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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.421 of the escape speed from Earth and (b) its initial kinetic energy is 0.421 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.) (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth? (a) Number Enter your answer for part (a) in accordance to the question statement 1.215 Units Choose the answer for part (a) from the menu in accordance to the question statement
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.421 of the escape speed from Earth and (b) its initial kinetic energy is 0.421 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.) (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth? (a) Number Enter your answer for part (a) in accordance to the question statement 1.215 Units Choose the answer for part (a) from the menu in accordance to the question statement
(a) At 0.421 escape speed, the radial distance is about 0.568 times Earth's radius.(b) At 0.421 escape kinetic energy, the radial distance is approximately 0.568 times Earth's radius.(c) The least initial mechanical energy requires launching at around 0.426 times Earth's radius.(a) Radial Distance at 0.421 Escape Speed:Escape Speed (v_esc):Substitute:G ≈ 6.674 × 10^(-11) N m²/kg²,M ≈ 5.972 × 10^24 kg, andR_E ≈ 6.371 × 10^6 m into :"v_esc = sqrt(2GM/R_E)"To find : "v_esc ≈ 11179 m/s".Projectile's Initial Speed (v):Calculate "v ≈ 0.421 * v_esc ≈ 4710 m/s".Radial Distance (r):Use "r = GM/v² * 1/R_E" with the known values to find "r ≈ 0.568 R_E".(b) Radial Distance at 0.421 Escape Kinetic Energy:Escape Kinetic Energy (K_esc):Calculate :K_esc = 1/2 * m * v_esc²"Using :m ≈ 5.972 × 10^24 kg" and "v_esc ≈ 11179 m/s" to get "K_esc ≈ 3.971 × 10^13 J".Projectile's Initial Kinetic Energy (K):Find "K ≈ 0.421 * K_esc ≈ 1.673 × 10^13 J".Radial Distance (r):Substitute values into "r = GM/v² * 1/R_E" to find "r ≈ 0.568 R_E".(c) Least Initial Mechanical Energy for Escape:Kinetic Energy (K):Using m ≈ 5.972 × 10^24 kg and v ≈ 4710 m/s, find K ≈ 7.030 × 10^13 J.Potential Energy (U):Using G ≈ 6.674 × 10^(-11) N m²/kg², M ≈ 5.972 × 10^24 kg, and r ≈ 0.426 R_E, find U ≈ -8.630 × 10^13 J.Total Mechanical Energy (E):Summing K and U, find E ≈ -1.600 × 10^13 J.Optimal Radial Distance for Minimum Energy:By setting dE/dr = 0, find the optimal radial distance r_opt ≈ 0.426 R_E.Multiple of Earth's Radius:0.426 times Earth's radius is the multiple of Earth's radius.The question probable may be:A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.421 of the escape speed from Earth and (b) its initial kinetic energy is 0.421 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.) (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?...