Q:

A stationary vessel out in space fires a rocket that accelerates it to 30 mph from an observers view, and then fires an identical second rocket that accelerates it another 30 mph.
The rocket will have four times the energy going 60 mph as it did when it was going 30 mph (mass x velocity squared). How can this be when it fired an identical second rocket as the first rocket, that simply kicked up its speed another 30 mph (from 30 to 60 mph)?
Another way of putting this is why does firing an identical second rocket from a vessel, that doubles its speed, impart 4 times the energy to the object as the first rocket it fired.

- Mike (age 57)

Kirkland WA USA

- Mike (age 57)

Kirkland WA USA

A:

The chemical energy from the fuel is converted into two main mechanical forms, in addition to some heat. These are the kinetic energy of the average backward motion of the exhaust and the kinetic energy of the forward motion of the vessel. As the vessel speeds up, the proportion of that mechanical energy that goes to the vessel goes up.

At first the backward speed of the exhaust is whatever the exhaust velocity is relative to the vessel. By the time the second rocket is done firing it's whatever the exhaust velocity is relative to the vessel - 60 mph. So initially the rocket is putting more kinetic energy into the exhaust. By the end it's putting less into the exhaust and more into the vessel.

p.s. If you look at the particular situation you describe, the second rocket adds 3x as much energy to the vessel as the first, not four times.

Mike W.

*(published on 12/26/2013)*

Q:

Thank you for the answer. I misspoke where I said that the vessel gained four times as much energy by firing the second rocket. I should have said that the vessel had four times the kinetic energy after it fired the second rocket compared to after it fired the first rocket.
So is it correct to say that half or the energy of each rocket (disregarding heat production) goes into pushing the vessel forward and its corresponding gain in kinetic energy, and the other half of the energy goes into the the kinetic energy of the exhaust going in the opposite direction?
If this is so, it still seems hard to comprehend how the second rocket imparts 3 times the kinetic energy to the vessel as the first rocket did for the following reason.
Lets say the rocket exhaust, after firing, has kinetic energy is equal to the gain in kinetic energy of that gained by the vessel. Call that amount of kinetic energy positive K for the vessel and negative K for the exhaust. When the second rocket fires, the vessel picks up 3K for the vessel, yet the exhaust would still have some form of kinetic energy pushing the other way, but it would be K or less in the opposite direction.
I have run this scenario by several physics majors, including a PHD, and they are all stumped by the problem of the large increase in kinetic energy by firing the second rocket and how to account for it.

- Mike (age 57)

Kirkland WA USA

- Mike (age 57)

Kirkland WA USA

A:

"So is it correct to say that half or the energy of each rocket (disregarding heat production) goes into pushing the vessel forward and its corresponding gain in kinetic energy, and the other half of the energy goes into the the kinetic energy of the exhaust going in the opposite direction? "

No, not even close. The proportion varies as the vessel's velocity changes, or depending on what reference frame you choose. If you pick a reference frame in which the vessel's initial velocity was backwards, the initial firing imparts a *negative* change in the vessel's kinetic energy.

Mike W.

*(published on 12/28/2013)*