Question

given that Earth's surface gravitaitional field strength has a magnitude of 9.80N/kg, determine the distance(as a multiple of Earth's radius re) above the Earth's surface at which the magnitude of the field strength is 3.20N/kg.

Answer 1

3.2 = Gm/r^2 (use earth's mass for m and the universal constant for G)

Answer 2

questions asking what times the radius of the earth will give you the distance.

FIRST, find the radius from the earth all the way ABOVE the earth surface
g=Gm/r^2..... r=Gm/g*square root whole equation*..... (6.67x106-11)(*mass of earth*)/3.20*square root* = 1.1x10^7

SECOND, find the distance by subtracting the radius above the surface to the radius to the equator (re) ... (1.1x10^7) - (6.38x10^6) = 4.8x10^6

THRID, you need to divide 4.8x10^6 by 6.38x10^6 = 0.75

final, there for the distance is 0.75 x 6.38x10^6 or 0.75re