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 0.440 of the escape speed from Earth and (b) its initial kinetic energy is 0.440 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

Potential energy , U = -GMm/r

Escape velocity v = 2GM/Re

(a)

Total energy of projectile is conserved

K+U = constant

initial speed is 0.440 of escape speed ,Let it reaches the radial distance r. at this distance kinetic energy become zero.

Let it reaches the radial distance r. at this distance kinetic energy become zero.

K_s + U_s = K_r + U_r

0.5*m*[0.440*2GM/Re]² - GMm/Re = 0 - GMm/r

0.1936 /Re - 1/Re = -1/r

1/r = 0.8064/Re

Radial Distance ,r =1.24*Re

(b)

initial kinetic energy is 0.440 of the kinetic energy required to escape Earth.

Let it reaches the radial distance r. at this distance kinetic energy become zero.

K_s + U_s = K_r + U_r

0.440*(0.5*m*v²) - GMm/Re = 0 - GMm/r

0.22*m( 2GM/Re)² - GMm/Re = -GMm/r

0.22/Re - 1/Re = -1/r

1/r = 0.78/Re

Radial distance , r = 1.28*Re

(c)

mechanical energy required at launch if the projectile is to escape Earth

E = 1/2*m*v² - GMm/Re

E = 0.5*m*(2GM/Re)² - GMm/Re

E = 0 J

Hence least mechanical energy required is Zero.