Question

3. A SpaceX StarshipTM single-stage rocket is launched vertically from the moons surface. It burns k kg of fuel per second, so its mass as a function of time is m = mo - kt. Assume constant acceleration of gravity g and a constant exhaust speed of vex (a) By integrating with respect to time, calculate the rockets velocity versus time (while it still has fuel).

Answer 1

(a) By This figure a f = Fy - mg - V fama - fx = Ver dm = K Ven =) So, Equaticm One become I ma = kVex - mg a= k Vex -g Im=mo-kt a = K vex k Vex Mo-kt - g VC) = - Vem [ lo m-kt)] - gt VE) = - Vex [ In mo – kt) - In mo] – gt VIE) = - Vexe en[ mo-K4] - gt vel a Vox beyen )

yle) a) (b) Vit) = so, form] = fueldt ye): ver In [mo:] - gre) dt 407 = -filen en (mat) dt - faldt yes = = kan en - (243. Let mo-kt mo = da -k dt mo - dt = -mo de . At £ = t =) a = morkt By Putling Values t o =) = mo markt Yt) = Movex Tlna dx - & gt² к lenne = alna - a 1. We know that a Therefore a y(t) = Movers Mo-kt 7 mo -X - 4 g 4²

lo-kt mo Sre) mouse please em (main) (n=1293-C0-13] | muon corak) (no titl - (1-44) + ]*g maken [. -*t) en (me ) - 1 + kit + ]y.ge YE) = Vex (m. -Kt) ln(mo-Kt) + Vent - įgt? = movex [mo { [" tkt tilat² mo / 23 -- yt) - Véut - £9t' – Vex (ma-ket) In (m.)