Lorentz Factor / Spacecraft to Star

Smythe

A star is located 4.2 light years from Earth.

At what constant velocity must a spacecraft travel from Earth if it is to reach the star in 3.0 years time?

I've been working on this for a couple of days, but I'm not entirely sure which equation to use.

Also, the fact that the distance is given in light years. Generally formulas require the distance input in metres, so what would I input into a formula for distance? If I input 4.2 it may indicate 4.2 metres!

Thank you

Morbert

Smythe wrote: »

I've been working on this for a couple of days, but I'm not entirely sure which equation to use.

Also, the fact that the distance is given in light years. Generally formulas require the distance input in metres, so what would I input into a formula for distance? If I input 4.2 it may indicate 4.2 metres!

Thank you

time = distance/speed

time = t = 3.0 years

distance = L = 4.2 lightyears / γ (Where γ is the usual factor. You divide by γ because the distance 4.2 will be length-contracted)

speed = v (unknown)

So

t = L / (γ v)

Solve for v

FISMA

Smythe,
Is this for a class? If so, which one.

If it takes light 4.2 years to get to the star and you're hoping to do it in three, we're obviously going faster than the speed of light.

That's something us Physics types don't like!;)

Smythe

If I solve for v I get;

v = (L0 * c) / ( (c * t)^2 + L0^2) )^0.5

L0 = 4.2
c = 3 * 10^8
t = 3

I get an answer of v = 1.4

I'm not sure if this is correct or not? If it is correct, what is meant by 1.4? 1.4 what exactly?

Morbert

Smythe wrote: »

If I solve for v I get;

v = (L0 * c) / ( (c * t)^2 + L0^2) )^0.5

L0 = 4.2
c = 3 * 10^8
t = 3

I get an answer of v = 1.4

I'm not sure if this is correct or not? If it is correct, what is meant by 1.4? 1.4 what exactly?

What units are you using? For time, you should be using the second. For distance, the lightsecond. This will set c = 1 and make the calculation much easier.

I.e.

L = 4.2 lightyears = 132451200 lightseconds
t = 3 years = 94608000 seconds
c = 1 lightsecond / second

I am also not sure where you got

v = (L0 * c) / ( (c * t)^2 + L0^2) )^0.5

You should get

t = L / (γ v)
v = L/ (γ t)
v^2 = L^2 / (γ t)^2 = L^2/t^2 (1 - v^/c^2)
v^2 (1 + L^2/t^2 c^2) = L^2/t^2

v = Square root of { ( L^2 / t^2 ) / (1 + L^2/(t^2c^2)) }

The answer I got was 0.814 (I.e. 81.4% the speed of light).

Smythe,
Is this for a class? If so, which one.

If it takes light 4.2 years to get to the star and you're hoping to do it in three, we're obviously going faster than the speed of light.

That's something us Physics types don't like!

It depends on which reference frame we consider (the question is a poorly worded one in this regard).

From the reference frame of earth, it is indeed impossible to reach the star in 3 years. But from the reference frame of the spaceship, due to time dilation, it can reach the star in 3 years.

This is a subtle aspect of relativity. If we wanted to build a ship fast enough to reach the edge of the observable universe in a couple of seconds, there is no physical principle stopping us (Admittedly, from an engineering standpoint, it is effectively impossible.) The only thing is, when we reach our destination, the rest of the universe will have aged billions of years.

amen

A long time since I did any physics but I would take the 4.2 light years and convert to metres (approx 9.46 PM per light year). Take the 3 years and convert to seconds(31,536,000 seconds per year).

speed = distance/time

speed = (9.46 PM *4.2) /(31536000*3)

speed = 419964485 m/s

speed of light = 299 792 458 m/s

so dividing speed/speed of light gives 1.4 light years.

You divide by γ because the distance 4.2 will be length-contracted

why ? I thought you only need to do this if you were trying to measure something travelling at a large portion of the speed to the light.

In this case the spaceship just needs to go from A->B so their should be no contraction (length, time etc involved)

Morbert

amen wrote: »

A long time since I did any physics but I would take the 4.2 light years and convert to metres (approx 9.46 PM per light year). Take the 3 years and convert to seconds(31,536,000 seconds per year).

speed = distance/time

speed = (9.46 PM *4.2) /(31536000*3)

speed = 419964485 m/s

speed of light = 299 792 458 m/s

so dividing speed/speed of light gives 1.4 lightyears.

Lightyears is a unit of distance, not speed. 1.4 would be 1.4 times the speed of light, which is physically impossible.

why ? I thought you only need to do this if you were trying to measure something travelling at a large portion of the speed to the light.

In this case the spaceship just needs to go from A->B so their should be no contraction (length, time etc involved)

To go from A->B in 3 years, when they are 4.2 lightyears apart, requires a considerable portion of the speed of light. 0.814 c to be more precise.

amen

Ahh I see what you are doing.