Gravity
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Card # 12-Acceleration due to Gravity (g): When a body is allowed to fall freely under gravity due to the gravitational pull of earth, then its acceleration is known as gravitational acceleration. If the mass of the earth = M, Radius of the earth = R, Mass of a body on earth = m Then the gravitational force = F = ( )/R2 m =( )/R2 = (GM)/R2
Variations in g:
The value of gravitational acceleration ( ) changes from place to place. At equator its value is least and as we go towards the poles its value continuously increases and becomes maximum at poles. Following are the other variations: 1) Variation in g with height: Gravitational acceleration on the surface of the earth = = (GM)/R2 Gravitational acceleration at height h, from the surface of the earth = = [GM) ] = 1 – (2h/R)] The relation between g and g’ is given 2) Variation in g with depth:
Gravitational acceleration on the surface of the earth = g = (GM)/R2 Gravitational acceleration at depth h = = (GM)/(R-h)2 – (h/R)] The relation between g and g’ is given by = Thus with increase in h, g decreases or g’ < g At the center of the earth h = R, hence g’ = 0 or the acceleration due to gravity is zero at the center of the earth. 3) Variation in g with latitude:
The shape of the earth is not perfectly spherical in shape. It is slightly flattened at the poles and it is bulged at the equators. Let g at poles = = (GM)/R2p at equators = = (GM)/R2e and we know
.
4) Variation in g due to spin motion of earth: Earth continuously executes spin motion about an axis passing through the two poles. It completes one rotation in 24 hours from west to east direction. Let the angular velocity of rotation of earth = ω, the mass of a body on the earth = m, the latitude of the position of the body = λ. Due to spin motion of the earth the body will execute a circular motion in a path of radius PA = r. Centripetal force = F (mv2)/r = mrω2 along AP [ ] This causes reduction in the weight of the body The effective weight of the body = (real weight of the body – component of the centripetal force along the radius OA)
At poles λ = 90 therefore g’ = g At equators λ = 0 therefore g’ = (g – Rω2).
Variations in g:
The value of gravitational acceleration ( ) changes from place to place. At equator its value is least and as we go towards the poles its value continuously increases and becomes maximum at poles. Following are the other variations: 1) Variation in g with height: Gravitational acceleration on the surface of the earth = = (GM)/R2 Gravitational acceleration at height h, from the surface of the earth = = [GM) ] = 1 – (2h/R)] The relation between g and g’ is given 2) Variation in g with depth:
Gravitational acceleration on the surface of the earth = g = (GM)/R2 Gravitational acceleration at depth h = = (GM)/(R-h)2 – (h/R)] The relation between g and g’ is given by = Thus with increase in h, g decreases or g’ < g At the center of the earth h = R, hence g’ = 0 or the acceleration due to gravity is zero at the center of the earth. 3) Variation in g with latitude:
The shape of the earth is not perfectly spherical in shape. It is slightly flattened at the poles and it is bulged at the equators. Let g at poles = = (GM)/R2p at equators = = (GM)/R2e and we know
.
4) Variation in g due to spin motion of earth: Earth continuously executes spin motion about an axis passing through the two poles. It completes one rotation in 24 hours from west to east direction. Let the angular velocity of rotation of earth = ω, the mass of a body on the earth = m, the latitude of the position of the body = λ. Due to spin motion of the earth the body will execute a circular motion in a path of radius PA = r. Centripetal force = F (mv2)/r = mrω2 along AP [ ] This causes reduction in the weight of the body The effective weight of the body = (real weight of the body – component of the centripetal force along the radius OA)
At poles λ = 90 therefore g’ = g At equators λ = 0 therefore g’ = (g – Rω2).
