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 0.367 of the escape speed from Earth and (b) its initial kinetic energy is 0.367 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

a) initial PE = - GMm / (Re)

so initial KE = GMm / Re then mass will be escaped from earth's gravity.

m v^2 /2 = GMm / Re

v = sqrt(2GM / Re)

initial speed = 0.367v = 0.367 sqrt(2GM / Re)

to find the height,

initial PE + initial KE = final Pe + final KE

- GMm / Re + m (0.367 sqrt(2GM / Re) )^2 /2 = - GMm /(Re +h) + 0

-GM / Re + 0.135GM/Re = - GM / (Re+h)

1 / Re+h = 1/Re - 0.135/Re = 0.865/Re

Re = 0.865Re + 0.865h

h = 0.156Re

b) initial KE = 0.365GMm/Re

so - GMm / Re + 0.365GMm / Re = -GMm / (Re + h)

1/ Re+h = 0.635/Re

Re = 0.635Re + 0.635h

h = 0.575Re

c) initial machanical energy required = GMm / Re

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