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 0.480 of the escape speed from Earth and (b) its initial kinetic energy is 0.480 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?

Answer 1

escape velocity of earth = 11.2 km/s

maximum height of projectile = v^2 * sin^2(theta) / (2g)

maximum height = (0.48 * 11200)^2 * sin^2(90) / (2 * 9.8)

maximum height = 1474560 m

radius of earth = 6371000 m

maximum height in multiple of radius of earth = 1474560 / 6371000

maxumum height in multiple of radius of earth = 0.2314

its kinetic enrgy = 0.48 * kinetic energy required to escape earth

0.5 * mv^2 = 0.48 * 0.5 * m * 112000^2

v = 7759.587 m/s

maximum height = 7759.587^2 * sin^2(90) / (2 * 9.8)

maximum height = 3071999.51074 m

maximum height in multiple of radius of earth = 3071999.51074 / 6371000

maximum height in multiple of radius of earth = 0.482

least mechanical energy required = 0.5 * m * escape velocity^2

least mechanical energy required = kinetic energy required to escape earth