An object is launched from the surface of the Earth with speed v1 at an angle θ = 30◦ to an axis perpendicular to the surface. The object’s low-orbit path is displayed in the figure below, reaching a maximum height (called apogee) of 3R⊕ from the Earth’s center, where R⊕ = 6371 km. The mass of the Earth is M⊕ = 5.972 × 1024 kg.a. Is angular momentum conserved relative to the center of the Earth? b. Is energy conserved in this motion? c. What is the projectile’s launch speed from the surface, v1, and its speed at apogee, v2? 3RV.

a. As the object is simply launched without any engines, it attains certain angular momentum. Ignoring the effects of the atmospheric drag, there are no non-conservative forces acting on this object. The only force acting on it is the force of the Earth’s gravity, which is conservative. This force acts radially inward on the object’s trajectory around the Earth. Since its tangential component is zero, it constitutes no torque. Since no torque (τ) acts on the object once it is launched, its angular momentum (L) is conserved. This can be expressed by using Newton’s second law for rotation as follows: τ=0dLdt=0 ∵dLdt=τL=constantb. Again, neglecting the effects of air resistance, there are no dissipative forces acting on the object once it is launched. If G, m, and r denote the gravitational constant, the object’s mass, and its distance from the Earth’s center, respectively, its gravitational potential energy (V) can be expressed as follows: V=-GM⊕mr The negative sign indicates that the object is bound to the Earth due to gravity. Thus, the object’s potential energy is purely dependent on its position (r) relative to the Earth. Hence, the system is conservative or its total energy is conserved.c. Let r1 and r2, and v1 and v2 denote the object’s distances from the Earth’s center and its speeds at launch and at apogee, respectively. Hence, r1 and r2 are R and 3R, respectively (R is the Earth’s average radius). The angular momentum conservation expression can be used to deduce the relationship between v1 and v2 as follows: mv1r1=mv2r2v2=v1r1r2=v1R3R=v13Let K1 and K2, and V1 and V2 denote the object’s kinetic and potential energies at launch and apogee, respectively. Use the above relation in the energy conservation relation and solve for v1 and v2 in terms of the constants as follows: K1+V1=K2+V212mv12+-GM⊕mR=12mv22+-GM⊕m3Rv122-v1322=GM⊕R-GM⊕3R ∵v2=v13v1=3GM⊕2R2v2=GM⊕6R2 ∵v2=v13Evaluate v1 and v2 by substituting appropriate constants given in the above expressions as follows: v1=36.67×10-115.972×1024 kg26371 km1 km103 m=9684.22 m/sv2=6.67×10-115.972×1024 kg66371 km1 km103 m=3228.07 m/s