Question

How great would the muzzle velocity of a gun on the surface of
the moon have to be in order to shoot a bullet to an altitude of 101
km?
Additional Information:
The mass of the earth is mE = 5.98*10^24kg
The mass of the moon is mm = 7.35*10^22 kg
The distance from the center of the earth to the center of the moon is 3.84*10^8m
The radius of the earth E = 6.38*10^6m
The radius of the moon is m = 1.74*10^6

Answer 1

seems that you have been given all the data to calculate moon gravity from earth gravity

i say this because you can say that for earth or moon or any large body that
\[F = mg = \frac{G \ M \ m}{r_s^2}\] where r_s is planet radius, so
\[g = \frac{G \ M}{r_s^2}\]or \[G = \frac{r_s^2 g}{m}\]
and G is the same for all planets so you can work out a g for the moon from earth's g, eg from
\[\frac {r_e^2 g_e} {m_e} = \frac {r_m^2 g_m} {m_m}\]
then use
\[v^2 = u^2 + 2 a x\]
where a is (-ve) your moon gravity

Answer 2

This is as I would have approached it, however I have 1 reservation:

you have to call upon your'knowledge' of ge =9.81m/s^2

this is NOT given in th equestion. If oyu are allowed to look up this value, or assume it, then you may as well look up gm instead.

There appears to be redundant information (distance of earth to moon). I can't see how to use this alone to get gm.

Can anyone solve using ONLY the given information?

Answer 3

@IrishBoy123

Answer 4

though i did note the "redundant" info, i though t it was just that. i had absolutely no context.

but, two other [small] things.

you can take:
\[m_m \omega^2 r_{me} = \frac{G m_e m_m}{r_{me}^2}\]
and use to calc GM allowing a direct calculation of g_e to be made

that uses the distance apart but needs one to "assume" knowledge of the period of the moon.

plus, 101km is really quite a height but i reckon grav on the moon only drops to about 1.4 at that distance. IOW i had thought maybe one could calc the combined grav potential on the bullet due to both moon and earth, in which case their distance apart might be germane; but that doesn't feel to me like it will go anywhere.

if you find a better answer, please share it here.

Answer 5

yeah - it thought of the mrw^2 opotion - but oyu are not given w

I don't think this can be solved without assuming ge

Answer 6

The answer comes out around 500m/s which is well within normal muzzle velocities.

The energy would equate to approx. 17km or over 10 miles altitude on earth.

I guess the air resistance which we conveniently ignore in most physics questions reduces the height actually achieved on earth very significantly!!