A projectile is shot directly away from Earth\'s surface. Neglect the rotation o

Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.
(a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is one-third of the escape speed from Earth?
multiplied by RE

(b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is one-fifth of the kinetic energy required to escape Earth?
multiplied by RE

(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?
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Explanation / Answer

Let M and R be the mass and radius of earth

The escape velocity ve = (2GM/R)

Since the terminal veocity v0 = (1/5) ve = (1/5) (2GM/R)

(a) Let h be the height from the surface of the earth

From the law of conservation of energy

(KE)1 = Change in potential energy

(1/2) m (v0)2 = (GMm/R) - (GMm/(R+H))

(1/2) m (1/25) (2GM/R) = (GMm/R) - (GMm/(R+H))

(1/25R) = (1/R) - (1/(R+H))

1/25 = H / R+H

R+H = 25H

H = R/24

Now R+H = (25/24) R

(b) According to the problem

(KE)i/(KE)e = 1/5

(1/2)Mp (v0)2 divided by (1/2)Mp (ve)2 is equal to 1/5

i.e, (ve)2 = 5(v0)2

2GM/R = 5GM/r

or r = 5R/2 = 2.5R

TOtal radius = r +R = 3.5R

(c) From the law of conservation of energy

(KE)i = Change in potential energy

The least mechanical energy of a particle to escape the earth

ME = (KE)i + (-change in potential energy)

= zero.

Hence ME = 0

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