A newly established colony on the Moon launches a capsule vertically with an initial speed of 1.453 km/s. Ignoring the rotation of the Moon, what is the maximum height reached by the capsule? km

Question

The answer key is 1040km. I have tried v^2/2g (1.4532km/s)^2/(2*.0162km/s^2) and get the answer as 65.1km but that is not the correct answer. Can somebody shows me how to get to the answer key?

Answer 1

g = 1.62 m/s^2 = 0.00162 not 0.0162 km/s^2
but I suspect you have other unit errors
1.453 km/s * 3600 = 5231 km/hr which is awful fast
I get 651 km using your data with the g repaired

Answer 2

our answer isn't even close to the answer key. Have other idea?

Answer 3

have you had calculus?

the launch speed and low moon gravity means that the capsule goes beyond the region where g is relatively constant

the force is related to the height by the universal gravitational equation ... f = G M m / (r + h)^2

force * height = energy ... ∫ de = ∫ f dh = 1/2 m v ^2

sorry, but my calculus is pretty rusty

Answer 4

I think Scott is aiming you in the right direction
F = G M m/r^2
potential energy = integral from infinity to r of F dr
take it at 0 at infinity so it gets negative as you move in
F dr = G M m int dr/r^2 = -G M m /r

so as you move from Ro the surface to R, the increase is

G M m (1/Ro - 1/R)
that = 1/2 m v^2

note that G M/Ro^2 = g

Answer 5

hey ... this works

the gravitational potential energy of a object at a distance r from a large body is ... G M m / r

the difference in the P.E. on the surface and at altitude is the K.E. at launch

(G M m / r) - [G M m / (r + h)] = 1/2 m v^2

solve for h ... might be easier to plug in numerical values at the beginning