y of the rocket, V = 5 km/s = 5 × 103 m/sMass of the Earth, Me = 6 × 1024 kgRadius of the Earth, Re = 6.4 × 106 mHeight reached by rocket mass, m = hAt the surface of the Earth,Total energy of the rocket = Kinetic energy + Potential energy= (1/2)mv2 + (-GMem / Re)At highest point h,v = 0And, Potential energy = -GMem / (Re + h)Total energy of the rocket = 0 + [ -GMem / (Re + h) ]= -GMem / (Re + h)From the law of conservation of energy, we haveTotal energy of the rocket at the Earth’s surface = Total energy at height h(1/2)mv2 + (-GMem / Re) = -GMem / (Re + h)(1/2)v2 = GMe [ (1/Re) – 1 / (Re + h) ]= GMe[ (Re + h – Re) / Re(Re+ h) ](1/2)v2 = gReh / (Re + h)Where g = GM / Re2 = 9.8 ms-2∴ v2 (Re + h) = 2gRehv2Re = h(2gRe – v2)h = Rev2 / (2gRe – v2)= 6.4 × 106 × (5 × 103)2 / [ 2 × 9.8 × 6.4 × 106 – (5 × 103)2h = 1.6 × 106 mHeight achieved by the rocket with respect to the CENTRE of the Earth = Re + h= 6.4 × 106 + 1.6 × 106 = 8 × 106 m.I HOPE this will HELP youIf not then comment me