Question

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 0.437 of the escape speed from Earth and (b) its initial kinetic energy is 0.437 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.)

Answer 1

Potential energy is given as GMm Escape speed is given as Total energy of the projectile is conserved. The final kinetic energy of the projectile is zero Initial speed is 0.437 of escape speed. That is, v, 0.437v Here, v is the escape speed. Apply law of conservation of energy GMim m(0.437v) 2GM (0.437-mGnGMm (0.437)-1)= 1-(0.437) = 1.24K.

Initial kinetic energy is 0.437 of the kinetic energy required to escape the earth. That is., -m,2 = (0.437 )-mv' Apply law of conservation of energy 2 GMin GMm (0.437 )-mv2 = 0.437)2GMGMm GMm ( 0.437)-1)= -- 1-(0.437) 1.78R