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
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
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
Hmmm, how much work does the earth do on the moon?
To calculate the ratio E/T, we first need to determine the energy required to propel the probe around the quarter circle turn.
Since the probe needs to maintain the same tangential speed of v = 20,000 m/s, we can consider it as a point mass moving in a circular path.
The centripetal force required to keep the probe moving in a circular path is provided by the ion thruster rocket. This force is given by the equation:
F = m * (v^2 / r)
where F is the force, m is the mass of the probe, v is the tangential speed, and r is the radius of the circular path.
Since the probe maintains the same tangential speed throughout the turn, the tangential speed is equal to the probe's initial speed, v = 20,000 m/s.
Now, we need to find the radius of the circular path. The quarter circle turn implies that the probe will turn by 90 degrees. The circumference of a quarter circle is given by πr/2, where r is the radius of the circular path.
Since the probe has a tangential speed of v = 20,000 m/s, the time taken to complete the quarter circle turn is given by:
t = (πr/2) / v
We can rearrange the equation to solve for r:
r = (2v * t) / π
Now, we can substitute the values into the equation for the centripetal force:
F = m * (v^2 / r)
= m * (v^2 / [(2v * t) / π])
= (π * m * v) / (2t)
The work done by the ion thruster rocket to move the probe around the quarter circle turn is equal to the force multiplied by the distance traveled, which is πr/2:
W = F * (πr/2)
= (π * m * v) / (2t) * (π * (2v * t) / π)
= (m * v^2) / 2
Thus, the energy required to propel the probe around the quarter circle turn, E, is equal to the work done:
E = (m * v^2) / 2
Now, we need to calculate the ratio E/T:
E/T = [(m * v^2) / 2] / (1/2 * m * v^2)
= 1
Therefore, the ratio E/T is equal to 1.
I hope this helps! Let me know if you have any further questions.