# Derivate the forumla for the acceleration due to gravity

1. Jun 11, 2004

### cutesoqq

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. Jun 11, 2004

### Gokul43201

Staff Emeritus
3. Jun 12, 2004

### 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."

4. Jun 12, 2004

### 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...

5. Jun 12, 2004

### Staff: Mentor

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?

6. Jun 12, 2004

### TeV

Yup,if the planet is a perfect sphere with radially symmetric mass distribution (density function in spherical coordinate system)

7. Jun 12, 2004

### dedaNoe

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?

Last edited: Jun 12, 2004
8. Jun 12, 2004

### pmb_phy

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.

9. Jun 12, 2004

### Staff: Mentor

What does that statement mean?
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.

10. Jun 12, 2004

### 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?!

11. Jun 12, 2004

### Staff: Mentor

Yes. This of course assumes a simplified model of the moon (or planet) of being spherically symmetric. (As TeV explained.) But close enough!

12. Jun 12, 2004

### Gokul43201

Staff Emeritus
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.

13. Jun 12, 2004

### HallsofIvy

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.

14. Jun 12, 2004

### 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.

15. Jun 12, 2004

### Janus

Staff Emeritus
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.

16. Jun 12, 2004

### 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 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. Jun 25, 2004

### jvicens

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)

18. Jun 25, 2004

### 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.

19. Jun 25, 2004

### 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.

20. Jun 25, 2004

### Nenad

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

21. Jun 27, 2004

### 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.

22. Jun 27, 2004

### zoobyshoe

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?

23. Jun 29, 2004

### 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)."

24. Jun 29, 2004

### zoobyshoe

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

25. Jun 29, 2004

### 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].

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.

Last edited: Jun 29, 2004
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