## Gravitational potential energy

**Gravitational field.**

### What is gravitational potential ?

**Potential at a point**

**∞**) so potential at infinity is zero, now you are able the compare any new point potential respect to infinity potential. But remember point does not have energy but whenever you bring a mass near to that point then energy is stored in mass.

So it is clear that energy is available in mass but potential at point or at location indicate how much energy will be available at that location in the body or mass, which is bring to that point. You must remember every point has some potential, If you bring a mass then energy is stored in mass, If you don’t bring the mass then energy is not stored, but potential is always there.

You generally listen the peoples use potential word like, there is potential in this business, what do you understand by using this word ? Well that person is indicating about money in that business, Hence money is form of energy in Physics.

So potential indicate about energy of at point or location but when you bring a mass.

#### Gravitational potential due to a mass M

#### Derivation of potential due to mass at a point

Ans you know that due to mass m potential formula is given as V = -Gm/r , where r is distance from mass to point p.

So first calculate the distance AP = BP = CP , now from triangle APcos30 = a/2

APx√3/2 = a/2 or AP = a/√3 , Hence BP = a/√3, CP = a/√3 now potential due to single mass m at point P will be V = -Gm/(a/√3) = -Gm√3/a .

potential is scalar quantity so due to all three mass potential will be same at point P

Hence total potential at point P will be three times Vtotal = -3xGm√3/a = -3√3Gm/a .

Now from question Q2 above picture, You have to find the work done during moving small mass m from point A to point B.

Ans Here m<

VB – VA = (work done from A to B)/m.

Hence (Work done from A to B) = m(VB- VA) , VB = -GM/4x , VA = -GM/2x so put the value to find work done.

(Work done from A to B) = m(-GM/4x – (-GM/2x)) = mGM(1/2x – 1/4x) = mGM/4x

(Work done from A to B) = mGM/4x .

See the numerical question below.

As from question this is ring so it is not a point mass, but whatever formula derived for potential is only applicable for point mass.

Remember in this case you can’t assume the whole mass of ring concentrated at centre of ring, because it is extended mass and formula for potential is inversely proportional to distance.

Hence to solve the problem you have to divide ring mass in many small masses as shown dm and after that for addition of this all masses, you know just integrate whole masses.

So potential at point P due to dm mass will be ∫ dVp = -G∫dm/(**√R²+x²** )

Now here R and x are also constant and ∫dm = m

So Vp = -Gm/(**√R²+x²) .**

Now suppose if this point is very very far distance just feel then ring will look like a point mass, because radius will not look so R→0 so in this case Vp = -Gm/x .

Now what will be potential at centre of ring, gravitational field will be zero but potential will not be zero, at centre of ring put x =0 then you will get Vp = -Gm/R .

So in this way you can solve numerical problem, If there may be two ring separated with some distance and ask the potential at axis of ring you can find at any point due to both rings.

### Relation between gravitation field (E) and gravitational potential (V)

In previous post i have discussed in details about gravitational field and link is given above in first paragraph, you can refer. Gravitational potential discussed in this post, Now you will see the relation between these two quantity. This is important relation you must remember it will also help you to solve numerical problems.

→

Gravitational field E is a vector quantity and gravitational potential V is a scalar quantity.

Gravitational field is property of a Mass whereas potential is a characteristics of a point or location. Now the relation between gravitational field and gravitational potential is given as

→

E = -dV/dr where dV/dr indicate differentiation of gravitational potential with respect to position vector. You can also write Ex = -𝜕v/𝜕x, Ey = -𝜕v/𝜕y , Ez = -𝜕v/𝜕z partial differentiation of potential along x, y, and z axis because gravitational field is a vector quantity.

Potential is not along x,y and z direction because you know potential is a scalar quantity so it has no any direction. Now you can check the above formula.

You know potential at a point P, x distance due to mass M Vp = -GM/x.

use Ex = -𝜕v/𝜕x = -𝜕(-GM/x)/𝜕x = GM𝜕(x⁻¹)/𝜕x = -GM/x² Ep = -GM/x here negative sign in field indicate direction is towards the mass M centre.

→

Hence E = -dV/dr is correct formula.

See from the figure E a gravitation field, you know that in a field if you place a mass then gravitational force mE will act on mass towards the field direction as shown.

Now to move the mass m from point A to B, you have to apply very little greater force in opposite direction Fext (external force) -mE negative sign indicate opposite direction.

Now work done by external force Wext = force*displacement = -mEr , this work done bringing mass m from point A to B.

Again recall potential difference VB – VA = (work done A to B)/m , now put the value of work done, so you will get.

VB – VA = -mEr/m = -Er , Now if i told you A is very very close to B then r will be dr and VB – VA will be dV , hence you can write .

dV = -Edr or E = -dV/dr hence it proved.

I will continue post with more concept. I hope you have enjoyed learning gravitational potential energy. I want your feedback through comment, like and share, learn and grow thanks for sharing .

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