2014-06-13 05:40:16 8 Comments

Why does gravitational force of the Earth decrease as we move towards the centre of the Earth? Where as inverse square rule says that distance is less than gravitational force is more.

### Related Questions

#### Sponsored Content

#### 2 Answered Questions

### [SOLVED] Why do we not account for the radius of the Earth when calculating the gravitational force between the Earth and an extra-terrestrial body?

**2017-07-18 11:42:57****Pancake_Senpai****140**View**0**Score**2**Answer- Tags: newtonian-gravity mass terminology planets distance

#### 5 Answered Questions

### [SOLVED] Near Earth vs Newtonian gravitational potential

**2016-10-14 15:52:35****Bobbie D****1333**View**13**Score**5**Answer- Tags: newtonian-gravity potential-energy earth conventions approximations

#### 7 Answered Questions

### [SOLVED] Distribution of Gravitational Force on a *non-rotating* oblate spheroid

**2014-11-05 07:01:45****user49111****2405**View**9**Score**7**Answer- Tags: newtonian-mechanics newtonian-gravity

## 3 comments

## @Frederic Brünner 2014-06-13 05:46:09

This is because as you get inside the earth, you are not only pulled towards the center but also attracted by matter "behind you", i.e. there is a force counteracting the motion towards the center. The result a decrease in the total acceleration as you move towards the center, and as you reach it, it should be zero (provided you manage to stay there without moving). To visualize this, here is a diagram showing the gravitational potential energy in the presence of a uniform sphere, which approximately represents the earth:

In the picture, $a$ represents the radius of the surface. As you can see, inside the sphere, the energy (and the force) decreases linearly as you approach the center. The inverse square rule only describes the behaviour of the field outside the sphere. Note that the linear behaviour is just valid in the approximation of uniform density, which is not realistic. This answer should however give an idea of what happens to the gravitational field inside a spherical body.

## @Stan Liou 2014-06-13 10:44:03

It's not actually linear because the density is not constant, but rather increases nearer the center, up to about a factor of $2$ compared to the average. However, conceptually this answer is completely correct: the density doesn't diverge at the center, so the force still goes to zero.

## @Frederic Brünner 2014-06-13 12:19:18

You are correct. As I thought I had pointed out clearly, my answer just addresses the uniform-sphere-approximation. I have added a remark to clarify this issue.

## @AkshayP 2014-06-13 05:51:46

When you are on the surface of Earth, we experience gravitational force from center of gravity (we can simplify and say the source is center of the Earth). When we move towards center of Earth, we experience gravitational force from part of Earth above us and part of Earth below us. So these two force cancel out each other. So at center of Earth you will be floating (hypothetically).

## @Lord_Gestalter 2014-06-13 05:51:33

Inside a arbitary rigid body the gravitational forces of the masses outside the sphere around the center of mass on which the measurement takes place (sphere, not center) compensate each other. (Long, incomprehensible sentence, I know. Hegel would be proud).

Therefore only the mass inside this sphere count for the point mass model.

While the force should increase at constant mass with $\frac{1}{r^2}$, the effective mass decreases with $r^3$ in bodies with homogenous mass densities. Therefore the force will decrease with $r^1$