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 REgives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.444 of the escape speed from Earth and (b) its initial kinetic energy is 0.444 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?

Please provide right answers with explanations and units plase thanks

Explanation / Answer

a)

Using energy conservation, we have
KEi + Ui = KEf + Uf
1/2 mv2 - GMm/RE = 0 - GMm/r

Using the escape speed from Earth,
v = ? GME / 2RE

so we get

1.2*m*(0.44)^2+ (- GMm/RE )= GMm/r

If V = (0.444) Ve at R = Re, then
(1/2) m (0.444)^2 Ve^2 - G M m/R = constant
= 0.098m Ve^2 - G M m/Re = GMm/R
When V = 0 (maximum value of R),
(0.197) Ve^2 = GM [(1/Re - (1/R)]
(0.197) G M/Re = G M [(1/Re - (1/R)]
R/Re = 1/1.197
R = 0.836Re
So the multiple of RE is 0.836

b)As the projectile

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