# 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

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

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