# Calculating energy needed for acceleration over time

1. Jul 19, 2008

### Norseman

First, to preface this, I have never taken any formal engineering courses, and I don't remember the physics class I took very well. I'm trying to figure out how much energy it takes to accelerate a mass of 2,000,000 kg at 1G for 153,778 seconds. In other words how much energy it takes to get a spaceship to Mars in 1 day 18 hours and 43 minutes, thus maintaining artificial gravity for nearly the entire the trip, except for a brief moment halfway through in order to turn the ship around. First, I determined that distance to Mars was 57,936,384 to 402,336,000 kilometers. I remembered that gravity was 9.8 meters per second per second. I wrote a simple python program to figure out how long the trip would take, by finding out how long it would take to accelerate halfway, and then doubling that value.

Code (Text):
half_distance = 57936384000.0 / 2
current_distance = 0
current_velocity = 0
time = 0
while current_distance < half_distance:
current_velocity += gravity
current_distance += current_velocity
time +=  1
print time * 2
That gave me a value of 153,778 seconds for the first distance, and I got a value of 405,240 seconds for the second distance. Next, I wanted to find out how much energy that trip would require, so I went to wiki and looked up the article on kinetic energy, and found that it was 1/2 the mass times the velocity squared. So, I ran that program again and got the peak velocities, which were 753,512.2 m/s for the first distance, and 1,985,676.0 m/s for the second. According to that formula, the energy required for the first distance is 1,000,000 * 753,512.2^2 joules, or about 567,780,635,548,839,870 joules. That's 567 petajoules. That's more energy than released by detonating a 100 megaton nuclear bomb, and more energy than the electrical power consumed by Norway in 2005. That's just to get to the halfway point. It needs to be done again to decelerate, and two more times if the trip were to return.

I guess this is possible, but I wonder about that velocity squared part. Let's say a spaceship accelerates at 1G for, say, 10 seconds in one direction. Does it take more energy to accelerate again at 1G in the same direction than it does to accelerate at 1G in the opposite direction? If so, why? Or, if not, then am I using the correct energy calculation here?

Last edited: Jul 19, 2008
2. Jul 19, 2008

### Staff: Mentor

There's an easy, sneaky way to do this: work = energy = force * distance.

3. Jul 19, 2008

### Norseman

Why is it force * distance? Shouldn't the energy needed each second to accelerate a mass be constant? Shouldn't it be force * time?

Last edited: Jul 19, 2008
4. Jul 20, 2008

### Defennder

You can do it with respect to their time or distance. The definition of work done is $$W=\int_L \vec{F} \cdot d\vec{r} = \int_L \vec{F} \cdot \frac{d\vec{r}}{dt} dt$$. Of course the problem here would be figuring out r(t) and that itself would depend on how you choose to accelerate the ship.

5. Jul 20, 2008

### rcgldr

Work is force times distance relative to the point of application of the force.

Rockets are different, there's nothing in empty space to push against, so they expell a part of their own mass (spent fuel) at high rate of acceleration backwards, which results in a forwards acceleration of the rocket. If there are no external forces to consider (like gravity), then momentum is conserved (fuel mass times fuel velocity + rocket mass times rocket velocity = constant). The result of burning fuel is the increase in kinetic energy of spent fuel and rocket. To maintain constant acceleration, the burn rate will decrease directly in proportion to the decrease in mass of the rocket as the spent fuel is eject, but it will still be a huge number to maintain 1g for a long period of time.