### 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 acceleration due to gravity**. So let’s find out the expressions for g on the earth surface first. These expressions will help us to find out the **variation of acceleration due to gravity with height** and **variation of acceleration due to gravity with depth. **We have added a few **Numerical Problems** as well at the end of this post. Now you will be able to answer this common query- **What is the value of acceleration due to gravity at the centre of earth**?

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

**Definition: Acceleration due to gravity of an object is its rate of change of velocity due to the sole effect of the earth’s gravitational pull or gravity on that object that directs towards the centre of the earth.**

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 with height

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 ratio 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 value of acceleration due to gravity falls.

In the** next section** we will see **how g varies with depth** below the surface of the earth.

## Variation of Acceleration due to Gravity with depth

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 value of acceleration due to gravity falls.

In the **next paragraph** we will **compare these 2 equations **to get a clearer picture.

## Equations – formula- 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) value of 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 acceleration due to gravity 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, the value of 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**

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