F GMem1r2m1ag GMer2g 6.673×1011m3kg sec25.95×1024 kg(6.367×106m)2g9.8msec2r 30000 m 6.367×106 m 6.397×106 mg 6.673×1011m3kg sec25.95×1024 kg(6.397×106m)2g9.7msec2Force and MotionNEWTON'S LAWS OF MOTIONRev. 0Page 3CP-03where:M=mass of the earth (5.95 x 10 kg)e24m=mass of the object1r=radius of the earth (6.367 x 10 m)6The mass (m ) of the object cancels, and the value of (g) can be determined as follows1since a=g by substituting (g) for (a) in the previous equation.If the object is a significant distance from the earth, we can demonstrate that (g) is not aconstant value but varies with the distance (altitude) from the earth. If the object is at analtitude of 30 km (18.63 mi), then the value of (g) is as follows:As you can see, a height of 30 km only changes (g) from 9.8 m/sec to 9.7 m/sec . There2 2will be an even smaller change for objects closer to the earth. Therefore, (g) is normallyconsidered a constant value since most calculations involve objects close to the surfaceof the earth.
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