A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth. (a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is one-third of the escape speed from Earth? (b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is two-fifths of the kinetic energy required to escape Earth U = -GMm/r K = 1/2mv^2 Radius of earth = 6.38e6 Mass of earth = 5.98e24 (a) Escape speed of earth = 11.2 km/s =11200 m/s U1 + K1 = U2 -GMm/r + 1/2mv^2 = -GMm/(r+h) m cancels out 11200/3 = 3733 (-6.67e-11 * 5.98e24) / 6.38e6 + 1/2 * 3733^2 = (-6.67e-11 * 5.98e24) / (6.38e6 + h) -55550537.32 = -3.98866e14 / (6.38e6 + h) h = 800236.5781 h / Re = 800236.5781 / 6.38e6 = 0.125 times the radius of the earth (b) U1 + K1 = U2 -GMm/r + 1/2mv^2 = -GMm/(r+h) but it asks for 2/5 the KE, so -GMm/r + (2/5)1/2mv^2 = -GMm/(r+h) m cancels out -GM/r + 1/5 v^2 = -GM / (r+h) Plugged in the same numbers as part (a), except for v, which is 11200 m/s I got h, divided by Radius of earth (6.38e6) and get 0.670 times radius of earth Did I make any mistakes? Because the answers are wrong, apparently
Well, at least the eq looks right for me. The only thing I could think of, is that h is counted also from the center of earth. So instead of .125, h is 1.125?