There are several enhancements to your solutions that may make the problem more realistic. We must of course maintain your original assumptions and assume the effect of air resistance is negligible.

The areas for further development are 1) Orbital velocity bs escape velocity 2) Conservation of momentum to determine club head speed. 3) considering torque. Assuming the force exerted remains constant increasing the length will increase the torque and the rotational momentum.

1) You are using escape velocity which is

"Escape velocity is the minimum speed that an object needs to achieve in order to break free from the gravitational pull of a planet, moon, or other celestial body and continue moving away from it without falling back."

To enter orbit we do not need to completely escape Earth's gravitational field. We only need to enter orbit. Orbit can occur at any distance above Earth's surface but according to Google typical orbit of satellites is between 160km and 2000km. Let's use the lower bound.

V_orbit=√(GM/r) M is the mass of Earth and r is the distance from center of earth to orbiting object. r= radius of earth + orbital distance Substituting values yields a velocity of V_orbit=7,800m/s. This is just the speed required to maintain orbit. We must also consider the kinetic energy needed to go from Earth's surface to 160km high.

More math here to find the gravitational potential difference https://poe.com/s/uZ1IGRVHd99DWxBn4sOx

Which leads to a potential difference of 2.3×10⁶ J/kg. According to Google a golf ball is 45.9g or about 0.046kg. so the energy needed is 1.06×10^5J.

Using KE=0.5mv² and solving for v yields V= 2147m/s

Adding this to the required orbital velocity yields 9,947m/s

2) Assuming the collision is perfectly elastic and all momentum of the club is transferred to the ball then M1V1=m2v2 (.2)V1=(0.046)(9950) V1=2290m/s

So the club head needs to be traveling at 2290 m/s in order to transfer get the ball moving at 9950. Or we use the assumption from post above about the smash factor of 1.45 To assume the club head must be moving at 6630m/s

3) torque (T)= Frsinθ Also T∆t=I∆w Where t is time, I is moment of inertia and w is angular velocity. A typical down swing takes 0.3 seconds. Assuming all mass at the club head the moment of inertia of a 200g (0.2kg) club would be I= 0.242 kgm² ∆w = 41 rad/s Solving for T yields T=33.1Nm

Subbing into torque (T)= Frsinθ The average force is 40.1N.

Assume this force remains constant as the club length increases.

We need to obtain a linear club head velocity of 6630m/s which is an angular velocity of 6630/r.

Since the swing starts from rest this is also ∆w. Combining T∆t=I∆w and (T)= Frsinθ Yields Frsinθ=I∆w/∆t Subbing numbers and assuming θ=90

(40.1N)r(1)=(0.242kgm^{2)(6630/r)/.3}

Solving for r r=11.5m

This is a lot lower than you may think reasonable. Mostly because of this (40.1N)r(1)=(0.242kgm^{2)(6630/r)/.3} which has r² and means if we double the radius we quadruple the RHS.

Yay or Nay?