Question

A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the projectile's speed two-thirds its initial value?

Answer 1

let mass of the projectile be m kg.

values used:

mass of moon=M=7.34767*10^22 kg

gravitational constant=G=6.674*10^(-11)

radius of moon=R=1737 km=1.737*10^6 m

2/3 rd of the initial velocity=1360*2/3=906.67 m/s

using conservation of energy principle:

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

(-G*M*m/R)+0.5*m*1360^2=(-G*M*m/(R+h))+0.5*m*906.67^2

==>0.5*(1360^2-906.67^2)=G*M*((1/R)-(1/(R+h))

==>1/(R+h)=4.7093*10^(-7)

==>h=3.8644*10^5 m

hence at a height of 3.8644*10^5 m the projectile would have a speed of 2/3rd of its initial value.