Height of projectile from earth?

Question

At the Earth's surface, a projectile is launched straight up at a speed of 10.0km/s. To what height will it rise? ignore air resistance.

i know that you have to take into consideration the gravitational force constant, as the projectile has such a high inital velocity so that the acceleration due to gravity wont b 9.8 ms-2

Answer 1

Use the conservation of energy principle.

The projectile starts at speed v and ends up at speed 0 (at its maximum height), so its change in KE is mv²/2.

The change in PE is not so obvious. For small changes in altitude it's mgh. But for larger changes (where g varies), You have to integrate (Force • distance) from the ground up to the height in question. This is:

ΔPE = Integral(Force•dr)
= Integral((GMm/r²)dr)
between the limits r0 and r1, where r0 is your distance from the center of the earth at the beginning (i.e. the earth's radius), and r1 is your distance from the center of the earth at your maximum height.

If you do that math you find:

ΔPE = GMm(1/r0 – 1/r1)

Since that equals the loss in KE, we have:

mv²/2 = GMm(1/r0 – 1/r1)

Or (cancelling the "m"):

v²/2 = GM(1/r0 – 1/r1)

You can simplify this further by noting these facts:

r1 - r0 = h (the max height above the surface)

GM/(r0)² = g (acceleration of gravity at the surface)

Then we have:

v²/2 = GM(1/r0 – 1/r1)
= GM(r1–r0)/(r1r0)
= GM(h)/(r1r0)
= (GM/(r1r0))(h)
= (GM/r0²)(r0/r1)(h)
= g(r0/r1)(h)
= g(r0/(r0+h)) (h)

From that point it should be simple to solve for h in terms of v, g, and r0 (I will leave that as an exercise for the reader. :-))