# An acceleration and distance question

1. Mar 21, 2016

### Gardenman

I'm writing a sci-fi novel where a spaceship is leaving Earth's orbit accelerating at 2 feet per second per second. I'm trying to figure out how long it'll take to go past the moon's orbit (238,000 miles.) My calculations show it to take about 9.8 hours. A fellow writer, who's a math nut says it'll take more like 20 years. One of us is very, very wrong. I'm hoping it's him. I try to stay reasonably accurate in my writing and I get annoyed at the insane acceleration they show in many sci-fi shows/novels. Those rates of acceleration would have the crew being nothing but bloody splatters against the back wall of the ship if they really took off that quickly.

Here's my math. V=A*T (assuming zero velocity initially.) So with an acceleration of 2 feet per second at the end of an hour (3600 seconds) you'd be going 7200 feet per second. With there being 5,280 feet per mile, you'd be going 1.3636 miles per second which scales up to 4909.09 miles per hour (3600 seconds per hour.) The speed would then increase by 4909.09 miles per hour each hour.

Now to find the distance I'm using the following formula d=vt+1/2 at squared.In this equation the v is the initial velocity which was zero, so that part of the equation can be dropped. Zero times anything is zero. That brings us to the 1/2 acceleration which is 2 feet per second) and half of two feet per second is one. So we end up with the distance being 1+ the time squared. So, the distance is basically the square root of the time squared. I have to convert the distance back to feet from miles since I kept acceleration in feet per second. So the 238,000 miles is multiplied by 5280 and you get 1,256640,000 feet. The square root of that is 35,449,11. So the time is 35,449.11 seconds. Divide that by 3600 seconds to get how many hours it is and you get 9.8 hours. This coincides pretty nicely with the speeds I'm calculating. To go 238,000 miles in 9.8 hours, you'd need an average speed of 24,285.7 miles per hour which you'd hit just before the five hour mark based on accelerating at 4909 miles per hour.

Is that math more or less right? I haven't done this type of math in about forty years, so I'm not overly confident in it, but that's the number I keep coming up with. Thanks. Later on I'll bug you guys with the time shift between people on Earth and those traveling quickly in the spacecraft when I get to that part of the book.

2. Mar 21, 2016

### drvrm

with your calculation every moving thing on earth can think of going past the moon.
may be i am not correct but leaving earth takes a lot of kinetic energy as you throw things on earth and the things come back after reaching a certain height. something is pulling you back and first one has to get out of its pulling force....

as these days people go past the moon or land on the moon (if you believe them) take their velocity/acceleration data to start the calculations-your friend was perhaps right at the spot.

3. Mar 21, 2016

### PeroK

If you walked at a constant speed of 2ft/s, it would take about 20 years to walk 238,000 miles. Approximately.

If you acclererate at 2ft/s/s, then you travel 238,000 miles in about 10 hours. Approximately.

4. Mar 21, 2016

### jbriggs444

Yes, it is right. The sanity-check you did was a good idea and was correct as well.

5. Mar 21, 2016

### Gardenman

Thanks. I keep coming up with that number, but everyone else seems to be assuming a constant speed of two feet per second as opposed to an acceleration of two feet per second. If the speed is a constant two feet per second and not accelerating it takes twenty years to reach the moon, but if you're accelerating at two feet per second the time drops a ton. At least by my math.

6. Mar 21, 2016

### ProfuselyQuarky

Tell us when your novel is published!

7. Mar 21, 2016

### pixel

Agreed. But some novels do address this. Charles Sheffield's The McAndrew Chronicles employs a "balanced inertia" drive that allows for large accelerations. Basically, gravity from a very dense, moveable mass is used to balance the g force due to the acceleration. It's all highly futuristic and probably not practical, as I think Sheffield himself discusses in the afterward, but at least the issue is addressed.