If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surf

Question

If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is 41% the escape speed? km

Answer 1

Escape speed is the speed at which the projectile could make it to infinity.

Conservation of energy:
PE at infinity: 0 (by convention) +
KE at infinity: 0 (because you had just enough speed at escape speed to get there)
=
PE at earth's surface (=-G(m object) (m earth) / r) +
KE at earth's surface (=1/2 (m object) v^2)

In this equation, mass of object cancels, so you can solve for the escape speed in terms of the mass and radius of the earth.

Now imagine you had only 41% of that speed.

PE at the max height (-G (m object) (m earth) / (r max))
+ KE at the max height (zero)
=
PE at the surface (-g (m object) (m earth) / (r earth))
+ KE at the surface (1/2 (m object) (41% v)^2)

You use the escape velocity you solved for in the first part and then use that in the second equation to solve for r max.

The question asks you for the difference between r max and r earth.

Answer 2

very simple approach
formula for finding escape speed is
v=square root(2gr)
we have o find r actually
v we have that is 11km/s(escape velocity)
40/100 *11=square root(2*9.81*r)
taking square on both side u can find distance
i h not myself complicated this question,just did it simply
hope it will help
g.ali
pakistan