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September 18, 2019

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


Acceleration due to gravity and Gravitational Attraction of the earth

Acceleration due to gravity (g) 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.

1> 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.

2> Then we will explore the variation of acceleration due to gravity with height and variation of acceleration due to gravity with depth. In other words we will see how height and depth from the surface of the earth influence the value of acceleration due to gravity.
So the goal of these sections will be to find out the variation of g with altitude and variation of g 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

Force of gravity (gravitational force value due to earth) 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 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) / R^2

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

=> g = GM / R^2 _________ (3)


Another expression of g on earth’s surface here:

If mean density of earth is p then 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:

g = (4/3) Pi R p G ________(4)


In the next 2 sections, we will discuss on variations of g . This includes
(a) How g changes with height
(b) Variation of g with depth

acceleration due to gravity-height
acceleration due to gravity-height

Variation of Acceleration due to gravity with height 

variation of g with altitude

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 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)

=> g1 = g (1 – 2h/R) ______(6)

This gives the formula for g at height h.

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.

acceleration due to gravity - depth
acceleration due to gravity – depth

Variation of Acceleration due to Gravity with depth

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.

acceleration due to gravity at a 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)  ______ (8)

This gives the formula for g at depth.

So as depth h increases, the value of g falls.

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

Equations – formula- Comparison

formula for g at height h – 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 – Variation of g with depth

g2 = g (1 – h/R) 

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 g 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 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.

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?

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.

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

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