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-fourth of the escape speed from Earth? (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? (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Explanation / Answer

solution:

our equation only works for the escape velocity. If you are not using the escape velocity, then you would need final K and U, right?

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) = (1/4)*Ve = (1/4)*sqrt(2*G*M/RE)

Now apply energy conservation:

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
(1/2)*m*Vi^2 + (-G*m*M/RE) = -G*m*M/Rf -eq[1]
We know that Vi = (1/4)*Ve, so....
Vi^2 = ((1/4)Ve)^2 = (1/16) Ve^2
Vi^2 = (1/16)*(2*G*M/RE)
Vi^2 = (1/8)*(G*M/RE)
Now plug this back into eq1
(1/2)*m*(1/8)*(G*M/RE) + (-G*m*M/RE) = -G*m*M/Rf
You an drop G and m ,Mout of all the parts of the equation.
(1/2)*(1/8)*(M/RE) + (-M/RE) = -M/Rf
(1/4)*(M/RE) + (-M/RE) = -M/Rf
-(3/4)*(M/RE) = -M/Rf
Rf/RE = 4/3

I have provided you basic equations please solve other part its very lengthy.

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