3 ms·

Thank you for this. Let's see, I think a million watts is a bit more than we need to 235 mL of water in 0.5 seconds. A calorie (not a Calorie) is the amount o

Thank you for this. Let's see, I think a million watts is a bit more than we need to 235 mL of water in 0.5 seconds. A calorie (not a Calorie) is the amount of energy it takes to raise 1 ml of water 1 deg C. Suppose 15C water (pipes run through the ground which is cool). https://www.google.com/search?q=(85+calories+*+235)+%2F+0.5+seconds https://www.google.com/search?q=(85+calories+*+235)+%2F+0.5+... Is that right? While a megawater later is overkill for your cup of tea, it is NOT overkill for ten cups of tea (in half a second)? For some reason, I always thought of a megawatt as BIGGER. Or, I guess I thought of boiling water as SMALLER. In any case, there's still something I don't get here. Obviously if you are trying to change the velocity of your spacecraft and all you have is a cup of water and a megawatt laser, you're not going to point the laser into space, you're going to point it at the cup. (Which happens to be roughly and conveniently conical.) Picture this: point your megawatt laser at your friend's spacecraft and her cup of water... She'll get delta-V when her water boils, but that's AFTER you already got a little (very little) recoil from firing the laser. I guess I don't understand the relationship between momentum and mass when mass is zero.

MereInterest 12y ago
You are correct, and I was wrong in the amount of energy necessary to bring the water to a boil. I was using wolfram alpha, "energy to boil a cup of water", which probably then gave me the energy needed to boil it all away, not just to bring it to a boil. It takes a tremendous amount of energy to heat things, overall. For example, compare the energy in a bullet to the energy in a warm cup of water. A 20g bullet moving at 400 m/s has 1600 joules of energy. That amount of energy, when used to warm a cup of water, would only warm it by less than 2 degrees Celsius. Let's start with your last statement, on the relationship between energy, momentum, and mass. The most common formula is E=mv^2/2. This works for objects with mass that are moving slowly. Now, by "moving slowly", I mean "relative to the speed of light". The fastest object made by man, the Juno spacecraft, moves at about 140,000 km/hr, which is a paltry 0.00013c. This formula still works here. If we want to know what happens at faster speeds, we need to dip into special relativity. There are some messy derivations, but one critical formula that comes out of it is E=sqrt(m^2c^4 + p^2c^2). If you imagine a ninety-degree triangle where the shorter sides are the mass of an object and the momentum of an object, then the long edge is the energy. Increasing either mass or momentum will increase the energy. Suppose we start dialing down the momentum in this equation. After some math, (http://en.wikipedia.org/wiki/Binomial_theorem http://en.wikipedia.org/wiki/Binomial_theorem ), we arrive at E=mc^2 + mv^2/2. This is the origin of the famous E=mc^2. We now have two terms, the rest energy, and the kinetic energy. Note that the other term is the more familiar form of kinetic energy, which is why it works in everyday life. Now, suppose we go the other route and set m=0. Then we get E=pc. Even though one side of the triangle (mass) has been reduced to zero, the other side (momentum) still gives it a non-zero energy. I apologize if that is more math than you had been hoping for, but eventually one always runs into some math. A few things on the spaceship. You are absolutely right on the thinking of it. As soon as the laser switches on, it starts applying force. If my friend's spacecraft is one light-second away, I don't need to wait one second for the laser to reach her before feeling anything, because the light itself carries momentum away. In fact, if I wanted, I could just fire the laser out into space. I get propelled one way, and to balance out the momentum, the photons are travelling in the opposite direction.

MereInterest 12y ago

You are correct, and I was wrong in the amount of energy necessary to bring the water to a boil. I was using wolfram alpha, "energy to boil a cup of water", which probably then gave me the energy needed to boil it all away, not just to bring it to a boil. It takes a tremendous amount of energy to heat things, overall. For example, compare the energy in a bullet to the energy in a warm cup of water. A 20g bullet moving at 400 m/s has 1600 joules of energy. That amount of energy, when used to warm a cup of water, would only warm it by less than 2 degrees Celsius. Let's start with your last statement, on the relationship between energy, momentum, and mass. The most common formula is E=mv^2/2. This works for objects with mass that are moving slowly. Now, by "moving slowly", I mean "relative to the speed of light". The fastest object made by man, the Juno spacecraft, moves at about 140,000 km/hr, which is a paltry 0.00013c. This formula still works here. If we want to know what happens at faster speeds, we need to dip into special relativity. There are some messy derivations, but one critical formula that comes out of it is E=sqrt(m^2c^4 + p^2c^2). If you imagine a ninety-degree triangle where the shorter sides are the mass of an object and the momentum of an object, then the long edge is the energy. Increasing either mass or momentum will increase the energy. Suppose we start dialing down the momentum in this equation. After some math, (http://en.wikipedia.org/wiki/Binomial_theorem http://en.wikipedia.org/wiki/Binomial_theorem ), we arrive at E=mc^2 + mv^2/2. This is the origin of the famous E=mc^2. We now have two terms, the rest energy, and the kinetic energy. Note that the other term is the more familiar form of kinetic energy, which is why it works in everyday life. Now, suppose we go the other route and set m=0. Then we get E=pc. Even though one side of the triangle (mass) has been reduced to zero, the other side (momentum) still gives it a non-zero energy. I apologize if that is more math than you had been hoping for, but eventually one always runs into some math. A few things on the spaceship. You are absolutely right on the thinking of it. As soon as the laser switches on, it starts applying force. If my friend's spacecraft is one light-second away, I don't need to wait one second for the laser to reach her before feeling anything, because the light itself carries momentum away. In fact, if I wanted, I could just fire the laser out into space. I get propelled one way, and to balance out the momentum, the photons are travelling in the opposite direction.