Question

A projectile is launched vertically from the surface of the Moon with an initial speed of 1230 m/s. At what altitude is the projectile's speed three-fifths its initial value?

Answer 1

g = acceleration due to gravity at moon = 1.6 m/s²

v_o = initial speed at the time of launch = 1230 m/s

v_f = final speed = (3/5) v_o = (3/5) (1230) = 738 m/s

r_o = initial distance from the center = 1.737 x 10⁶ m

h = height above the surface

r_f = final distance from the center = (1.737 x 10⁶ ) + h

m = mass of the projectile

M = mass of moon = 7.35 x 10²² kg

Using conservation of energy

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

(0.5) m v_o² - GMm/r_o = (0.5) m v_f² - GMm/r_f

(0.5) v_o² - GM/r_o = (0.5) v_f² - GM/r_f

(0.5) (1230)² - (6.67 x 10^-11)(7.35 x 10²²)/(1.737 x 10⁶) = (0.5) (738)² - (6.67 x 10^-11)(7.35 x 10²²)/((1.737 x 10⁶ ) + h)

h = 3.6 x 10⁵ m