Gravity on a flat earth?

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Sir i am getting 4241870.638 m .Please tell is my answer correct???
HERE IS THE PROCEDURE
Use Gauss’ Law of gravitation
Consider the cylinder of height 2t and radius d orientated such that its radial axis (z-axis) is perpendicular to the slab(here i treat new earth as slab).

First, we consider the field outside of the slab (|t| >T/2){Here T is the thickness of the slab)

2(πd^2)gout = Φ = −4πGMenclosed

Menclosed = V ρ = πd^2Tρ{ρ is the density of slab}
gout = −2πGρTeˆz

HERE gout is a vector quantity.

So put the values of the ρ,G,g we get the value of T....Show more

Consider mass m on the surface of the earth at a distance x from it.
Perpendicular dropped from m on the earth's surface is at O.
Consider a thin annular ring of the earth's surface of radius r and width dr.
A small element of this annular ring of length dl having thickness t will have mass
= t (dl) (dr) (5515) kg
The force of attraction on mass due to this small element of the earth,
dF (due to the small element dl of the annular ring)
= Gm * [t (dl) (dr) (5515)] / (r^2 + x^2)

All horizontal components of this force around the ring will cancel each other and the vertical components will add up
=> dF (due to the entire mass of the annular ring)
= Gm * [t (2π r) (dr) (5515)] / (r^2 + x^2) * x / √(r^2 + x^2)

Integrating between limits r = 0 to r = ∞,
Total force due to the earth
= 5515 πGmt x ∫ 2r / (r^2 + x^2)^(3/2) dr ... [r = 0 to r = ∞]
= - 11030 πGmt x / √(r^2 + x^2) ... [r = 0 to r = ∞]
= - 11030 πGmt x [0 - 1/x]
= 11030 πGmt

Equatting to mg
=> 11030 πGmt = mg
=> t = g / (11015 πG)

Plugging g = 9.81 and G = 6.67428 x 10^-11
t = (9.81) / (11030 π * 6.67428 x 10^-11) m
= 4241.696 km.

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EDIT:

As t is not small compared to x, revising the steps above,
The force of attraction on mass due to this small element of the earth,
dF (due to the small element dl of the annular ring)
= Gm * [t (dl) (dr) (5515)] / [r^2 + (x + t/2)^2]

All horizontal components of this force around the ring will cancel each other and the vertical components will add up
=> dF (due to the entire mass of the annular ring)
= Gm * [t (2π r) (dr) (5515)] / [r^2 + (x + t/2)^2] * (x + t/2) / √[r^2 + (x + t/2)^2]

Integrating between limits r = 0 to r = ∞,
Total force due to the earth
= 5515 πGmt (x + t/2) ∫ 2r / [r^2 + (x + t/2)^2]^(3/2) dr ... [r = 0 to r = ∞]
= - 11030 πGmt (x + t/2) / √[r^2 + (x + t/2)^2] ... [r = 0 to r = ∞]
= - 11030 πGmt (x + t/2) [0 - 1/(x + t/2)]
= 11030 πGmt

Equatting to mg
=> 11030 πGmt = mg
=> => t = g / (11015 πG)

Plugging g = 9.81 and G = 6.67428 x 10^-11
t = (9.81) / (11030 π * 6.67428 x 10^-11) m
= 4241.696 km.

[Note: the answer remains the same even for x << t.]

[I shall recheck the steps and calculations tomorrow.]...Show more

Less than 1 mm -- infinity is a BIG number.
Not to mention infinity squared!

Of course, on your thin, flat earth; the gravity would pull us sideways....Show more