G 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 0.462 of the escape speed from Earth and (b) its initial kinetic energy is 0.462 of the kinetic energy required to escape Earth

a.)r=4RE/4

b.)r=2RE

c.)ZERO

Explanation:

a.) given v= 0.462 which is V= 0.466Ve

Since the projectile is shot directly away from earth surface the speed it escape with;

v=√2GM/REeqn(1)

M is the Earth's Mass

RE is the Radius

From the Law of conservation of Energy

K+U=0eqn(2)

K₁+U₁=K₂+U₂eqn(3)

where K₁ and U₁ is initial kinetic and potential energy

K₂ and U₂ are final kinetic and potential energy

Kinetic Energy (K.E) decrease with time as the projectile moves up and there is decrease in Potential Energy (P.E) , it will let to a point where K.E will turn to zero i.e K₂=0

U₂=K₁U₂ eqn(3)

From Gravitational Law

U₁= -GMm/RE (5)

U₂= -GMm/r (6)

Where "r" is the distance

v= 0.462√2GM/RE

v= √GM/2RE

GM/4RE - GM/RE = -GM/r

r= 4RE/3

b.) "r" is calculated by this equation;

K₁=0.466

K₁= 1/4MVe²eqn(9)

substitute eqn(1) into eqn (9) then

1/4m2GM/RE=0.466GM/2Re

GM/2RE - GM/RE =-GM/r

r=2RE

c.)The potential energy and kinetic energy is the same in terms of their size both in different directions, while the potential energy face outward, the kinetic energy face inward therefore the least initial mechanical energy

required at launch if the projectle is to escape is ZERO

object 2 because all the other ones are wrong

no sign of dia of fuel cell