Question

what will be value of g at the centre of the earth

Answer 1

Hi!

Leaving behind everything, lets start from the start and answer your question.

So, First of all,

What is gravitation or gravitational force & acceleration due to gravity ? Many people get confused between gravitational force and acceleration due to gravity. Let me make it clear to you at first that they both are different. Gravitation if a force (S.I. Unit is Newton (N)) and acceleration due to gravity is acceleration (S.I. Unit is m/s²). They both are different things at all!

→Gravitation is a force that exists between any two matters (masses).

→Acceleration due to gravity is acceleration produced in a freely falling body in vacuum.

In the above lines, the term free fall is used.

What is free fall? A object is said to be freely falling if it's moving towards the center of a planet only by the gravitational force of the planet and no other force is acting upon the object- not even air resistance.

Acceleration due to gravity on earth on an average is always 9.8 m/s².

Here, the term on an average is used because acceleration due to gravity (g) is equal to (G*M)/(R)² , where G is the universal gravitation constant, M is the mass of the planet and R is the radius of the planet(we'll not go into that details....); and since the earth is not a perfect sphere, the radius varies at the equator and the poles, and, hence, the value of g (acc. due to gravity) also varies.....

In the above paragraph, it is said that :

$acceleration \: due \: to \: gravity \: (g) \\ \\ = \frac{[universal \: gravitation \: constant \: (G) \times mass \: of \: the \: planet \: (M)]}{[radius \: of \: the \: planet \: (R)]^2} \\ \\ \\ OR, \: g = {(G \times m)} \div {(r^2)} \: ....................................$

The value of universal gravitation constant (G) is defined in newton's universal law of gravitation and it's value is $6.67 \times {10^{-11}}$ N ms² / kg ².

Since, g = Gm/r² , if radius is increased, the value of g will decrease and vice-versa.

Now, read the last paragraph very carefully.........

If an object is placed above the surface of the earth, The value of g will be less than 9.8 m/s². Now, you will think that if an object is placed under the surface of the earth, the value of g will increase. But, this is false. It will still decrease. And, at the core (center of the earth), it's value will be 0 m/s².