# Why does the gravitational force decrease below the Earth's surface?

1. Mar 21, 2017

### Steven Hansel

1. The problem statement, all variables and given/known data
Why does gravity force decreases below earth's surface?

2. Relevant equations
F = G (m1 m2)/r^2
F = force of gravity
G = Universal gravity constant
m1 and m2 = mass of the objects
r = distance between two objects

g = GM/r^2
g = gravity of the object
G = universal gravity constant
M = mass of the object
r = distance/radius between two objects

a = g

g1 = g0 (1-h/r)
g1 = gravity below earth surface
g0 = gravity in the earth surface
h = height of the object
r = distance between two objects

3. The attempt at a solution
I'm really confused on why does gravity force decreases below earth's surface. Gravity becomes stronger when the radius/distance become shorter between two objects. Now, an object is getting shorter radius/distance to the planet why the gravitational force decreases? Isn't it supposed to be stronger? I looked up at the internet and it said "The body will be attracted by the mass of the Earth which is enclosed in a sphere of radius (R - h)" Does this mean that gravity decreases because there are mass in that enclosed sphere?
So, will the gravitational force increases if i create a really deep hole below earth surface and put an object into the deep hole?

Last edited: Mar 21, 2017
2. Mar 21, 2017

### Bandersnatch

Can you write down the gravitational force equation and list all the variables in it?

3. Mar 21, 2017

### Steven Hansel

Done!

4. Mar 21, 2017

### Bandersnatch

Ok, consider an uniform solid sphere (a ball) composed of some material of density ρ. The sphere has mass M and radius R. Can you express the mass of the sphere in terms of its radius and density?

5. Mar 21, 2017

### Steven Hansel

Msphere = 4/3π.r^3.p
i think replace M with volume of a sphere and density of the sphere? into the equation of g = 4π.r^3.p/3.(r-h)^2
Correct me if i'm wrong thanks!

6. Mar 21, 2017

### Bandersnatch

so far so good.
Forget h. We're looking at gravitational acceleration at the surface of the sphere.
You forgot G and the factor of 1/3 from the volume equation. The equation should read:
$g=G \frac{4 \pi R^3 \rho}{3 r^2}$
Where R is the radius of the sphere and r is the distance to the centre of the gravitational field. When you're above the surface of a sphere, these two are different. But they're equal when standing on the surface of a sphere, which lets you write:
$g=CR$ where C are all the constants grouped together.

Is everything clear so far? Can you see what happens if you dig down?

7. Mar 21, 2017

### Steven Hansel

Wow, thanks! that really give me an insight, i was confused at the sphere part but now i understand. thanks again!

8. Mar 21, 2017

### PeroK

If you put an object in a very deep hole, then the Earth's mass above the object will be trying to pull it out of the hole, and no longer be pulling it towards the centre.

9. Mar 21, 2017

### Steven Hansel

That really helped me more, because of the radius decreasing as we go to the core of the earth, the mass of earth/radius into the core also decreases. thanks a lot!

10. Mar 21, 2017

### Staff: Mentor

That's not a good description as it contradicts the shell theorem. The object should feel no net force from any of the planet's mass above it. It only feels force from the mass below it, and said mass is less the deeper the object is.

11. Mar 21, 2017

### BvU

For the benefit of Steven:
It looks as if PerOk and Gneill disagree. They don't. They just have a different idea of 'above'. For P it's in a direction away from the object and away from the center of the earth towards the surface and for GNeill it's everything that's further away from the center of the earth than the object. PeroK: dome, Neill: shell.

Makes a difference.

The nice thing is that indeed the 'above' from PerOk and his 'below' (but outside the sphere with a radius equal to the distance from the object to the center of the earth) exactly compensate each other: no net force as Gneill states.