².

 

[robphy]

In my experience, I see lots of students apply "g" as the acceleration down an incline, the acceleration in an Atwood machine, etc... possibly because these masses involved do move "due to gravity" [although the value of acceleration is not only due to the force of gravity].

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 agree with JohnDubYa that "acceleration due to gravity" is a poor term.
A better term might be "freefall acceleration".

As he points out above in a recent post in this thread, the g appearing in the definition of weight is identified as an "acceleration" only after using Newton's Second Law.

The best term is probably "gravitational field [of the earth]" in complete [classical] analogy to "electric field [of a point charge]".

F_{Grav}=m\frac{GM}{r^2}=mg
F_{Elec}=q\frac{kQ}{r^2}=qE

Like \vec E, the "gravitational field" \vec g is a radial vector field, which is approximately constant in a small enough region of space. As has been discussed above, the magnitude g=9.8 m/s^2 when r is equal to the radius of the earth.

One can continue to find analogies in defining a "potential" and a "potential energy". Indeed, one can apply Gauss' Law to both vector fields. Wouldn't the use of the "gravitational field" \vec g be a good introduction to next semester's electric field?

By the way, this does not mean that you have to introduce Newton's Law of Gravitation right away. Simply use the term "gravitational field" for "g". If you need to describe the "freefall acceleration" when first discussing kinematics, maybe a_{freefall}=9.8\ m/s^2 is a better notation. Some bright student might recognize that a_{freefall} has the same units and value as g (unlike E). You can use that as a hook to mention relativity, if you wish.

 

[JohnDubYa]

Freefall acceleration would certainly be better than "acceleration due to gravity" and light-years better than the God-awful "acceleration of gravity" (as if gravity accelerates).

As with robphy, I find calling g the gravitational field constant a nice way to introduce students to terminology they will find in the second semester.

By the way, this is not merely a matter of semantics. Words mean things. Poor terminology can turn a good student into a confused student, so instructors should take great care in phrasing physics concepts.

 

[Doc Al]

As with robphy, I find calling g the gravitational field constant a nice way to introduce students to terminology they will find in the second semester.

By the way, this is not merely a matter of semantics. Words mean things. Poor terminology can turn a good student into a confused student, so instructors should take great care in phrasing physics concepts.

I like it and will begin using it immediately! Thanks, JohnDubYa. (Yes, terminology does matter.)

I'll probably start calling g the "gravitional field strength/constant at the Earth's surface".

 

[zoobyshoe]

By the way, this is not merely a matter of semantics. Words mean things.

This is a disturbing couplet of sentences to read from someone concerned about avoiding confusion. "Words mean things." suggest that semantics has nothing to do with the meanings of words. But of course it does.

Poor terminology can turn a good student into a confused student, so instructors should take great care in phrasing physics concepts.

I agree, absolutely. Physics has specific and rigorous terminology in many cases, which helps to keep things clear. Once in a while it comes up that people aren't all on the same page about terms, though. To the extent that textbooks have perpetuated confusion, it should be corrected, but I'm not sure to whom you would appeal for that.

 

[chris_tams]

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."

Surely g is an accelleration due to gravity! How can it not be, when gravity causes something to accelerate at g. If it isn't an acceleration, how is it a field constant? It isn't, because g isn't constant throughout a field?

 

[ArmoSkater87]

cutesoqq, its this simple, g = Gm/r^2, u can think of g as the acceleration due to gravity of any planet. See for urself by pluging in the values for Earth, and u will get 9.8 m/s^2

 

[JohnDubYa]

RE: "Surely g is an accelleration due to gravity!"

This would imply that if an object is placed inside the Earth's gravitational field, it will accelerate at g. Not necessariy so, but the confused student will do this everytime.