# What is Gravity?

Gravity is the force by which a planet or other body draws objects toward its center. As a result of this gravity when an object is released from a distance above the surface of the earth the object moves towards the center of the earth.

Anything that has mass also has gravity. Objects with more mass have more gravity and objects with less mass have less gravity. Gravity also gets weaker with distance. So the closer objects are to each other the stronger their gravitational pull is.

Newton clearly stated that any two objects in this universe attract each other with a certain amount of force but he could not explain what caused this attractive force i.e he told us about the strength of gravity but he never found the source of gravity.

This concept was fulfilled by Einstein's general theory of relativity. Einstein said that in reality gravity is not a force but it is actually the same thing as acceleration.

According to this theory, matter causes curvature in space-time. Wherever there are matters in the universe there will be curves in spacetime. More matter will create more curvature and less matter will create less curvature in spacetime.

For this curvature, the objects in the universe attract each other. this is why planets are orbiting the sun. This curvature is the source of gravity.

Now let's understand the concept of gravity with an example. Suppose we have a piece of fabric a large ball and two smaller balls.

Now if we put the big ball on the fabric then what will happen. it will push the fabric inwards. Again if we put two small-sized balls on that fabric, then you will see that both those balls also push the fabric inward but less than the big one.

As the big ball has pushed the fabric more inward, so it will pull the small ball toward its center. Einstein said that this is why the objects in the universe attract each other.

But remember space-time fabric is a 4th-dimensional concept whereas 3 dimensions of space and 1 dimension of time are involved in it. But the simplest way to demonstrate this is with a 2-dimensional fabric. Original space-time fabric looks like this -

The concept of gravity will be more clearly understood if you study Einstein's general theory of relativity more deeply.

## Acceleration due to gravity

The acceleration gained by a freely falling object, due to the gravitational force, or gravity is called acceleration due to gravity. The symbol we use to represent the acceleration due to gravity is small g.

Measurement of acceleration due to gravity

According to Newton's second law of motion (From F = ma), if m is the mass of an object which is in acceleration g then the force acting on it is

$F=mg$

And we also know from Newton's law of gravitation that if two masses and M are separated by a distance r, then the gravitational force acting between them is

$F=\frac{GMm}{r^{2}}$

By comparing these two equations we get

$mg=\frac{GMm}{r^{2}}$

$or\;g=\frac{GM}{r^{2}}$

In the case of any point on the surface of the earth r = R (The radius of the earth R = 6400km). Then the formula for g goes like this

$g=\frac{GM}{R^{2}}$

This is the relation between G and g

Where, G  = The universal gravitational constant = 6.67×10 -11 Nm2/kg2.

M = Mass of a very large body like Earth = 5.98 × 10 24 kg.

R = The radius of the earth = 6.38×10 m.

So the value of acceleration due to gravity on the earth is

$g=\frac{(6.67\times 10^{-11}N.m^{2}/kg^{2})(5.98\times 10^{24}kg)}{6.38\times 10^{6}m}$

or g ≈ 9.8 m/s2

## Variation in Acceleration due to gravity

Acceleration due to gravity is not a constant quantity. Its value changes for various reasons. For example. As the earth is elliptical the value of g varies in different parts of the surface of the earth. The value of g varies at different heights from the surface of the earth.

Similarly, the value of g varies at different depths from the surface of the earth. It also varies in different parts of the earth's surface due to the rotational motion of the earth.

Variation in acceleration due to gravity with heights

Let the earth be a sphere of radius R and mass M. We know the acceleration due to gravity on the surface of the earth

$g=\frac{GM}{R^{2}}\to (1)$

We will determine the formula for the acceleration due to gravity at a point above the surface of the earth. So consider any point at the height of h above the surface of the earth.

The acceleration due to gravity at that point is

$g_{h}=\frac{GM}{(R+h)^{2}}\to (2)$

Now dividing equation no (2) by equation no (1) we get

$\frac{g_{h}}{g}=\frac{\frac{GM}{(R+h)^{2}}}{\frac{GM}{R^{2}}}$

$Or\;\frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}}$

$Or\;g_{h}=\frac{R^{2}}{(R+h)^{2}}g$

This relation clearly shows that gh＜ g. This means that as we continue to move above the surface of the earth the acceleration due to gravity decreases.

Now

$g_{h}=\frac{R^{2}}{R^{2}(1+\frac{h}{R})^{2}}g$

$Or\;g_{h}=\frac{1}{(1+\frac{h}{R})^{2}}g$

$Or\;g_{h}=(1+\frac{h}{R})^{-2}g$

Using the Binomial Theorem we get

$g_{h}=g(1-\frac{2h}{R})$

This is the formula for variation in acceleration due to gravity with heights.

Variation in acceleration due to gravity with depth

Again let the earth be a sphere of radius R and mass M. We know the acceleration due to gravity on the surface of the earth

$g=\frac{GM}{R^{2}}\to (1)$

If 𝜌 is the density of the earth and V is the volume of the earth then the mass of the earth

$\rho=MV$

$Or\;M=\frac{4}{3}\pi R^{3}\rho \;\;(\because V=\frac{4}{3}\pi R^{3})$

Now substituting the value of M in equation no (1) we get

$g=\frac{G}{R^{2}}\times \frac{4}{3}\pi R^{3}\rho$

By solving this equation we get

$Or\;g=\frac{4}{3}\pi R\rho G$

Now we will determine the formula for the acceleration due to gravity at a point below the surface of the earth.

So consider any point inside the earth, below the earth’s surface at depth d. In which, if an object is placed it will only experience the force due to the portion of the earth of radius (R-d).

Now again let M' be the mass of the earth of the portion of radius (R-d). Then the acceleration due to gravity at that point

$g_{d}=\frac{GM'}{(R-d)^{2}}\to (3)$

Substituting the value of M' in equation no (3) we get

$g_{d}=\frac{G}{\left ( R-d \right )^{2}}\frac{4}{3}\pi (R-d)^{3}\rho$

$Or\;g_{d}=\frac{4}{3}\pi (R-d)G\rho \to \left ( 4 \right )$

Now dividing equation no (4) by equation no (2) we get

$\frac{g_{d}}{g}=\frac{\left ( R-d \right )}{R}$

$\therefore\;g_{d} =\left ( 1-\frac{d}{R} \right )g$

This is the formula for variation in acceleration due to gravity with depth.

It clearly shows that the value of acceleration due to gravity decreases with depth. At the center of the earth where d = R the acceleration due to gravity is zero i.e

$g_{d} =\left ( 1-\frac{R}{R} \right )g =0$