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