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 0.704 of the escape speed from Earth and (b) its initial kinetic energy is 0.704 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.
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a)
Initial speed is 0.704 v = u ( say).
1/2 m [v^2 - u^2] = GMm/ r
1/2 v^2 [1 - 0.704 ^2] = GM/ r
Replacing 1/2 v^2 by GM/R and canceling GM on both sides
[0.5044] /R= 1/ r
r/R = 1.982.
--------------------------------------...
b)
Given: 1/2mu^2 = 0.704 of 1/2 m v^2
1/2 m v^2 - 1/2mu^2 = GMm/ r
[1 - 0.704] 1/2 v^2 = GM/ r
Replacing 1/2 v^2 by GM/r and canceling GM on both sides
[ 1 - 0.704] /R = 1/ r
r/R = 3.38

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