how high above the surface of the Earth would an object need to be for its weight to be half its weight on the surface?

The mass of an object at any point will remain the same( unless its velocity approaches that of light, of course )So, for the weight to reduce to half its value, the acceleration due to gravity has to reduce to half its value on earth.

now, the relation between acceleration due to gravity and distance from the center of the earth is given by

g = 1/R2
where R is the radius of the earth which is 6400 kms (approx)

let 'h' be the height at which the weight of the rocket will be half its initial value.

so, gh / g = R2 / (R+h)2

but gh = g/ 2

substituting it in the above expression yields h=2649.6 kms (approx)

so, at this height, the weight of the rocket will be half its initial weight.