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Giancoli 6th edition solutions by video




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6th Edition, Chapter 5, Problem 35

Giancoli physics 6th Edition answers: Choose a 6th Edition annual or monthly subscription to access all 1675 solution videos.

Chapter Number of 6th edition solutions for this chapter
Rotational Motion - spinning top Chapter 5 - Circular Motion; Gravitation 65 out of 65
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Video transcript for this Giancoli 6th edition physics answer:
So we’re told that at some position the acceleration due to gravity ‘g2’ equals a regular surface of the earth ‘g’ which is nine point eight divided by ten so where’s the acceleration due to gravity one tenth that that it has one the surface of the earth. So ‘g2’ is: ‘G’ times mass of the earth ‘mE’ divided by whatever that distance is which we have to find, ‘r2’ squared and that equals ‘G’ times mass of the earth ‘mE’ over radius of the earth ‘rE’ squared times one tenth. ‘G’ times ‘mE’ over ‘rE’ squared is ‘g’ on the earth’s surface so since the numerators are the same the denominators must also be the same including that ten there so we have: ‘r22’ equals ten times radius of the earth squared. Solving: ‘r2’ equals the square root of ten times the radius of the earth squared or we can write this as the square root of ten times the radius of the actual earth which is the square root of ten times six point three eight times ten raised to the power six meters which gives us two point zero times ten raise to the power seven meters. So at that distance from the center of the earth the acceleration due to gravity will be one tenth what it is on the surface of the earth.

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