Chapter 13, Problem 040 A projectile is shot directly away from Earth\'s surface

Question

Chapter 13, Problem 040 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 0.799 of the escape speed from Earth and (b) its initial kinetic energy is 0.799 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? (a) Number (b) Number (c) Number Units Units Units

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.
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a)
Initial speed is 0.799 v = u ( say).
1/2 m [v^2 - u^2] = GMm/ r
1/2 v^2 [1 – 0.799 ^2] = GM/ r
Replacing 1/2 v^2 by GM/R and cancelling GM on both sides
[0.361599]/R= 1/ r
r/R = 2.765
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b)
Given: 1/2mu^2 = 0.799 of 1/2 m v^2
1/2 m v^2 - 1/2mu^2 = GMm/ r
[1 - 0.799] 1/2 v^2 = GM/ r
Replacing 1/2 v^2 by GM/R and cancelling GM on both sides
[ 1 - 0.799] /R = 1/ r
r/R = 4.975

c) the mechanical energy = 0 for the “escape condition.

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