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 Earth.

(a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is one-half of the escape speed from Earth?
   RE

(b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is one-half of the kinetic energy required to escape Earth?
RE

(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?
J

Explanation / Answer

Lets call Ve the escape velocity

(1/2)*m*Ve^2 + (-G*m*M/RE) =0

Ve = sqrt(2*G*M/RE)

So Initial velocity (Vi) = (0.5)*Ve = (0.5)*sqrt(2*G*M/RE)

Now it seems like you want this equation

Ki + Ui = Kf + Uf
(I am calling the final radial distance Rf)
(1/2)*m*Vi^2 + (-G*m*M/RE) = (1/2)*m*Vf^2 + (-G*m*M/Rf)
We know that Vf = 0, I think?
(1/2)*m*Vi^2 + (-G*m*M/RE) = -G*m*M/Rf [EQ1]
We know that Vi = (0.5)*Ve, so....
Vi^2 = ((0.5)Ve)^2 = (0.25) Ve^2
Vi^2 = (0.25)*(2*G*M/RE)
Vi^2 = (0.5)*(G*M/RE)
Now plug this back into EQ1
(1/2)*m*(0.5)*(G*M/RE) + (-G*m*M/RE) = -G*m*M/Rf
You can drop G and m out of all the parts of the equation.
(1/2)*(0.5)*(M/RE) + (-M/RE) = -M/Rf
(0.25)*(M/RE) + (-M/RE) = -M/Rf
-(0.75)*(M/RE) = -M/Rf
we can drop the M
-(0.75)*(1/RE) = -1/Rf
Rf/RE = 1.33


b) Initial KE = (1/2) Escape KE
(1/2) m Vi^2 = (1/2)*(1/2) m Ve^2
Vi^2 = (1/2)*Ve^2 = (1/2)*(2*G*M/RE)
Vi^2 = (1)*(G*M/RE)

Now lets use EQ1 from part a)
(1/2)*m*Vi^2 + (-G*m*M/RE) = -G*m*M/Rf
and plug in our value for Vi^2
(1/2)*m*((1)*(G*M/RE)) + (-G*m*M/RE) = -G*m*M/Rf
Now this time, lets pull out m and M and G out of both sides
(1/2)(1)*(1/RE)) + (-1/RE) = -1/Rf
-(1/2)*(1/RE) = -1/Rf
- 2*RE = - Rf
Rf/RE = 2

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