## Gravitational Field and Intensity

### Gravitational field

### Gravitational field intensity/strength

#### Gravitational field intensity unit

#### Principle of superposition gravitational field

**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.

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