/sci/ - Science & Math

-_-

you could at least have the decency to give us SOME numbers... if you want us to do the dirty work you could at least google something.
Also the distance will change depending on where the planets are in their orbits

can light even travel from pluto to mars in one day?

>>3245055

Sorry, I checked. It takes light 12 minutes to reach Mars and 5.3 hours to reach Pluto.

depends on how close they are.
at closest distance thats about 4.2 billion kilometers.
speed of light is 3X10^5 kilometers a second.

after a simple division, i get 140000 seconds.

thats about a 39 hour trip.

..altho ive been pwned in math threads before, so i'd have someone verify it who isn't considered fucking retarded by the masses.

>>3245055
Also..

We need to know relative to WHO is the 1 day. I assume you mean relative to the crew on board the spaceship?? right?
Because it would be a different answer if you are on the Earth watching the ship travel there.

>>3245060
derp, possibly 14000 seconds, not 140000 (i must have clicked an extra zero)
so, 3.9 hours?

3.9/24 means you would have to go about 16% the speed of ight to get to pluto in one day.

>>3245073
You aren't taking into account length contraction though are you?

Relative to the crew the distance will be lessened.

>>3245092
wat?

>>3245092
<span class="math">\sqrt{1-0.16^2} \approx 0.987[/spoiler]
No one gives a shit about 1.5% difference.

>>3245092
that won't be significant at those speeds. More pressing would be the need to accelerate/decelerate. What forces would the travelers be subjected to? Unless you have some sort of inertial dampener, going like this would have turned you into gravy.

>>3245059
That's more like it. So then the difference is 5.1 hours for light, which means that the distance is <span class="math">5.504\times 10^{12}m[/spoiler]
This is the distance relative to observers on the Earth.

Now assuming
>>3245061
you want it to be relative to people on board the ship, then we use length contraction:
<span class="math">d'=d\sqrt{1-\frac{v^{2}}{c^{2}}}[/spoiler]
now the velocity is:
<span class="math">v=\frac{d'}{1day}=\frac{d'}{86400s}[/spoiler]
<span class="math">d'=(86400s)v[/spoiler]

so:
<span class="math">(86400s)v=d\sqrt{1-\frac{v^{2}}{c^{2}}}[/spoiler]
then just solve for v:
<span class="math">v=\sqrt{frac{c^{2}d^{2}}{c^{2}(86400s)^{2}+d^{2}}}[/spoiler]

think that's right...

So then:
<span class="math">v=6.231\times 10^{7}[/spoiler]

but you asked for the percentage of the speed of light which is just dividing by c=299792458m/s

<span class="math">\frac{v}{c}\times 100%=20.79%[/spoiler]

>>3245096

Relativity.

This distance actually shrinks while you're at relativistic velocity. Or you and your ship increase in length, it's the same thing.

>>3245130
sorry the stupid math tags are all fucked..

it's right though... trust me :p

>>3245059
That's more like it. So then the difference is 5.1 hours for light, which means that the distance is <span class="math">5.504\times 10^{12}m[/spoiler]
This is the distance relative to observers on the Earth.

Now assuming
>>3245061
you want it to be relative to people on board the ship, then we use length contraction:
<span class="math">d'=d\sqrt{1-\frac{v^{2}}{c^{2}}}[/spoiler]
now the velocity is:
<span class="math">v=\frac{d'}{1day}=\frac{d'}{86400s}[/spoiler]
<span class="math">d'=(86400s)v[/spoiler]

so:
<span class="math">(86400s)v=d\sqrt{1-\frac{v^{2}}{c^{2}}}[/spoiler]
then just solve for v:
<span class="math">v=\sqrt{frac{c^{2}d^{2}}{c^{2}(86400s)^{2}+d^{2}}}[/spoiler]

think that's right...

So then:
<span class="math">v=6.231\times 10^{7}[/spoiler]

but you asked for the percentage of the speed of light which is just dividing by c=299792458m/s

which is just 20.79% the speed of light, the required speed to travel that distance (measured from Earth) in exactly 1 day (observed by passengers on the ship).

>>3245059
That's more like it. So then the difference is 5.1 hours for light, which means that the distance is <span class="math">5.504\times 10^{12}m[/spoiler]
This is the distance relative to observers on the Earth.

Now assuming
>>3245061
you want it to be relative to people on board the ship, then we use length contraction:
<span class="math">d'=d\sqrt{1-\frac{v^{2}}{c^{2}}}[/spoiler]
now the velocity is:
<span class="math">v=\frac{d'}{1day}=\frac{d'}{86400s}[/spoiler]
<span class="math">d'=(86400s)v[/spoiler]

so:
<span class="math">(86400s)v=d\sqrt{1-\frac{v^{2}}{c^{2}}}[/spoiler]
then just solve for v:
<span class="math">v=\sqrt{\frac{c^{2}d^{2}}{c^{2}(86400s)^{2}+d^{2}}}[/spoiler]

think that's right...

So then:
<span class="math">v=6.231\times 10^{7}[/spoiler]

but you asked for the percentage of the speed of light which is just dividing by c=299792458m/s

which is just 20.79% the speed of light, the required speed to travel that distance (measured from Earth) in exactly 1 day (observed by passengers on the ship).

>>3245108
proabaly right.

There is a 4.7% difference in our answers, but I think that is just because we used different numbers.
I used the distances the OP stated here:
>>3245059

so that probably accounts for most of the difference.

considering that the actual distance is 32.86AU, that's the 18% of the speed of light

>>3245198
There is no actual distance, it changes with orbit.

18%, then? And to make the journey in four days or so, it would be 4.2% or so, right?

>>3245226

I mean the distance right now: http://www.wolframalpha.com/input/?i=pluto+distance+from+mars

In a 1 day travel any change is negligible for this calculation..

>>3245258

Right, i think it would be about 4.7%

>>3245276

Thanks bro

>>3245280

you're welcome.

Came to see if relativity was accounted for
>>3245061

Leaving satisfied

>>3245044
Isn't Pluto like half a lightyear from the Sun?

>>3245307

Come to this thread.
See the answer is about 0.15c.
Notice that depending on the time answer may change by about 20%.
See a bunch of kids talking about realtivity, when it gives about 2% correction.
That's /sci/ for you.

>>3245313

no