Careful measurements of local variations in the acceleration due to gravity can
ID: 2007505 • Letter: C
Question
Careful measurements of local variations in the acceleration due to gravity can reveal the locations of oil deposits. Assume that the Earth is a uniform sphere of radius 6370 km and density 5500 kg/m3, except that there is a spherical region of radius 1.5 km and density 842 kg/m3, whose center is at a depth of 3 km. Suppose you are standing on the surface of the Earth directly above the anomaly with an instrument capable of measuring the acceleration due to gravity with great precision. What is the ratio of the acceleration due to gravity that you measure compared to what you would have measured had the density been 5500 kg/m3 everywhere? (Hint: Think of this as a superposition problem involving two uniform spherical masses, one with a negative density.)
Ratio = 1 ?
Explanation / Answer
the acceleration due to gravity that we measure :
g = GM / r^2 ;
then actual mass
M = 5500 * ( 4 / 3 ) ( 6370 e 3 ) ^3 - 5500 * ( 4 /3 ) ( 1.5 e 3 ) ^3 + 842 * ( 4 /3 ) ( 1.5 e 3 )^3 ;
had the density been constant ,
M = 5500 * ( 4 /3 ) ( 6370 e 3 ) ^3 ;
g( new) / g ( uniform ) =
[ 5500 * ( 4 / 3 ) ( 6370 e 3 ) ^3 - 5500 * ( 4 /3 ) ( 1.5 e 3 ) ^3 + 842 * ( 4 /3 ) ( 1.5 e 3 )^3 ] /
[ 5500 * ( 4 /3 ) ( 6370 e 3 ) ^3 ]
= [ ( 6370 e 3 ) ^3 - ( 1.5 e 3 ) ^3 + ( 843 / 5500 ) ( 1.5 e 3 ) ^3 ] / [ ( 6370 e 3 ) ^3 ] ;
= [ ( 6370 e 3 ) ^3 - ( 0.846 ) * ( 1.5 e 3 ) ^3 ] / [ ( 6370 e 3 ) ^3 ] ;
= [ ( 6379 )^3 - (0.9457 * 1.5 )^3 ] / [ 6370 ^ 3 ] ;
= [ a^3 - b^3 ] / a^3 ; = 1 - ( b / a ) ^ 3 = 1 - ( 0.9457 * 1.5 / 6379 ) ^3 = 1- ( 1.0999983913 e -11 ) = 1
so the ratio is nearly 1 .