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 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 0.262 of the escape speed from Earth and (b) its initial kinetic energy is 0.262 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.) (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Explanation / Answer

1/2mv^2=mMG/R=mgR [escape kinetic energy]

ve=(2MG/R)^1/2=(2gR)^1/2 [escape speed]

(a)

Maximum height is

h=v^2/(2g)

if v=kve

h=k^2ve^2/(2g)

h=k^2R

h=(0.262)^2R

h = 0.068644R [R=Earth radius]

(b)

if 1/2mv^2=kmgR

v^2=2kgR

then the maximum height is

h=v^2/(2g)=kR

h = 0.262R [R=Earth radius]

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