In this post, we will discuss *how acceleration due to gravity changes with height and depth*. As we gain altitude or height with respect to earth’s surface we find a *variation of g with altitude* . Similarly deeper below the earth surface we notice a *variation of g with depth*. g here stands for the acceleration** due to gravity**.

1> First we will derive the **expression** of * Acceleration due to gravity (g) on the earth’s surface*. We need these expressions to carry on the next step given below.

2> Then we will derive the expression of g at a

**height**h above the earth’s surface. This will show the

*variation of acceleration due to gravity with height*.

3> After this, we will derive the expression of g at a

**depth**h below the earth’s surface. This will show us the

*variation of acceleration due to gravity with depth*.

We will also find out the value of

*acceleration due to gravity at the centre of the earth*.

**Formula FlashcardA quick reckoner for readers in a hurry. **

## Derive the equation of Acceleration due to gravity on the earth’s surface

In this section let’s find out the equation of g on earth’s surface.

Force of gravity (gravitational force value due to earth) acting on a body of mass m on the earth surface is expressed as:**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}

Now, we get the equation or formula of g on earth’s surface as follows*:*

Acceleration due to gravity on earth’s surface is represented as

**g = GM / ** R^{2} ** ______(3)**

[expression of g on earth’s surface]

Let’s find out ** another expression of g on the **earth’s surface using the density of the

*earth*If the mean density of the earth is *p* then the mass of the earth is expressed as:

M = volume X density = (4/3) *Pi* R^{3} . p

(Pi = 22/7)**g = G.( (4/3) Pi R ^{3} p) / R^{2} **

*So we get the second expression or formula for g on earth’s surface:*Acceleration due to gravity of earth is represented as**g = (4/3) Pi R p G _____(4)**

[2nd expression of g on earth’s surface]

In the next 2 sections, we will discuss variations of g. This includes

(a) **How g changes with height **

(b) **Variation of g with depth**

## Derive the expression for Variation of g with height.

This section covers * variation of g with altitude*. 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 the 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)

so,

**g1/g = (1 – 2h/R)**

**Variation of g with height** is represented with this equation: **=> g1 = g (1 – 2h/R) ______(6)**

g1 is acceleration due to gravity at height h.

### What is the variation in acceleration due to gravity with altitude?

g1 = g (1 – 2h/R)

This gives the *formula for g at height h*.

So as altitude h increases, the value of acceleration due to gravity falls. This describes the *variation of g with height* or altitude.

### What is the change in the value of g at a height h?

g1 = g (1 – 2h/R)

This g1 gives the *formula for g at height h*.

So the value of g will decrease by this amount: 2hg/R, at a height h above the earth’s surface. R is the radius of the earth.

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

## Derive the expression for Variation of g 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, **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) **

**Variation of g with depth** is represented with this equation:

=> **g2 = g (1 – h/R) ______ (8)**

g1 is acceleration due to gravity at a depth h

### What is the variation in acceleration due to gravity with depth?

**g2 = g (1 – h/R)**

This gives the ** formula for g at depth** h.

**So as depth h increases below the earth’s surface, the value of g falls.**

### What is the change in the value of g at a depth h?

**g2 = g (1 – h/R)**

This gives the ** formula for g at depth** h.

**So at a depth h below the earth’s surface, the value of g falls**by this amount: gh/R. R is the radius of the earth.

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

## How Acceleration due to gravity changes with height and depth? Equations – formula- Comparison

Let’s summarize *how acceleration due to gravity changes with height and depth*.

### Formula for g at height h showing Variation of g with altitude

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

### Formula for g at depth h showing Variation of g with depth

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

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

That means acceleration due to gravity is maximum at the surface of the earth.

2) We also noticed that, g1 < g2

**And that means:**

1) value of g falls as we go higher or go deeper.

2) But it falls more when we go higher.

3) It is also clear that acceleration due to gravity is maximum at: the earth’s surface.

**What is the value of acceleration due to gravity at the centre of the 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 equation 8 above.

Therefore from equation 8,

the** value of g** at the centre of the earth

is g (1- h/R) = g (1 – R/R)

=g (1-1) = 0

So we can see, the value of *g* at the centre of the earth equals zero.

## Numerical Problems

**Q1: What is the value of g 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}

See Solution

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

See Solution

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

See Solution

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

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