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| Expression for Acceleration due to Gravity |
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| Consider an object of mass m lying on or near the surface of the Earth. Let Me be the mass of the Earth and Re be its radius i.e., Re is the distance between the object and the centre of the Earth. |
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| According to Newton's law of gravitation, the force of attraction (F) between the Earth and the object is |
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| A Body of Mass m Lying on the Surface of the Earth |
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| According to Newton's second law of motion this force produces an acceleration (g) in the object. |
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| F = ma (a = g) |
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| F = mg |
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| Substituting the value of F in equation (2) we get, |
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| From equation (3) it is very clear that acceleration due to gravity does not depend on the mass m of the object. It only depends on the mass of the Earth (Me) and the distance from the centre of the Earth to the object. |
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| Gravitational Constant (G) = 6.6734 x 10-11 Nm2/kg2 |
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| Mass of the Earth (Me) = 6 x 1024 kg |
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| Radius of the Earth (Re) = 6.4 x 106 m |
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| The value of acceleration due to gravity on Earth is 9.8 m/s2. This means that when a body falls freely towards the Earth its velocity increases at the rate of 9.8 m/s during its motion. Similarly when an object is projected vertically upwards its velocity decreases at the rate of 9.8 m/s and eventually the velocity becomes zero. The height at which the velocity of an object moving against gravity becomes zero is described as the maximum height attained by the object. When the velocity becomes zero the object starts falling down, with an acceleration as if released from that height. |
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| The acceleration produced in a body moving against gravity is also 9.8 m/s2 but oppositely directed therefore g = - 9.8 m/s2. |
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| The expression for acceleration due to gravity is |
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| Acceleration due to gravity is inversely proportional to the square of the distance between the centre of the Earth and the object. The acceleration due to gravity produced in an object on the surface of the Earth is dependent on the radius of the Earth. Earth is not a perfect sphere (slightly bulging out at the equator) its radius decreases as we move from the equator to the poles. At the equator and at sea level its value is about 9.78 m/s2 and at the poles it is 9.83 m/s2. Its mean value is taken as 9.8 m/s2 for all calculations. |
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| Similarly the value of g decreases as we go up to the top of a mountain or higher up in the air. |
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| Distance of the Object from the Centre of the Earth |
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| The value of g inside the Earth is directly proportional to the distance from the centre of the Earth. Hence, g decreases as we go down into the Earth till it becomes zero at the centre of the Earth. |
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| We know that the earth attracts every particle towards the centre. A body can be considered to be made up of number particles. As the size of the body is small when compared to that of the earth, the gravitational pull acting on these particles can be regarded to be parallel to each other as shown in the figure. |
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| Centre of Gravity of an Irregular Lamina |
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| A single force passing through a fixed point called the centre of gravity of the body can replace these parallel forces acting vertically in the downward direction. The resultant force is equal to the weight of the body. |
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| Thus centre of gravity is the point through which the weight of the body acts irrespective of the position of the body. |
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| For bodies which are of regular shape and which have uniform density, the centre of gravity lies at the geometrical centre of the body. |
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| One of the important applications of Newton's law is to estimate masses of binary stars. A binary star is a system of two stars orbiting round their common centre of mass. |
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| Any irregularity in the motion of a star indicates that it might be another star or a planet going round the stars. This regularity in the motion of a star is called a wobble. |
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