Here, ve is the escape velocity and is the smallest velocity needed to escape the field of gravitational attraction. Comparing equations 6 and 7, we see that, for a given r, the escape velocity is a factor of v2 larger than the velocity necessary to maintain a circular orbit. Thus, if a satellite is on a circular orbit with velocity vc It should be noted that a satellite in a parabolic trajectory has a total specific energy, e, equal to zero. This means that when r increases, the kinetic energy is transformed to potential energy such that, at infinity, the residual velocity is equal to zero. Hyperbolic Trajectory(e> 1) For 0oo-tcos(1/e), we have r- 00. Hence, the trajectories are open. Moreover, if the velocity u, at a given r is known, the residual velocity is simply voo =U-Ve=U-V2u/r=v2E. For a given energy level, the eccentricity of the orbit is determined by h. An important parameter for hyperbolic orbits is the turning angle, a, which is the angle through which the velocity changes along the trajectory as the bod travels from -oo to oo. The turning angle is given by 8=2(0oo-),or 8=2sin(1/e). Below we show several hyperbolic trajectories which have identical terminal velocities for different values of the eccentricity (and turning angle =1.1 e=1.5 e=2 Example Different orbits as a function of v We consider the problem of launching a satellite at an altitude d with an initial velocity vo, along the direction tangent to the earth's surface. We consider the different trajectories that are obtained as we vary the magnitude of vo 0c≡√(R+d, the trajectory will be a circle(e=0).Fort=toe≡√2/(R+d the trajectory will be parabola(e 1). For vo >voe, the trajectory will be a hyperbola, whereas forHere, ve is the escape velocity and is the smallest velocity needed to escape the field of gravitational attraction. Comparing equations 6 and 7, we see that, for a given r, the escape velocity is a factor of √ 2 larger than the velocity necessary to maintain a circular orbit. Thus, if a satellite is on a circular orbit with velocity vc, the necessary ∆v to escape is (√ 2 − 1)vc. It should be noted that a satellite in a parabolic trajectory has a total specific energy, E, equal to zero. This means that when r increases, the kinetic energy is transformed to potential energy such that, at infinity, the residual velocity is equal to zero. Hyperbolic Trajectory (e > 1) For θ∞ → ± cos−1 (1/e), we have r → ∞. Hence, the trajectories are open. Moreover, if the velocity v, at a given r is known, the residual velocity is simply v∞ = v − ve = v − p 2µ/r = √ 2E. For a given energy level, the eccentricity of the orbit is determined by h. An important parameter for hyperbolic orbits is the turning angle, δ, which is the angle through which the velocity changes along the trajectory as the body travels from −∞ to ∞. The turning angle is given by δ = 2(θ∞ − π), or δ = 2 sin−1 (1/e). Below we show several hyperbolic trajectories which have identical terminal velocities for different values of the eccentricity (and turning angle). Example Different orbits as a function of v0 We consider the problem of launching a satellite at an altitude d with an initial velocity v0, along the direction tangent to the earth’s surface. We consider the different trajectories that are obtained as we vary the magnitude of v0. For v0 = v0c ≡ p µ/(R + d), the trajectory will be a circle (e = 0). For v0 = v0e ≡ p 2µ/(R + d), the trajectory will be parabola (e = 1). For v0 > v0e, the trajectory will be a hyperbola, whereas for 4