# Gravitation above and below the earth's surface

1. Oct 29, 2014

### erisedk

1. The problem statement, all variables and given/known data
The value of acceleration due to gravity at a point P inside the earth and at another point Q outside the earth is g/2 (g being acceleration due to gravity at the surface of earth). Maximum possible distance in terms of radius of earth R between P and Q is

Ans: R/2 ( 2*root2 + 1 )

2. Relevant equations
Acceleration due to gravity at a depth d under the earth = g(1-(d/R))
Acceleration due to gravity at a height h above the earth = g(1-(2h/R))
g-acc. due to gravity at surface

3. The attempt at a solution
g(1-(d/R))=g(1-(2h/R))
d=2h

g/2 = g(1-(2h/R))
g/2 = g(1-(d/R))
d=R/2

h=d/2=R/4

So, d+h= 3R/4

In all these steps, not once did I think about the maximum possible distance.
And my answer is also wrong.

2. Oct 29, 2014

### OldEngr63

Your expressions for the acceleration of gravity above and below the surface of the earth are incorrect.

Look up Newton's Law of Gravitation to see how it varies with the radius.

For the gravitation under the surface, consider only the mass inside the sphere below the test point. The outer shell cancels out in that calculation (although showing this is a bit hairy!).

3. Oct 30, 2014

### erisedk

They're correct. d and h do not indicate the distances from the center of the earth. In case an object is placed at a depth d under the earth's surface or a height h above the earth's surface, it's distance from the center of the earth is R-d and R+h respectively.

4. Oct 30, 2014

### Orodruin

Staff Emeritus
Your expressions are series expansions of the true expressions and assume that h and d are small in relation to the Earth radius. This is no longer true if you want the gravitational acceleration to be half of that on the Earth surface.

5. Oct 30, 2014

### OldEngr63

I say again, look up Newton's Law of Gravitation.

6. Oct 30, 2014

### haruspex

One of them is correct, though - yes?

7. Oct 30, 2014

### Orodruin

Staff Emeritus
Yes, although I will let the OP figure out which of them for him-/herself.

8. Oct 31, 2014

### erisedk

I got (for above the surface)
g/2 = GM/(R+h)^2
For below the surface,
g/2 = GM/(R-d)^2

So, substituting GM/R^2 for g, I get

2R^2 = (R-d)^2
or R^2 = d^2 - 2Rd
&
R^2 = h^2 + 2Rh

How do I find h+d in terms of R with the pesky +2Rh and -2Rd terms?

9. Oct 31, 2014

### Orodruin

Staff Emeritus
This equation states that gravity would be increasing as you go down, which is not true. This would be true if all of the Earth's mass was collected in a point at the center. This is not the case, but you rather need to use a method that will give you the gravitational constant inside a spherical mass distribution.

Edit: On a side-track. It is probably easier to figure out the radii rather than to figure out the height and depth relative to the Earth surface.

10. Oct 31, 2014

### erisedk

Oh ok.
M= p(density)*4/3 pi R^3
So, mass that contributes to acceleration due to gravity at a distance r(b) [r(b)<R] from the center of the earth is (M*r(b)^3)/R^3
So, g/2 = GM r(b) / R^3
&
g/2 = GM / r(a)^2
where r(a) is the distance above the earth from the center of the earth.

Substituting g as GM/R^2, I get
r(b) = R/2
and r(a) = +R*(root 2) or -R*root 2

Distance between the two points is r(a) - r(b) = R/2 [2*root 2 - 1] or R/2 [2*root 2 + 1]

Since I require the maximum possible distance, the answer is R/2 [2*root 2 + 1].

Thank you so much everybody :D

11. Oct 31, 2014

### Orodruin

Staff Emeritus
Just one remark: Radius is by definition a positive number. However, the directions are free. Your solutions correspond to the max and min of the possible distances, where the directions are opposite/the same. Any distance in between these two are allowed for some angle between the directions.