A relativity paradox on gravitational force Imagine an infinite number of point masses $m_0$ moving with velocity $V_0$ on $y$-axis towards $+y$ (with equal distance $L_0$ between each two consecutive masses).
Scenario 1:
A mass $m_1$ is moving on line $x=1$ with velocity $+V_1$ ($V_1$ towards $+y$). This mass is feeling gravitational force $F_1$ from the train of masses $m_0$.
Scenario 2:
Mass $m_1$ is moving on line $x=1$ with velocity $-V_1$ ($V_1$ towards $-y$). This mass is feeling gravitational force F2 from the train of masses $m_0$.
Suppose $V_0$>$V_1$.
I understand $F_1 < F_2$, because when moving in the -Y direction m1 sees relative velocity $V_0 + V_1$ which is greater than when it's moving in +Y direction ($V_{\text{rel}} = V_0 - V_1$. So, when moving in $-y$ direction, due to higher relative velocity, m1 sees more length contraction, so mass density per length is more, therefore gravitational force $F_2$ is greater than $F_1$.
Now, how does a stationary observer at $(x,y)=(0,0)$ would see this? would he/she agree $F_1<F_2$?
 A: You appear to be trying to apply Newton's laws of gravity to this situation. But Newtonian gravity is not compatible with special relativity, which is why general relativity had to be developed, and in any situation involving significant relativistic effects GR must be used rather than Newton's laws.
The spacetime curvature created by the bodies depends on the stress energy tensor, which is independent of the reference frame. In particular, it is independent of the relative velocity of any observer or negligible test mass.
A: Gravity can't be put to work in Special Relativity with such simplicity as Electrodynamics. If you were dealing with charges instead of masses, you'd be asking about a classical problem in Electrodynamics which is dealt with, for example, in the book by Griffiths. With gravity, the problem is more subtle.
Once relativistic effects are taken into account, the gravitational field can't be interpreted exactly as a force, which distinguishes it considerably from the electromagnetic problem. Instead, gravity is understood under the framework of General Relativity as spacetime curvature. In other words, gravity is not a force, but rather an effect of the shape of spacetime. Different observers attribute different coordinates to spacetime, just like different people might choose to attribute different coordinates to the Earth's surface. Nevertheless, any two observers will always agree on the shape of spacetime (or of the Earth).
The apparent paradox in your question is due to the fact you are using Special Relativity outside its domain of validity. SR works fine for situations in which gravity is negligible, but once you want to consider gravitational physics, GR is the way to go. Within GR, gravitation is nothing more than the curvature of spacetime and hence there is nothing to be worried about: by construction all observers will agree on what is gravity.
