Question

An interplanetary probe speeding in deep space suddenly needs to make a left turn.

It is presently traveling at v=20000 m/s , and needs to make a quarter circle turn, continuing at the same tangential speed. It is equipped with an ion thruster rocket capable of ejecting ions at V=50000 m/s . Moreover, it is a very advanced model that gets its ions from outer space so that the weight of the probe remains unchanged during its' turn.

It takes a certain amount of total energy E to propel the probe around this quarter circle turn. If the kinetic energy of the probe is T=1/2(mv²) [(which remains unchanged throughout)], what is the ratio E/T? Answer to 3 decimal places

Thanks

Answer 1

force=mass*v^2/r

so the thrusters stop movement in the original direction, and recreate it in the perpendicular direction.

workdone=1/2 mv^2 + 1/2 mv^2=mv^2

work done E=force*distance
= mv^2/r * r+ mv^2/r *r=2mv^2

E/T=4

check my thinking

Answer 2

probe mass = m
probe speed = v = 5*10^4
ion mass = M
ions/second = q
ion speed = V = 2*10^4
ion momentum change/second = q M V = force

force needed = mv^2/R
so
mv^2/R = q M V

kinetic energy of ion = (1/2)MV^2
kinetic energy/second of ion flow
= q(1/2)MV^2

Time to go 1/4 turn = 2 pi R/4v = pi R/2v
so total energy of ion flow = q pi R MV^2/4v
BUT we know from the centripetal force that
qMV = mv^2/R
so
(mv^2/R)(RV/4v) = E = mvV/4
E/T = (mvV/4)/ (mv^2/2) = (1/2)V/v
= (1/2 )5/2 = 5/4
whew, check my arithmetic

Answer 3

Hmmm, how much work does the earth do on the moon?