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

Question

asked Feb 21, 2022 99.6k views

1 Answer

Lundahl · Answer 1 · 2022-02-25T20:49:03+0000

Answer:

The height (h) will be:
(3)/(4)R =h

Step-by-step explanation:

The scape speed equation is given by:

$v_(scape)=\sqrt{(2GM)/(R)}$

Now, the speed of the missile is

v_(missile)=(1)/(2)v_(scape)

$v_(scape)=(1)/(2)\sqrt{(2GM)/(R)}$

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

E_(i)=E_(f)

(1)/(2)mv_(scape)^(2)-mgR =-mgh

(1)/(2)(2GM)/(4R)-gR =-gh

(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.

(1)/(4)gR-gR =-gh

(1)/(4)R-R =-h

Therefore the height (h) will be:

(3)/(4)R =h

I hope it helps you!