Last updated on April 16th, 2021 at 09:42 am

In this post, we will discuss and derive **the formula for acceleration due to gravity at a depth h**. As the depth from the earth’s surface is increased, the value of g falls. We will discuss and derive the equation that describes the change in the value of g due to depth from the earth’s surface.

**The formula for acceleration due to gravity at depth h**

The formula for acceleration due to gravity at **depth h** is expressed by the formula:**g2 = g (1 – h/R). **

Here g2 is the acceleration due to gravity at depth h and R is the radius of the earth.

g denotes acceleration due to gravity on the earth’s surface.

For example, considering g = 9.8 m/s^2 on the earth’s surface, g2 at a depth of 1000 meter from the surface of the earth becomes 9.7984 m/s^2.[check with online calculator]

**Derive the Formula for acceleration due to gravity at depth h**

Let’s say, a body of mass m is resting at point A, **where A is at a depth of h from the earth’s surface.**

The distance of point A from the center of the earth = R – h,

where R is the radius of the earth.

**Mass of inner sphere** = (4/3). Π. (R-h)^{3}. ρ

Here ρ is the density. and Π is 22/7.

Now at point A, the gravitational force on the object of mass m is

F = G M m/ (R-h)^{2}

= G. [(4/3). Π. (R-h)^{3}. ρ] m/(R-h)^{2}

= G. (4/3). Π. (R-h). ρ. m

Again at point A, the acceleration due to gravity **(say g2) = F/m = G. (4/3). Π. (R-h). ρ** _________________ (7)

Now we know at earth’s surface, **g = (4/3) Π R ρ G** ( see the proof here: **g formula on the surface of the earth using density**)

Taking the ratio, again,

**g2/g ****= [G. (4/3). Π. (R-h). ρ ] / [(4/3) Π R ρ** **G] ****= (R-h) / R = 1 – h/R.**** => g2 = g (1 – h/R) **

** The formula for acceleration due to gravity at depth h** is represented with this equation:

=>

**g2 = g (1 – h/R) ______ (8)**

g1 is acceleration due to gravity at a depth h

**Use our online calculator to test the equation. Click the image below for the calculator page.**