Is there a force that a body experiences as it approaches light speed?

Suppose there is a force that resists further acceleration. It becomes measurable at $v_+$. Working in 1-dimension for brevity, you can accelerate along $+z$ until you reach $v_+$. At that point, reverse thrusters and retrace you steps back to the original position.

Now fly in the $-z$ direction until you reach $v_-$.

You can now find the absolute rest frame for your flat universe. Defining the rapidity as:

$$ omega = tanh^{-1} v $$

then the average rapidity :

$$ omega_+ + omega_- $$

transforms under a $u$ boost along $z$ to:

$$ (omega_+ - tanh^{-1} u) + (omega_- - tanh^{-1} u) $$

which can be set to zero:

$$ (omega_+ - tanh^{-1} u) = -(omega_- - tanh^{-1} u)$$

$$ tanh^{-1} u = frac 1 2(omega_+ +omega_-) $$

So that:

$$ u = tanhbig(frac 1 2 (tanh^{-1}||v_+|| - tanh^{-1}||v_-||) big)$$

boosts you to a preferred rest frame, which relativity prohibits.

The real point of this exercise is to address the part that reads:

"As it approaches light speed, does some force actively resist further acceleration?"

which completely misses the point of relativity: you can't "approach the speed of light", no matter how fast some external observer measures your velocity, it has absolutely no bearing on your observables, and if it did, there would be an absolute rest frame in spacetime.

No, there isn't a force that resists further acceleration.

Suppose the galaxy is precisely mapped, and the crew knows the distance between stars in the route.

As times passes and the ship velocity increases, a naive ship navigator could think that eventually it is above light speed, by dividing the last travelled path by the time between starting and final point.

But the distance between the stars is contracting from the ship frame of reference, as the velocity increases. And the velocity, calculated by the length after correction divided by ship's time, never reaches light speed.

The relativistic momentum is given not by $mv$ but by $p=gamma mv$. If we copy over the Newtonian definition of force as (assuming we're in 1D space so I'm being cavalier about vectors for the moment):

$$f=frac{dp}{dt} = mfrac{dgamma}{dv}frac{dv}{dt}v+mgamma frac{dv}{dt}$$

we find (after some calculations, using $gamma = (1-beta^2)^{-1/2}$ where $beta=v/c$) that:

$$f = m gamma^3 frac{dv}{dt}$$

This is (one form of) Newton's 2nd law in 1D special relativity (with caveats, see below). To answer the OP's question then: the force can remain constant whilst the 'acceleration' $dv/dt$ goes to zero if $gamma^3$ increases accordingly. And indeed, since $gamma$ increases without limit as $vrightarrow c$, this will be the case.

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Important points skipped above:

1. Whilst the momentum $vec{p}$ and the force $vec{f}$ are both 3-vectors under rotations, only the momentum forms the space-like part of a 4-vector under Lorentz transformations (rotations+boosts). As such, $vec{f}$ is not that 'natural' a quantity in SR.

2. Some people claim that there is a 'relativistic mass' $m_{text{rel}}=gamma m$. But they run into the problem that it is still not true that $f=m_{text{rel}} a$.

3. The formula $vec{f}=mgamma^3 vec{a}$ is not correct. It is only correct in 1 dimension. In 2 and more, there are different equations for the component parallel and perpendicular to the force.

4. It's therefore usually best to work with the true relativistic quantities which are the four-momentum $P=(E/c,vec{p})$ and the four-force $F=dP/dtau$.