Gravitational field and field strength | Definition , formula , Derivation and solved examples

Gravitational field and field strength

Basically Gravitational field is the space around a mass or system of mass where any other mass ( test mass ) experiences gravitational force . Whenever any test mass is moved from one point to another point in gravitational field then gravitational force will be work on the particle in whole journey due to that , some work is also done by this gravitational force .

Mathematically , we can denote gravitational field by two physical quantities at any point . First one is a vector quantity, known as gravitational field strength or intensity of gravitational field and generally it is denoted by E . Gravitational field strength is the gravitational force (a vector quantity) experienced by a body of mass 1 unit in gravitational field . Sometimes in normal use we called it gravitational field .

The other physical quantity is gravitational potential. It is represented by V. This is related to the work done (a scalar quantity) by gravitational force in moving the test mass from one point to another point in the gravitational field .

Basically ” The force experienced ( both in magnitude and direction ) by a unit test mass placed at a point in a gravitational field is called the gravitational field strength or intensity of gravitational field at that point . Thus,

E= F/m

SI unit of E is N/kg .

Derivation of Intensity of Gravitational field

Let’s consider M is the mass of a body an it’s centre of gravity is at point O . Let F denote the gravitational force of attraction experienced by a test mass m , when placed at a point P in the gravitational field of body of mass M . As denote in the below figure the distance between the center of mass of the bodies of mass m and M is ‘ x ‘ .

Fig. 25.1

According to Newton’s law of gravitation, the force between the particle will be
F = GMmo/x²

As we discuss above that gravitational field strength is basically the gravitational force on body of unit mass in gravitational field . So , Intensity of gravitational field at point P will be

E = F/mo = [ GMmo/x² ]/mo = GM/x² … ( i )

In vector form we can write it as ,

E ( Vector ) = – Gm/x² X̂. …( ii )

Here the negative sign represent that the gravitational intensity is of attractive force . From Eq. (ii) , we can easily say that as we will increase the distance ( x increases ) then gravitational intensity will be decrease .

when x = ∞( infinity ) , E = 0.
It means that at the intensity distance from the body , value of gravitational field strength will zero .

If at point P , test mass is free to move then it will be accelerate due to the gravitational force of attraction F . Let acceleration produced of test mass due to the gravitational force is ” a ” ,
then

F = moa or a = F/mo …( iii )

From Eqs. (ii) and (iii) , E = a
It means that the acceleration of test mass placed at a point in gravitational field will be equal to the intensity of gravitational field at that point .

In case of earth , the gravitational field due to earth at point P that lies on (near) the surface of the earth , then x = R. Now

E = GM/R² = g

Here g denotes the acceleration due to gravity at the surface of earth .

Unit of intensity of gravitational field in SI system is Nkg-1 or ms-2 and in CGS system, it is dyne.g-1 or cm.s-2

Dimensional formula of gravitational intensity

E = F/mo = MLT-2/M = [ M⁰LT-2 ]

Gravitational field strength due to a point mass

Suppose, a point mass M is placed at point O . We want to find the intensity of gravitational field E at a point P, that is at distance r from point O , Magnitude of force F acting on a particle of mass m placed at P is

F= GMm/r²

So , for Intensity of gravitational force we have to find the force per unit mass .

E= F/m

or E = GM/r²

The direction of the force F and E will be same , that is from P to O .

Gravitational field strength due to a Hollow sphere

Figure 25.2 shows a hollow sphere of mass M and radius R. If we think about the gravitational field in its surrounding, the direction must be along the arrows shown in its surrounding . Every mass placed in its surrounding must experience the gravitational force toward the centre of the shell . Thus we can state that it is because of its symmetrical geometrical shape and its uniform mass distribution .

Fig. 25.2

Thus we can state that in the case of a hollow spherical shell we can consider that , its whole mass is concentrated at its centre when we have to find gravitational field at a point that located outside the body .

Thus the gravitational field strength at different points due to a hollow spherical shell can be given as shown in figure 25.3

Fig. 25.3

For outer points

gout = GM/x² [ Behaving as a point mass ]

For points on surface

gs = GM/R² [ Behaving as a point mass ]

For inner points

gin = 0 [As no mass is enclosed within it]

If we plot a variation graph for values of g with distance from centre, it is shown in figure 25.4

Fig. 25.4

Gravitational field strength due to a Solid sphere

As we see in case of hollow sphere such a way also In case of a solid sphere the direction of gravitational field at the nearby points is radially inward as shown in figure 25.5 . So in case of solid sphere also we can consider it to be a point mass for outer points .

Fig. 25.5

In case of solid sphere , the expression of gravitational field strength at the surface and at outer point is same as in case of hollow sphere .

For outer point

gout = GM/x²

For point on surface

gs = GM/R²

For a point that is inside the sphere , now gravitational field strength will be nonzero because there are mass content inside . To determine gravitational field at interior points at a distance x from its centre , we consider an inner sphere of radius x as

shown in figure 25.6 . Let the mass of this sphere of radius x is m , on the surface of which a point P exist where we wish to find the gravitational field , strength. The mass m is given as

m= [ M/(4πR³/3) ] × (4πx³/3)

m = Mx³/R³

Fig. 25.6

Now we can say that the given solid sphere is divided in two parts . First one is an inner solid sphere that have radius x and another is the outer shell that have outer radius R and inner radius x . As shown in above picture , here at point P gravitational field exist only due to the inner sphere and due to outer shell field will be zero . We’ve discussed in previous section ( in case of hollow sphere ) that, no gravitational field exist at interior points due to outer shell .

Thus , in net gravitational field strength at P we have to calculate field only due to the inner sphere of radius x , we can obtained it by considering the inner sphere of radius x as a point mass at the centre . So gravitational field at P will be

gin = Gm/x² = GMx/R³

In this case the graph of variation of g as a function of distance from centre of sphere is shown in figure 25.7

Fig. 25.7

Gravitational field due to a uniform circular ring at a point on its axis .

Figure 25.8 shows a ring of mass M and radius R placed in YZ plane with centre at origin . Here we wish to find the gravitational field strength at a point S on its axis at a distance x from its centre .

Fig. 25.8

To find this we consider an element of length dl on ring as shown in figure. The mass dm of this element can be given as

dm = M/2πR …( i )

Let the gravitational field strength at point S due to the element dm is dM then it is given as

dE = Gdm/(x² + R²) = GMdx/2πR(x²+R²)

This elemental gravitational field strength dE has two rectangular components , one along the axis of ring , that is dEcosθ and other will be perpendicular to the axis of ring , that is dEsinθ . Here when we integrate the result for the complete ring , dEsinθ component will be cancelled out by symmetry and dEcosθ will be summed up to give the net gravitational field strength at S .

Thus resultant gravitational field strength at point S is given as

E = ∫dEcosθ

= ∫ (GMxdx)/[2πR( x²+R²)3/2]
( limit of integration is from 0 to 2π )

E = GMx/(x²+R²)3/2

Fig. 25.9 Graph of variation of intensity of gravitational field for circular ring along axis

The direction of the electric field is towards the centre of the ring . It is zero at the centre of the ring and maximum when ,
r = R/(2)1/2 (can be obtained by putting dE/dr =0 )
Thus , E-r graph is as shown in Fig. 25.9

The maximum value is

Emax = 2GM/3(3)1/2

Solved examples

  • Problem 1. Two bodies of masses 1 kg and 100 are arranged such a way that the distance between these two particle is 1 m . At what distance along the line joining the resultant gravitational field intensity be due to both the masses will be zero?

Solution- Let there is a point P on the line joining , where gravitational intensity due to both the masses is zero . Let distance of point P from mass 1 kg is x . Let x be the distance of point P from the 100 kg body where the resultant gravitational intensity is zero .
For this condition , gravitational field intensity at P due to body of mass 1 kg should be equal in magnitude and opposite in direction of due to body of mass 100 kg.
Hence , we can write

G×1/x² = G×100/(1-x)²

100x² = (1-x²)

10x = 1-x

11x = 1 or x = 1/10

Means at distance 1/10 m from body of mass 1 kg , Gravitational intensity will be zero .

  • Problem 2 . A uniform sphere of mass ‘ M ‘ and radius ‘a’ is placed directly above a uniform ring of mass ‘ m ‘ and of equal radius . The centre of the ring is at a distance ( 3 )1/2a from the centre of the sphere . Find the gravitational force applied by the sphere on the ring .

Solution- The gravitational field at any point on the ring due to the sphere will be same if we consider solid sphere as a point mass of mass M placed at the centre of the sphere . Thus, the force on the ring due to the sphere is also equal to the force on it by a particle of mass M placed at that point . According to Newton’s third law ( action – reaction ) force due to particle on ring will be equal to the force on the particle by the ring . Now the gravitational field due to the ring at a distance d=(3)1/2 on its axis will be :

g = Gmd/(a²+d²)3/2
= (3)1/2Gm/8a²

The force on the sphere of mass M placed here is

F = Mg

= ( 3 )1/2GMm/8a²

  • Problem 3. If uniform solid sphere of radius ‘ a ‘ and mass ‘ M ‘ is surrounded symmetrically by a uniform thin spherical shell of equal mass and radius 2a . Find the gravitational field
    (a) At a distance 3a/2 from the centre ,
    (b) At a distance 5a/2 from the centre.

Solution –

( a ) The point at distance 3a/2 from the centre P1 and P2 is at 5a/2 distance a from the centre . As the spherical shell has radius of 2a so , point P1 is inside the cavity of the thin spherical shell , the field here due to spherical shell is zero. For solid sphere P1 is external point . so , field due to the solid sphere is

g = GM/(3a/2)²

g = 4GM/9a²

This is also the resultant field. The direction is towards the centre.

( b ) The point P2 is at distance 5a/2 means it is outside the sphere as well as the shell . So , we can replaced both ( solid sphere and spherical shell ) by single particles of the same mass at the centre . The gravitational field due to both body at point P2 is

g = GM/(5a/2)² = 4GM/25a²

The resultant field is g = 2g’= 8GM/25a² towards the centre .

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