Lesson_Text_4.5 - 1 Lesson 4.5 Newtons Law of Gravitation...

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1 Lesson 4.5 Newton’s Law of Gravitation 1. The Law of Gravitation. The law of gravitation states that every object attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. If r is the distance between the centers of two masses m 1 and m 2 , then this law can be written in mathematical form as: 1 2 2 m m F r This can be written as an equation using a constant as: 1 2 2 m m F G r = …(i) Here G is a constant and is called the universal gravitational constant. It has a value: G = 6.67 x 10 -11 N m 2 .kg -2 . Consider a mass m kg on the surface of the earth of mass M e kg. If the radius of the earth is R e , then the force exerted by the earth on the object can be expressed as: 2 e e M m F G R = …(ii) Here R e is the radius of the earth. We assume that the object of mass m is very small compared to the earth so that the distance between the center of the earth and the object is the radius of the earth. If the object is at a height h above the surface of the earth, then the equation becomes: 2 ( ) e e M m F G R h = +
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2 But the force with which an object is being pulled by the earth is called its weight and the weight of an object is given by W = mg where g is the acceleration due to gravity on the surface of the earth. Replacing F by W = mg in equation (ii) above, we have: 2 e e M m mg G R = Canceling m from both sides, we have: 2 e e M g G R = …(iii) For an object placed at a height h above the surface of the earth the value of g is: 2 ( ) e e M g G R h = + This means that the value of g decreases as we go away from the surface of the earth. Example 1: Find the value of g on the surface of the earth using standard values of R e and M e , the mass of the earth. Solution: If we use the mass of the earth as 5.98 x 10 24 kg and the mean radius of the earth as R e = 6380 km = 6.38 x 10 6 m, the value of g on the surface of the earth can be calculated as 24 11 6 2 2 5.98 10 6.67 10 (6.38 10 ) 9.8 . m g s - - × = × = × Example 2: Find the value of g at a height 200 km from the surface of the earth. Solution: Here we use the formula: 11 24 2 6 2 2 6 6.67 10 5.98 10 ( ) (6.38 10 0.2 10 ) 9.2 . e e M g G R h m s - - × × = = = + × + ×
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3 2. Gravitational Field fig. 1 The statement of the law of gravitation implies that an object placed outside the earth’s surface experiences a force towards the earth. This is true no matter where the object is placed, it will experience a force towards the earth. This is equivalent to saying that that there is a gravitational force field around the earth. The gravitational force field is a vector quantity that has a magnitude and direction. The direction is towards the center of the earth as indicated by the arrows in the figure above where this force field is represented by drawing vectors. The magnitude or strength of the force field is greater where the vectors are closer (at the surface of the earth), and
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Lesson_Text_4.5 - 1 Lesson 4.5 Newtons Law of Gravitation...

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