How about launching a HPR with a 3-axis gyroscope for stability and launch it off the ISS in the same direction of the ISS path? Wouldn't the combined speed of the ISS orbital velocity plus the the HPR rockets's added velocity cause the path of the rocket to slowly spiral upwards as it goes around the Earth, eventually escaping orbit?
A rocket can't "spiral slowly upwards" without producing continuous thrust. Once the motor stops burning, the rocket will be in free-fall and will "fall back" into orbit around the earth if it hasn't reached escape velocity.
As has been mentioned a couple of times, the rocket would have to increase it's speed by a value called "delta-v" to leave earth orbit. Delta-v varies based on where the rocket is and where it's going - see the chart at:
https://en.wikipedia.org/wiki/Delta-v_budget#Earth.E2.80.93Moon_space_.E2.80.94_high_thrust
To leave low-earth orbit and reach the moon, requires an increase in speed of about 3 kilometers/second. As I showed previously, an ideal G motor can't accelerate it's own mass to that speed. Since the specific impulse of larger HPR motors is about the same, a non-clustered/staged high-power rocket wouldn't be able to do it.
And it would a really large clustered/staged HPR rocket to reach that speed.
I just played with the values I used for the "ideal" G motor - total impulse of 160Ns, thrust duration of 10 seconds, mass of 0.07 Kg. This ideal G motor could reach delta-v by staging from a cluster of two motors to a single motor.
But, when I change the mass of the motor to something more realistic, like the weight of a G138 reload in its case, the powers-of-two rocket can't be accelerated to more than about 1.7 kilometers/second no matter how many stages are added. A ten-stage rocket with 512 motors in the first stage reaches about 1.7 kilometers/second. A twenty-stage rocket with more than 500,000 motors in the first stage will reach a speed which is only about .05 kilometers/second faster.
A larger HPR motor, like an O motor, might be able to reach 3 kilometers/second using the scenario above because the ratio of the mass of the propellant to the total mass of the motor is larger. But it's trivializing the problem to answer it by saying that you could just use a bigger motor.
-- Roger