The Gravitational Field Strength

The Gravitational Field Strength

1. Every mass particle is surrounded by a space within which its influence can be felt. This region or space is said to be occupied with gravitational field.
2. Each point in the field is associated with a (vector) force which is experienced by a unit mass placed at that point and is called the gravitational field strength.
3. If a test mass m at a point in a gravitational field experiences a force F.
4. (11.4)

Field Due to Sphere

1. For an external point (r > R) a sphere (solid or hollow) behaves as if whole of its mass is concentrated at its centre, i.e.,
2. [for r > R](11.6)
3. In case of a spherical shell, for an internal point, i.e., (r < R) field is zero, i.e.,
4. Iin = 0
5. In case of a spherical volume distribution of mass (i.e., a solid sphere) for an internal point (r < R), the portion of the sphere that lies outside the radius r will not contribute to the field [as the field inside a spherical shell is zero]; so

∴(11.7)

i.e., intensity inside a solid sphere varies linearly with distance from the centre. So it is minimum (Ic = 0) at the centre and maximum at the surface (Is = GM/R2). This is shown graphically in the figure.

Acceleration due to gravity ‘g’

If in Newton’s law of gravitation one body (say earth) is taken as ‘reference body’, the force with which the reference body attracts any other body towards its centre is called force due to gravity (of reference body) and the phenomenon ‘gravity’. Now if the body is free to move, force of gravity (of reference body) in accordance with Newton’s II law will produce an acceleration in it. The acceleration produced in a body by the force of gravity of reference body (usually earth) is called acceleration due to gravity and is represented by ‘g’.

1. If the reference body has mass M and radius R, the force on a body of mass m at the surface of reference body by Newton’s law of gravitation will be
2. F = (GMm/R²)
3. and if ‘g’ is acceleration due to gravity of reference body on its surface, by Newton’s II law,
4. F = mg
5. ∴g = GM/R²) (11.8)
6. This is the relation between g and G and shows that g depends on reference body.
7. It is independent of mass, shape, size, etc., of ‘falling body’, i.e., a given reference body produces same acceleration in a light and a heavy falling body.
8. Variation in ‘g’

1. With altitude
   1. As for an external point a spherical distribution of mass behaves as if the whole of its mass were concentrated at the centre, i.e., g = I = (GM/r2)
   2. so at the surface of earth
   3. g = (GM/R²)
   4. and for a height h above the surface of earth
   5. g′ = [GM/(R + h)²] [as r = R + h]
   6. 
2. With depth
   1. As in case of spherical distribution of mass for an internal point
   2. g = I = (GM/R₃)r
   3. So at the surface of earth g = (GM/R₂)
   4. and for a point at a depth d below the surface,
   5. 
   6. So with increase in depth below the surface of earth ‘g’ decreases and at the centre of earth it becomes zero.

1. ∴( r = R cos λ)
2. org′ ≈ g - ω₂R cos₂λ(11.11)
3. At equator λ = 0^o i.e. g′ ≈ g - ω₂R
4. At the poles λ = 90^o i.e. g′ = g
5. Note that the vector g′ is not exactly towards the center of earth.