### Acceleration due to gravity – how it changes with height and depth

##### October 18, 2017

# Gravitational Attraction of the earth and Acceleration due to gravity

In this post at first we will work on the expression of **Acceleration due to gravity** or **gravitational acceleration (g) **on the earth’s surface. We will take help of Newton’s Law of Gravitation as well as Newton’s second Law of Motion. Then we will explore how height or depth from the surface of the earth influence the value of this **acceleration due to gravity**. So let’s find out the expression for g on the earth surface first.

## Acceleration due to gravity – on the earth’s surface

Force of gravity acting on a body of mass m on the earth surface = F = GMm/R^2 ____________ (1)

Where R is the radius of the earth (considering it a homogenous sphere)

and M here is the total mass of earth concentrated at its centre. G is the gravitational constant.

Now, from Newton’s 2^{nd} Law of Motion,

F = mg.___________________(2)

Just to recapitulate, as a body falls downwards it is continuously acted upon by a force of gravity. The body thus possesses an acceleration, called Acceleration Due to gravity(g).

From equation 1 and 2 we can write

mg = (GMm) / R^2

=> **g = GM / R^2 _______________ (3)**

Another expression of g here:

If mean density of earth is p then mass of the earth = M = volume X density = (4/3) Pi R^3 p

(Pi = 22/7)

g = G.( (4/3) Pi R^3 p) / R^2

**g = (4/3) Pi R p G ________________ (4)**

In the next 2 sections, we will discuss how this g varies as we go (a)higher from the surface and go (b) deeper from the surface.

## Variation of Acceleration due to gravity when height or altitude from the surface is changed

At a height of h from the surface of the earth, the gravitational force on an object of mass m is

F = GMm/(R+h)^2

Here (R + h) is the distance between the object and the centre of earth.

Say at that height h, the gravitational acceleration is g1.

So we can write, mg1 = GMm / (R+h)^2

=> g1 = GM/(R+h)^2 _________________ (5)

Now we know on the surface of earth, it is

g = GM / R^2

Taking ration of these 2,

g1/g = R^2 /(R+h)^2

= 1/(1 + h/R)^2 = (1 + h/R)^(-2) = (1 – 2h/R)

=> g1/g = (1 – 2h/R)

**=> g1 = g (1 – 2h/R) ___________________ (6)**

So as altitude h increases, the acceleration due to gravity falls.

## Variation of Acceleration due to Gravity when depth changes

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.**

Distance of point A from the centre of the earth = R – h,

where R is the radius of the earth.

Mass of inner sphere = (4/3). Pi. (R-h)^3. p

Here p is the density.

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

F = G M m/ (R-h)^2 = G. [(4/3). Pi. (R-h)^3. p] m/(R-h)^2 = G. (4/3). Pi. (R-h). p. m

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

Now we know at earth’s surface, acceleration due to gravity = g = (4/3) Pi R p G

Taking the ratio, again,

g2/g = [G. (4/3). Pi. (R-h). p ]/ [(4/3) Pi R p G] = (R-h) / R = 1 – h/R.

=> **g2 = g (1 – h/R) _________________________ (8)**

So as depth h increases, the acceleration due to gravity falls. In the **next paragraph** we will compare these 2 equations to get a clearer picture.

## Comparison

**g1 = g (1 – 2h/R) ___________________ (6) at a height h from the earth’s surface**

**g2 = g (1 – h/R) _________________________ (8) at a depth h from the earth’s surface**

1) Now from eqn 6 and 8 we see that both g1 and g2 are less than g on the earth’s surface.

2) We also noticed that, g1 < g2

**And that means:**

1) acceleration due to gravity falls as we go higher or go deeper.

2) But it falls more when we go higher.

**What is the value of g at the centre of earth?**

At the centre of the earth, depth from the earth’s surface is equal to the radius of the earth.

So h = R for the equation 8 above.

Therefore from equation 8, the value of * acceleration due to gravity* at the centre of the earth = g (1- h/R) = g (1 – R/R) =g (1-1) = 0

**So we can see, acceleration due to gravity at the centre of the earth equals zero.**

## Numerical Problems

**Q1: What is the value of acceleration due to gravity at a height 4 miles above the earth’s surface? Diameter of the earth is 8000 miles. g at the surface of the earth = 9.8 m/s^2.**

**Q2: At what depth under the earth’s surface, the value of acceleration due to gravity will reduce by 1% with respect to that on the earth’s surface?**

**Q3: If the value of acceleration due to gravity at a small height h from the surface equals the value of acceleration due to gravity at a depth d, then find out the relationship between h and d.**

###### Link to the Solutions of these problems:**Solution to this physics problem set**

Dear Reader, hope you have liked the post!

Now it’s your turn to *share it among others*. Please use the social media buttons to share.

###### Suggested Read from this Physics BLOG:

**Gravitation: Why a stone fall towards the earth but..**

**Free fall versus fall against air drag, Terminal Velocity**

## 7 Responses “Acceleration due to gravity – how it changes with height and depth”