Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius R_E gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.484 of the escape speed from Earth and (b) its initial kinetic energy is 0.484 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 speed on earth = V = sqrt(2GM/R_e)

a) If projectile speed v = 0.484*R_e = 0.484 * sqrt(2GM/R_e)

Initial energy on the earth Ei = (1/2)mv² - GMm/R_e

At the maximum point final energy Ef = -GMm/r

From energy conservation, Ef = Ei

-Gmm/r = -GMm/R_e + (1/2)m*0.484² * (2GM/R_e)

-1/r = -1/R_e + 0.484²/R_e

-1/r = -0.7657/R_e

r = 1.306R_e

b) If projectile speed v = 0.484R_e = 0.484sqrt(2GM/R_e)

Initial energy on the earth Ei = (1/2)mv² - GMm/R_e

At the maximum point final energy Ef = -GMm/r

From energy conservation, Ef = Ei

-Gmm/r = -GMm/R_e + (1/2)m * (2GM/R_e) * 0.484

-1/r = -1/R_e + 0.484/R_e

-1/r = -0.516/R_e

r = 1.938R_e

c) Least energy E = (1/2)mvE² - GMm/R_e

E = (1/2)m*2GM/RE - GMm/R_e

E = -GMm/2R_e