When a falling meteor is at a distance above the Earth\'s surfaceof 3.40 times t

Question

When a falling meteor is at a distance above the Earth's surfaceof 3.40 times the Earth'sradius, what is its free-fall acceleration caused by thegravitational force exerted on it?

My questionis what changes in the problem becuase everytime I try and do whatthe solution did I get 1.2475E-14. I dont understand what im doingwrong. Is the radius of the earth correct?

My questionis what changes in the problem becuase everytime I try and do whatthe solution did I get 1.2475E-14. I dont understand what im doingwrong. Is the radius of the earth correct?

Explanation / Answer

Height h = 3.4 R where R = Radius of earth = 6.38 * 10 ^ 6 m So, free fall accleration   g = GM / ( R + h ) ^2 where G = Gravitational constant = 6.67 * 10 ^ -11 N m ^2 / kg ^ 2            M= Mass of earth = 5.98 * 10 ^ 24 kg plug the values we get g = ( 6.67 * 10 ^ -11 ) ( 5.98 *10 ^ 24 ) / [ R + 3.4 R ] ^ 2                                     = 5.061 * 10 ^ -1                                      =0.5061 m / s ^ 2

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