A satellite is circling the moon (radius 1700 km) close to the surface at a speed v. A projectile is launched vertically up from the moon's surface at the same initial speed v.
How high will the projectile rise? (Take m and M as the masses of the satellite and the moon, respectively.)

 Rmax = 3.4 10⁶ m

Explanation:

For this exercise we will use the concept of energy

Initial. On the surface of the luma

    Em₀ = K + U

    Em₀ = ½ m v² - G m M / R_moon

Final. At the furthest point

    Emf = U

    Emf = - g m M / R_max

    Em₀ = Emf

    ½ m v² - G m M / R_moon = - G m M / R_max

    ½ v² + G M (-1 / R_moon + 1 / R_max) = 0           (1)

Let's use the fact that tells us that the speed of the rocket is equal to the speed of a satellite that rotates around the moon near the surface, let's use Newton's second law

        F = m a

Acceleration is centripetal

       a = v² / r

        r= R_moon

       G m M / R_moon² = m v² / R_mon

        G M / R_moon = v²

We substitute in 1

     ½ G M / R_moon + G M (1 / R_max - 1 / R_moon) = 0

    1 / R_max = 1 / R_moon (1- ½)

     R_max = R_moon 2

     Rmax = 2 1700 103

     Rmax = 3.4 10⁶ m

that would be b i just took the test : )

answer: the correct answer is   air pressure at the top of mount everest is higher than at the bottom of the grand canyon.

hope this : )

explanation: