Question

A missile is launched upward with a speed that is half the escape speed. What height (in radii of Earth) will it reach?

Answer 1

Answer:

The height (h) will be: $\frac{3}{4}R =h$  

Explanation:

The scape speed equation is given by:

$v_{scape}=\sqrt{\frac{2GM}{R}}$

Now, the speed of the missile is

$v_{missile}=\frac{1}{2}v_{scape}$

$v_{scape}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

Using the conservation of energy, we can find the maximu height of the missile.

$E_{i}=E_{f}$

$\frac{1}{2}mv_{scape}^{2}-mgR =-mgh$

$\frac{1}{2}\frac{2GM}{4R}-gR =-gh$

$\frac{GM}{4R}-gR =gh$

Let's recall that g = GM/R², using the equivalence principle. When R is the radius of the earth and M is the mass of the earth.

$\frac{1}{4}gR-gR =-gh$

$\frac{1}{4}R-R =-h$    

Therefore the height (h) will be:

$\frac{3}{4}R =h$    

I hope it helps you!