Consider a rocket that is in deep space and at rest relative to an inertial refe
ID: 2240993 • Letter: C
Question
Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket's engine is to be fired for a certain interval. What must be the rocket's mass ratio (ratio of initial to final mass) over that interval if the rocket's final speed relative to the inertial frame is to be equal to (a) the exhaust speed (speed of the exhaust products relative to the rocket) and (b) 4.00 times the exhaust speed? (The exhaust products from the rocket escape at a constant velocity relative to the rocket)
Explanation / Answer
It's actually a weird MOMENTUM problem.
p = p' {I will use prime notation: the prime (') after any variable means the FINAL mass or velocity, AFTER the exhaust has separated from the rocket.}
You can consider the rocket at rest (all that jazz about inertial frame of reference), and so its momentum must always equal zero. Originally, the rocket isn't releasing thrust. So its momentum, as I will refer to it, is equal to M*V which we know is equal to 0. (p = mv)
Now once the thruster starts firing, we have two objects to look at - the mass of the thrust, and its velocity, and now the newfound velocity and mass of the rocket. So,
-mv + M'V' = 0
where
M' = (M-m)
Get it? (The negative sign is because the velocity is in an opposite direction.)
What you're looking for is M'/M. The variable is v (we defined m and v as the mass and velocity of the thrust ejected).
In the first scenario, -v = V'. In the second, -4v = V'.
Now, just do the math.
If -v = V', let's substitute -v in that final equation.
-mv + M'v = 0
-mv = -M'v
m = M'
M'/m = 1 (the ratio is one!)
Now, with -4v:
-mv = -4M'v
m =4M'
m/4= M'
1/4 = M'/m (the ratio is fourth!)