Hello students here we are going to discuss the acceleration due to gravity at depth ‘d’ below the surface of earth. We know that the acceleration due to gravity depends upon mass and radius of planet. Here we are going assume the earth as sphere of uniform density.
Derivation of formula for Acceleration due to gravity below the surface of earth at depth ‘d’
Let’s find the formula for the acceleration due to gravity at depth ‘d’ below the surface of earth.
Consider a body of mass m situated on the surface of the earth of mass M radius R and of uniform density ρ.
On the surface of the earth the acceleration due to gravity is given as,
g = GM/R2 …………….(1)
Since the earth is considered to be a homogeneous sphere of uniform density, then
Density of earth=( mass of earth)/(volume of earth)
∴ mass of earth = volume of earth × density of earth
∴ M = 4/3 πR3ρ
On substituting in equation (1)
If body is reaches at depth d below the earth’s surface, let gd be the acceleration due to gravity at depth d, then
Where (R-d) is radius of the earth at depth d.
Let Md and Vd be the mass and volume at depth d then,
mass of earth at depth = volume of earth at depth× density of earth
Equation (5) gives the formula for acceleration due to gravity at depth ‘d’ below the surface of earth.
Points to remember:
- We can conclude that the value of acceleration due to gravity decreases as body moves below the surface earth.
- At d = 2h, then gh = gd
- At centre of the earth d = R gd = 0
Numerical based in the Acceleration due to gravity below the surface of earth at depth ‘d’
Ex 1: Find the acceleration due to gravity below surface of earth at depth equal to 1/3rd of radius of radius. (Use g=10 m/s2 , R= 6400 km)
Solution:
Here, d=R/3.
The value for acceleration due to gravity at height ‘h’ is given as,
Ex 2: Find the depth at which the acceleration due to gravity is equal to acceleration due to gravity of height of 3200 km.
Solution:
Here, d=?.
The value for acceleration due to gravity at height ‘h’ is given as,