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 0.677 of the escape speed from Earth and (b) its initial kinetic energy is 0.677 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.)

Answer 1

a)let initial speed is v.

given that v=0.677*escape speed

=0.677*sqrt(2*G*M/r)

where G=universal gravitational constant

M=mass of earth

r=radius of earth

initial energy at the surface =(-G*M*m/r)+0.5*m*v^2

=(-G*M*m/r)+0.5*m*0.677^2*2*G*M/r

=(G*M*m/r)*(0.677^2-1)

-0.54167*G*M/r

at maximum distance from earth , speed of the projectile is zero.

if the distance is k*r,

then final energy=initial energy

==>-G*M*m/(k*r)=-0.54167*G*M/r

==>k=1/0.54167=1.8461

hence it will travel to a distance of 1.8461 times radius of earth.

part b:

as kinetic energy is proportional to square of speed, intial speed=sqrt(0.677) times escape speed

given that v=sqrt(0.677)*escape speed=0.8228*escape speed

=0.8228*sqrt(2*G*M/r)

where G=universal gravitational constant

M=mass of earth

r=radius of earth

initial energy at the surface =(-G*M*m/r)+0.5*m*v^2

=(-G*M*m/r)+0.5*m*0.8228^2*2*G*M/r

=(G*M*m/r)*(0.8228^2-1)

-0.323*G*M/r

at maximum distance from earth , speed of the projectile is zero.

if the distance is k*r,

then final energy=initial energy

==>-G*M*m/(k*r)=-0.323*G*M/r

==>k=1/0.323=3.096

hence it will travel to a distance of 3.096 times radius of earth.