How Acceleration due to gravity changes with height and depth?

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 Flashcard
A quick reckoner for readers in a hurry.

how acceleration due to gravity changes with height and depth
If you want to study in details including derivation and explanation,
then please read on.

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/R2 ____________ (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 2nd 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) / R2

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 / R2 ______(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 R3 . p
(Pi = 22/7)
g = G.( (4/3) Pi R3 p) / R2

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 / R2
Taking ratio of these 2,

g1/g = R2 /(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.

variation of g with depth

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/s2
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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