Question

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average density from the attached density profile for a spherical shell (which does not have to be very thin, for example, a shell from the inner core-outer core boundary to the middle depth of the outer core). Do not make the shell contains a large density contrast. Calculate enough points to make the g profile. Note that only the mass at radius less than r contributes to the gravitational acceleration. (15 points)

Answer 1

Acceleration due to gravity of the earth, g
`$=(GM)/(R^2)$`

`$g=(G(4/3 \pi R^2 \rho))/(R^2)=G(4/3 \pi R \rho)$`

Acceleration due to gravity at 1000 km depths is :

`$g=G\left((4)/(3)\pi (R-d) \rho\right)$`

`$g=6.67 * 10^(-11)\left((4)/(3)* 3.14 * (6371-1000) * 5.5 * 10^3\right)$`

`$= 822486 * 10^(-8)$`

`$=0.822 * 10^(-2) \ km/s$`

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

`$g=G\left((4)/(3)\pi (R-d) \rho\right)$`

`$g=6.67 * 10^(-11)\left((4)/(3)* 3.14 * (6371-2000) * 5.5 * 10^3\right)$`

`$= 673552 * 10^(-8)$`

`$=0.673 * 10^(-2) \ km/s$`

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

`$g=G\left((4)/(3)\pi (R-d) \rho\right)$`

`$g=6.67 * 10^(-11)\left((4)/(3)* 3.14 * (6371-3000) * 5.5 * 10^3\right)$`

`$= 3371 * 153.86 * 10^(-8)$`

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

`$g=G\left((4)/(3)\pi (R-d) \rho\right)$`

`$g=6.67 * 10^(-11)\left((4)/(3)* 3.14 * (6371-4000) * 5.5 * 10^3\right)$`

`$= 153.84 * 2371 * 10^(-8)$`

`$=0.364 * 10^(-2) \ km/s$`

= 3.64 m/s