I'm in middle school and I was gone the day my teacher explained gravitational force and gravitational energy. Can someone tell me the difference between them?

  • $\begingroup$ Are you perhaps referring to potential energy when you say gravitational energy? $\endgroup$ – Kane Billiot Jan 5 '18 at 3:03
  • $\begingroup$ Potential and kinetic energy, yes. $\endgroup$ – Julieta Jimenez Jan 5 '18 at 3:05
  • $\begingroup$ So the main difference between force and energy is that force is the mass of an object multiplied by its acceleration, while energy is the force exerted on an object multiplied by the distance it traveled due to said force (this is also the definition of work). $\endgroup$ – Kane Billiot Jan 5 '18 at 3:10
  • $\begingroup$ I think that's the main idea you missed when you were out of class. $\endgroup$ – Kane Billiot Jan 5 '18 at 3:12
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    $\begingroup$ Ok, you're using two different terms. Gravitational energy is not a gravitational field. $\endgroup$ – Jakob Lovern Jan 5 '18 at 3:17

A force could be defined as a push or a pull. Fields can be thought of as all the possible forces between two objects when you vary one object's position. The canonical example in my mind is that of iron filings around a magnet; the way they align around the poles shows the direction of the magnetic force at that point. You'll eventually get to learn about vector fields and suddenly this will make a ton more sense.

Now, about force vs. energy:

Force, in general, can be thought of as mass * acceleration. If you push on a 4 kg object with 40 newtons of force, it'll accelerate at $10$ m/s$^2$. Now, energy is an effect of force, the joule is defined as N*m. In other words, energy is what happens when force moves things.


the gravitational field is the agency by which the gravitational force is transmitted. this field exists independent of the presence of masses in it which could experience the force.


From what I read,your concepts would be solidified if you could somehow visualise your perspectives regarding fields and energy,and fortunately the concept of field lines is just the way to do that.You can use the following page for a basic diagrammatic intuition.


Energy,regardless of the force nature,can be viewed as a value per say,(a scalar actually,which means the number assigned is independent of any directional parameters),which you can associate with a body depending on its position,orientation by virtue of its existence(think elastic bands and dropping apples).It is this value,which obeys a conservation law and is either stored as potential energy like,when you think of the work that can be done by a ball above ground level and kinetic energy which is being dissipated as heat as the ball falls down.The initial potential energy,i.e.the value of the number assigned,depends on how you view the object itself,or rather the frame of reference,like you could observe the ball falling from the ground itself or from the top of the Eiffel tower. In each case ,you'll see that you have to assign different numerical values along with change of sign(based on from where you measure height difference ).However the dissipated kinetic energy always depends on motion only,instead of orientation and hence it is always positive(it is proportional to the Square of the velocity).

Fields are an altogether different entity.Checked the diagram above in the link?

That's how Faraday understood it,a collection of arrows throughout space.It is quite helpful to think of it as such,even thought the arrows are fictitious themselves.Quantifying field is very hard physically in fact,but I'll try.

How I see it is as follows.A field is a quantity which determines the efficacy of a force at a given distance.You cannot talk of a field like you can about force,the question you could perhaps ask instead is:How strong is so-and-so field at a particular distance?

Example:You must know the formula for calculating gravitational force.Try calculating the force between yourself and the earth when

i) you are on the top of a mountain

ii)when you are on the moon.

The second force is much smaller. But ever wondered why that small force has no effect(you would not be falling through space towards the earth if and when you visit our satellite.)?

You can say that the intensity of Earth's gravitational field near the moon is lesser than at the top of a mountain.In Faraday's diagram you would see that the density of arrows per unit space is very low in such conditions of low field strength.The concept helps to find the source and sink of a force also depending on the nature of the force itself.Remember that arrows congregate near the sink and propagate outward from a source.To find the field parameters,divide the effective force by the quantifying parameter (mass of object observed for gravitational,charge for electric,etc etc).And remember that it's a vector,so take care of the directions.Also there's the small matter of permeability constants which depend on the nature of the space the field exists in.For your purpose,there are none for gravitational fields.If you knew vectors in detail,you could observe that a field is the gradient of potential,but that's for another time.

Hope this helped.


Force is a pull or push simply but in case of gravitational force it is simply a pull . But field is upto which extent gravitational force can be experienced.

  • $\begingroup$ I don't think this is an accurate characterization of the gravitational field. $\endgroup$ – ZachMcDargh Jan 5 '18 at 3:36
  • $\begingroup$ So.what according to you $\endgroup$ – Shashank Mishra Jan 5 '18 at 3:41
  • $\begingroup$ What you want exactly $\endgroup$ – Shashank Mishra Jan 5 '18 at 3:42
  • $\begingroup$ The gravitational field tells us the gravitational force on a hypothetical test particle. Since it does not even have the same units as force, it is not the extent to which force can be experienced or applied. $\endgroup$ – ZachMcDargh Jan 5 '18 at 3:48
  • $\begingroup$ You are telling mathematical answer. "field is upto which extent gravitational force can be experienced"you can not say wrong to it $\endgroup$ – Shashank Mishra Jan 6 '18 at 3:46

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