Inside the Earth is the gravitaional acceleration smaller or greater as compared to that on its surface ?

Question

Theoretically, it should be greater but then how come the value of g is zero at the center of earth. Please give exact answer.

Answer 1

'g' is the symbol for the acceleration caused by a gravitational force, which in turn is caused by mass. Thus the Earth has mass, sufficient to cause a gravitational force 'G', which in turn acts on all mass around it producing an acceleration 'g'.

Now, at the surface of the Earth, all of the Earth's mass is beneath you, hence why you feel the effect of gravity acting downwards and pulling you down. As you approach the center of the Earth the value of 'g' gets smaller.

At the exact center the Earth's mass is no longer all beneath you, rather it is distributed around you equally in all directions. Therefore, there is no nett force pulling you in one direction, rather an overall force pulling you in all directions at once ! As there is no single direction of action of gravity at this location, the acceleration force 'g' is zero.

Answer 2

Smaller. The mass term in the inverse square law for a spherically symmetry body should be the mass of the earth *interior* to the radius r equal to your distance from the center. Newton proved this himself, and it follows from Guass's law. Assuming earth's density D were uniform, for example (although it's really not), the interior mass would increase in proportion to r^3 (from D 4 pi r^3 /3) until you reach the surface. Divide by r^2, then, and you get g increasing in proportion to r. That is, it drops off linearly as you go down, unit it's zero at the center.

Answer 3

When you dig into the surface of earth, the mass of sphere whose radius is equal to (Radius of earth - the depth through which you have dug the hole) plays role in the force due to gravity.

This is because, when you dig, the force increases with square of distance (consider equation G*m1*m2/r^2 => when r decreases the force increases by r^2) Perhaps this is what makes you think that it should increase.

However as you dig, the mass of the remaining sphere (R-depth) decreases by cube of the distance. Hence the gravitational acceleration, in fact decreases.

When you move away from surface of earth to a height h then the gravity will be (G*m1*m2/(r+h)^2) which is again less than g.

The accn due to gravity is thus, maximum at the surface of earth.

Answer 4

Theoretically it should not be greater. And presto its not.

Its quite complex to calculate gravity inside a body - you have to do volume integrals. However, for a uniform sphere it is simple - you can use Gauss' theorem. This shows that the force of gravity will be basically what it would be at that radius if you removed all of the material overhead - its effect just cancel. So as r falls, the mass of the Earth falls as r^3 - and gravity with it. At r = 0, m = 0.

Answer 5

At the center of the earth,g=0.

As u go down from the surface,acceleration due to gravity decreases.

Answer 6

It is the same. Acceleration due to gravity is constant and in the direction of the centre of the earth. The direction is always towards the centre of the earth.

But at the centre it is zero, because past the centre the value is same but direction changes, if at one side is considered +ve then at the other side it is -ve.

You wrote "Theoretically, it should be greater but then how come the value of g is zero at the centre of earth. Please give exact answer".

You are and the people who gave you different answers are mixing up gravitational force with gravitational acceleration. Acceleration value is constant at 32ft/sec x sec.

It is the gravitational force that change. According to the laws of gravitational force.
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Laws of gravitational force states that, it is directly proportional to the mass of the object, it acts from the centre of the mass ie centre of gravity, and it is inversely proportional to the square of distance between the centre of the masses.
This means that as the distance increase the force decreases. So if the distance is d then it is dxd and dxd = Fourth power of d (d.d.d.d).

Answer 7

Look.........we can derive an expression for the acceleration of the body of mass 'm' situated below the earth surface at a height 'h'

then,
g'=g[1-h/R]

And when the body is situated at the centre of the earth.....then the value of g is 0 because the depth at centre of earth is h-r
then
g'=g[1-R/R] Therefore g=0

Answer 8

The law of universal gravitational states that gravitational force can be calculated by following formula :

F = G ( m1xm2 )/ r²

Since 'g' is acceleration due to gravity, therefore we can write that acceleration due to gravity at any point on earth as g = Gxm1/r² where G is constant, r is its radius on that point and m1 is mass of earth if its radius is r.

As we go inside the earth radius r decreases but at the same time the mass also decreases as the mass of earth at that point should be calculated by the decreased radius.

At the centre of the earth its mass is zero, and hence g is zero.

Answer 9

The gravity of the earth is greatest at the surface, it is smaller inside the earth.

Imagine taking a series of spheres outwards from the core of the earth, each will contain more mass than the previous one, meaning that gravity steadily increases. At the centre, your sphere contains no mass, hence no gravity.

Answer 10

According to the theory,
g'= g(1-2d/R)
where
g'=the changed gravity due to certain parameters
g=original gravity of earth
d=depth
R=radius of earth

At the center of the earth,
d=R
=>g'= g(1-R/R)
=>g'= g(1-1)
=>g'= g(0)
=>g'= 0

Thus the value of accleration due to gravity is zero at the center of the earth and maximun at the surface.