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-fifth 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?

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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