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 0.584 of the escape speed from Earth and (b) its initial kinetic energy is 0.584 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

If v is the escape velocity and R is the radius of earth,
v = [2GM/R]
v^2 = [2GM/R]
1/2 v^2 = GM/R-------------1.
--------------------------------------...
a)
Initial speed is 0.584 v = u ( say).
1/2 m [v^2 - u^2] = GMm/ r
1/2 v^2 [1 - 0.584 ^2] = GM/ r
Replacing 1/2 v^2 by GM/R and canceling GM on both sides
[0.658944] /R= 1/ r
r/R = 1.517579
--------------------------------------...
b)
Given: 1/2mu^2 = 0.584 of 1/2 m v^2
1/2 m v^2 - 1/2mu^2 = GMm/ r
[1 - 0.584] 1/2 v^2 = GM/ r
Replacing 1/2 v^2 by GM/r and canceling GM on both sides
[ 1 - 0.584] /R = 1/ r
r/R = 0.416

c) Least KE required to make the mass to escape right from the pull of the earth
Here we use KE + PE = 0 as it has escaped from the hold
OR 1/2 m Ve^2 - GM m / Re = 0
Hence least mechanical energy ie 1/2 m Ve^2 =1.9231*10^24

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