Couple of questions about Gravitational field of an infinite plane 
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*Is it possible to find the gravitationaal field without using infinite integrals or Gauss's law? I would like to know if so because I haven't learnt doing infinite integrals or using Gauss's law yet.

*I also don't logically understand how an infinite plane can produce a finite field at a point. 
WE can easily find the gravitational field due to infinite rod with linear mass density $\lambda$ at a perpendicular distance $R$ from rod as $$2G\lambda /R.$$ 


*Since an infinite plane can be considered a collection of infinite rods, and since each of those rods apply a non differential field, shouldn't the sum be infinite?

*How would I integrate the fields of the rod for all rods? 

*Do I need to know infinite integrals here too? 
 A: Well you can find approximate values by reasoning but for the exact value you'll need at least to know limits.
Let's reason for a moment as you are proposing, but slightly different. Since the plane is infinite you can consider the point in the center of the plain, and at distance R from it. Now think of the plain as a collection of rings of very fine radial thickness dr. Each portion of the ring acts on the point with equal force to its opposite portion in the ring, and their vertical components will add up while their horizontal (parallel to plane) will annihilate. The contribution from all of them will be $$2\pi r dr \sigma \frac{G}{r^2+R^2} \sin{\theta(r)}$$ where $$\sin{\theta(r)}=\frac{R}{\sqrt{r^2+R^2}}$$ and $r$ is the distance to the center of the ring (projection of point in $R$ on the plane).
So the expression for gravitational pulling of each ring is: $$2\pi\sigma G\frac{Rrdr}{(r^2+R^2)^{3/2}}$$ and the total pull of the plane will result from adding all these contributions from increasingly larger radius $r$ rings.
The expression shows that adding contributions from ever larger radius rings ($r$) will be ever smaller, so you can see how infinitely distant rings will have infinitely small contribution, hence the finite value. 
If you're interested in the exact result, is $$2 \pi \sigma G $$ but I wouldn't know how to arrive to the result without at least using series and limits.
A: Here is an attempt to answer this question from a general relativistic point of view.
To enable a comparison with a spherical body, let the height above the plane be 'r', the 2 horizontal coordinates x and y. Time as measured by a distant observer will be 't' as in the Schwarzschild metric.
Consider a hollow spherical arrangement of test particles which are held in place above the infinite plane. (cf. "The Meaning of Einstein's Equation" by John C. Baez and Emory F. Bunn). Initially the particles are at rest with respect to each other and also with respect to the massive plane. At the center of the sphere is an observer.
The sphere of particles and the observer are now released and allowed to fall freely.
From Einstein's equation, the Ricci tensor is zero.
In particular, Rtt = 0 = Rrttr + Rxttx + Rytty i.e. the sum of the 3 tidal accelerations as measured by the observer in the sphere, who is now an inertial observer, is zero.
By symmetry, it would seem that Rxttx and Rytty are zero since the is no preferred horizontal direction for the particles to accelerate. Thus Rrttr also = 0.
By similar symmetry considerations, Rrr, Rxx, Ryy are also zero, and the space-time is flat.
Thus the geometry is just that of the uniformly accelerated observer of Minkowski space-time.
The gravitational force (weight) that a stationary observer would feel would thus simply decrease as 1/r, where in this case, r is proper radial distance.
This is of course vastly different from the Newtonian case, where the force (weight) would be independent of the height above the plane.
