## Gravitational Field and Intensity

Today topic is gravitational field and its intensity, You will learn in this post, what is field and how this field is created and what is strength or intensity of this field. This is new concept of gravitation and interesting,In previous topic you have learned in details about Gravitational force, that concept of point mass will be also use here to understand better about gravitational field, so you can refer first Gravitational force introduction. Now lets start and learn some new concept for gravitational field.

Gravitational field and electric field are similar, Gravitational field concept will apply in electric filed. Only difference is that in gravitational field mass will be source of field but in electric field charge will be source of electric field.

Gravitational field is always attractive in nature but electric field attractive as well as repulsive depend upon charge nature positive and negative. you will learn latter about electric field, here you will learn best concept for gravitational field.

### Gravitational field

See the picture above, Suppose a room is completely vacant, now you released a very small mass then that mass automatically start to moves towards a particular direction as shown in picture.

Now you were start thinking how this mass is moving, so that nothing is pushing the mass but it is moving, the thing which is invisible but effect are showing mass is moving, Then i told you that there is an imaginary field, Someone has made this field due to which this small mass is pulling, After that you remove the screen and found that Oh, There is a big mass, Now you clearly think that definitely big mass is pulling the small mass by law of gravitation.

Hence this big mass has created gravitational field. Since the big mass will not seen but its effect you seen this is the imaginary field. This is a hypothetical concept story imaginary field but in actual not like this happen.

It is important if there is a field then there will be a source for that field, In this case big mass is source for gravitational field. small mass on which you do experiment is called test mass .

Condition for test mass is limit lt m→0 .

Test mass is taken small because it can also create its own field which will disturb main source field, which will create inaccuracy for field strength measurement.

Hence gravitational field is a space around a body, in which gravitational force can be experienced. There are two method to apply forces one is through contact which are called contact forces and other is without contact which acts in nature, which are called field forces.

Now field do two important work, first whenever any mass body comes in field region then it inform the source mass that someone has entered in the field and second it attract that entered mass in its field. for practical example when a criminal live in a particular area then he create a field in that area, people of that area fear from him, whenever he moves people don't want to see and meet him.

When other criminal enter in this area then first criminal came to know that somebody has entered in my area and start fighting and small criminal finally captured by big criminal like big mass overcome on small mass.

### Gravitational field intensity/strength

Gravitational field strength is define as force per unit mass Eg = F/m, understand the concept field is not a measurable quantity, actually field strength is measured, field is just a feel, which a small mass experience due to a big mass field created.

Gravitational field is a vector quantity so it has also direction hence the direction of field will be same as direction of force acting on small mass ,If you want to measured the field strength, then just put a small mass at that location and check the force experienced by that small mass and just divide that force with small mass you will get the strength of the field at that location.

→ →

Mathematically gravitational field is express as Eg = F/m , Now see the picture below to understand better concept.

See the above first picture the Earth and same mass of 1kg experienced different forces of attraction by the Earth at different location, But question is why ? the same mass of Earth and the same mass of 1kg but different location, 1kg mass experienced different force.

You will get the answer of why wait ? but first you notice that gravitational field depend upon the location, As you see changing the location force value on 1kg mass is changing.

You know from previous post gravitational force between two masses F = GMem/r² .

now from above formula Eg = F/m = GMem/mr² = GMe/r².

Eg = GMe/r² N/kg which clearly shows that gravitational field depend upon r position as the distance will increase from the source mass Me field value will decrease.

Now you see second Earth image for answer why ?

Gravitational field work just like spring, see the image of ball near the Earth, Three spring are pulling that ball towards the Earth so its gravitational field strength at this location will be more. Similarly when the same ball is moved some distance above the Earth then see only two spring is pulling the same ball at different location, So here less field strength at this location.

Again move the same ball some more distance away from the Earth then see only one spring is pulling the same ball hence more less attraction, So here field strength will be more less, Hence in this way field strength varies according to location. Now i hope you have understand the concept of gravitational field and its strength.

#### Gravitational field intensity unit

As you have seen the definition of gravitational field strength Eg = F/m now put the unit of force in this equation, unit of force is kg*m/s² .

Eg = kg*m/s²*kg = m/s² wow this is unit of acceleration due to gravity g , So gravitational field strength is acceleration.

Hence remember field strength is always calculate per unit mass at any location in space.

#### Principle of superposition gravitational field

Principle of superposition state that if there is more than one mass at a location and you want to find the gravitational field strength at a point, due to all masses then vector sum of all fields due to each masses, this will give the total field strength at that point.

See the picture below, find field strength.

**How to find gravitational field strength at a point**

From the first question to find out gravitational field strength at point P due to mass m1 and mass m2, You will put a unit mass at point P and you will see, this unit mass is attracted by m1 and m2 as shown in figure.

Hence direction of gravitational field is clear by mass m1 and m2, Now you know that gravitational field is a vector quantity hence it will add as vector sum, So total field at point P will be like this.

→ → →

Eg = E1 + E2 at point P, You know magnitude of E1 = Gm1/r₁² , similarly magnitude for E2 = Gm2/r₂² .

Now if you want to write in vector form to sum the actual value of E1 and E2 then take component of E1 along x and y axis, Hence in vector we can write as.

→

E1 = -E1cos45i^ + E1sin45j^ ( minus due to negative direction of x component)

→

E2 = E2cos45i^ +E2sin45j^ now for total field at point

→ → →

Eg = E1 + E2 = (E2cos45 - E1cos45)i^ + (E1sin45+E2sin45)j^ , now put the magnitude of E1 and E2

→

Eg = (Gm2/r₂²*1/√2 - Gm1/r₁²*1/√2)i^ + (Gm1/r₁²*1/√2+ Gm2/r₂²*1/√2)j^

Now after putting all values of G, m1 and m2 you can find the total field strength at point P in vector form, for magnitude take square root of i^ and j^ vector.

Now for second question equilateral triangle, find field strength at point P centre of the triangle as shown in figure.

As you see in this case three masses of same magnitude are placed at vertices of the triangle. Now do the same process put a unit mass at the centre of triangle and analyse all the three masses will attract that unit test mass placed at the centre.

Hence the direction of field will clear to you as shown in figure. Now total field due to all three masses will be vector sum of three field .

→ → → →

Eg = E1 + E2 + E3 now it is given that all three masses are same and distance from centre to all masses are same hence field due to all masses will be also same without doubt. So E1 = E2 = E3 , now all three field are in different direction so it will cancel to each other, Hence net field at centre of the triangle will be zero.

→ → → →

So Eg = E1 + E2 + E3 = 0

You can prove it also first you take base side two masses see the x component magnitude are same and in opposite direction, so it will cancel. now its y component vertically downward and its magnitude 2Ecos60 = E (downward) and by third mass top vertices E upward hence both will cancel each other, So total field at point P at triangle centre will be zero.

Hence it is clear that whenever symmetrical mass distribution is given and asked to find the field at centre point of that geometry shape then field will be zero at centre point.

Now you can also check this for square and ring , put the same mass on the square corner and find the field at centre of square it will be zero try yourself this case.

### Gravitational field intensity is applicable for point mass only

You must remember that gravitational field intensity formula Eg = GM/r² is only valid for point mass, not valid for distributed masses, How you will find gravitational field strength for distributed masses. See below some question, Which will clear your concept.For first question Q1, suppose x distance from m field strength will be zero, then field due to m will be E1 = Gm/x², now field due to 4m will be E2 = G4m/(12-x)² , since field will be zero its mean that vector sum of both field will be zero.

→ →

E1 + E2 =0 or E1 = E2 hence equate both field magnitude

Gm/x² = G4m/(12-x)²

(12-x)²/x² = 4 or (12-x)/x = ±2 if take positive value then (12-x) = 2x or 3x = 12

x = 12/3 = 4 hence x = 4cm.

Now take negative value of x, then expression will be (12-x) = -2x or x = -12cm now see for this case x = -12 which will lie left of m at this point both the direction of E1 and E2 will be in right direction , so E1 and E2 will add, Hence this value is not possible.

Hence x = 4 value is lie in between them so final value of x will be x = 4 at this point field strength will be zero.

Now for second question Q2, Which is not a point mass it is distributed over the semi circle, here you can not apply directly point mass formula because it is distributed mass, so for this you can do one thing divide this semi circle in small dm mass then you can apply formula for point mass because dm mass will be very small like point mass.

Now see here Eg at a point mass dm as shown Eg = Gdm/R² now along y axis it is symmetric so its x component will cancel each other Ex = Gdmcosϴ/R² , but its y component will not cancel so Ey = Gdmsinϴ/R² , now to add this types of all small masses you have to integrate this small dm.

∫ dEy = ∫Gdmsinϴ/R² = G/R²∫sinϴdm , now for dm in terms of ϴ we can use mass per unit concept .

dm/Rdϴ = M/𝛑R or dm = Mdϴ/𝛑 now put this value in main equation

G/R²∫sinϴdm = G/R²∫sinϴ*Mdϴ/𝛑 = GM/𝛑R²∫sinϴdϴ , limit will be from 0 to 𝛑

𝛑

GM/𝛑R²∫sinϴdϴ

0 𝛑

Ey = - GM/𝛑R²⎡ cosϴ⎤ = - GM/𝛑R²(-1-1) = 2GM/𝛑R²

0

Ey = 2GM/𝛑R² direction will be vertically upward this will be gravitation field at point P.

Now i hope you have enjoyed to learn gravitational field and intensity, I will continue post with more best concept for you, I want your feedback, through comment, share and likes learn more and grow, thanks for sharing.

Dated 9th Dec 2018

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