Value of g(9.8 m per second square) decreases going above sea level then it should increase as we move below?

Question

earth surface by digging it.What do you say?

Answer 1

No. The formula for g is GM/r2, where G is the gravitational constant, M is the mass of, here, the earth, and r is the distance. However, this assumes that all of the mass of the Earth is at one point. As long as you are outside the Earth, it all works out OK, ans so g decreases with distance r as you would exepct. However, as soon as you dig into the Earth, some of the Earth's mass will be outside, so only the sphere of mass within the radius you are at now counts (if I remember correctly, teh "shell" outside where you are cancels itself out) To take it to the extreme, if you were in the centre of the Earth, there would be no downward force due to Gravity, because all of the Earth's mass is "above" you and it pulls equally in all directions.

Answer 2

(Before my English makes you doubt, I'm a Russian)

The first answerer is right, what I can add is some math explaining the matter.

Newton's Law of Universal Gravitation:
F = G*m1*m2/r^2, G - the gravitational constant, m1 and m2 masses of the interacting bodies, r - distance between the centers of the bodies. The law works for spheric simmertical or point bodies.

Let's now consider the case of a small body being placed underneath the Earth surface. According to the Gauss theorem we should take into accont only that part of total mass which is inside a sphere with radius equal to the distance from the center of Earth to the body. Let's denote that mass as m1,
small body mass as m2,
total Earth mass as M,
Earth radius as R,
mean Earth density as rho,
distance from the center of the Earth to the body as r,
gravitational acceleration on the surface as g0,
gravitational acceleration underneath the surface as g.

Earth volume V = 4*Pi*R^3/3,

rho = M / V = 3*M / 4*Pi*R^3,

m1 = rho *4*Pi*r^3 / 3 = M *r^3 / R^3.

m2*g = G*m1*m2 / r^2,

g = G*m1 / r^2.

Substituting m1 we get:

g = G*M*r / R^3.

Now let's get rid of G and M, they are hard to remember:

m2*g0 = G*M*m2 / R^2,

g0 = G*M / R^2,

G*M = g0*R^2.

Finaly we get:
g = g0*r / R.
It is lineary decreasing when we moving to the center of the planet. On the surface it becomes equal to g0. As we now see, gravitational acceleration is maximum on the Earth surface.

One additional remarc. Due to the Earth density is not unifom (deeper layers are more dense), g is actually increasing when depth is small, but this effect is very weak.

Answer 3

No.

When you go below sea level you have some of the gravity of the earth's mass pulling in a direction against the downward vector.

When you get to the center of the earth all of the gravitational attractions should cancel each other out and you should be in equilibrium.
.

Answer 4

This can be mathematically proven using Newton's Law of Universal Gravitation.
Mathematically proved by,

G= Universal Gravitational Constant
F= Gravitational Force of Attraction
m1= mass of earth
m2= mass of object
W= weigth of object
r(sq)= squared distance from the center of the earth to
the object

F= G(m1)(m2)/r(sq)

Since F=W, where W=(m2)g

Therefore, (m2)g= G(m1)(m2)/r(sq)

g=G(m1)/r(sq)

From the end equation, it is proven that g is inversely proportional with the distance from the object and the centre of the earth.So, what you are saying is true( the credit also goes to Newton).

Answer 5

It does increase at poles as compared to equator but rigth at the center of earth, it is zero.