Special Relativity WS#2 -- Earth-Pluto SuperShuttle line time dilation

[relativelnr00]

Homework Statement

The new earth-Pluto SuperShuttle line boast that it can take you between the two planets (which are about 5.0 h apart in distance) in 2.5 h according to your watch on board the shuttle. Assume that the shuttle travels at a constant velocity.

a.) What time interval must the synchronized space station clocks register between the shuttle’s departure and its arrival at Pluto if the advertisement is true?

b.) What is the shuttle’s cruising speed?

Homework Equations

∆Sab = √∆t^2ab - ∆x^2ab (?)
u=x/t (?)

The Attempt at a Solution

a) I believe the time interval would be coordinate time, as the space stations are in different locations, but are in the same inertial frame, although I'd appreciate a second opinion.

b) Stuck on this part mainly. Does cruising speed equate to velocity in this case? The formula ∆Sab = √∆t^2ab - ∆x^2ab won't work since t is less than x in this case, thus resulting in an imaginary number which can't be the answer.

Anyone have any advice? I'm not looking for the whole solution to be given to me, but I'm at a standstill in terms of my understanding of how the problem works...

 

[Chestermiller]

b) Stuck on this part mainly. Does cruising speed equate to velocity in this case? The formula ∆Sab = √∆t^2ab - ∆x^2ab won't work since t is less than x in this case, thus resulting in an imaginary number which can't be the answer.

The Δt in this equation is the time interval as reckoned from the Earth's frame of reference, not the shuttle's frame of reference.

Chet

 

[relativelnr00]

The Δt in this equation is the time interval as reckoned from the Earth's frame of reference, not the shuttle's frame of reference.

Chet

Chet, would that mean that the time interval would in fact be 5 hours? I attempted to plot a worldline graph, but it doesn't seem to make sense if I use 5 hours for the t axis instead of 2.5 hours...

 

[Chestermiller]

Chet, would that mean that the time interval would in fact be 5 hours? I attempted to plot a worldline graph, but it doesn't seem to make sense if I use 5 hours for the t axis instead of 2.5 hours...

No. What is this same ΔS as reckoned from the shuttle's frame of reference where time and distance are t' and x', respectively? What is Δx' for an observer on the shuttle?

Chet

 

[relativelnr00]

No. What is this same ΔS as reckoned from the shuttle's frame of reference where time and distance are t' and x', respectively? What is Δx' for an observer on the shuttle?

Chet

Would Δx in this case be the 5 hours? As it mentions in the problem, the planets are 5 hours apart in distance and I'm on the shuttle measuring by my watch. I'm sorry if I'm missing any obvious parts of the problem...we only recently began learning about these types of problems and they make almost no sense to me.

Also, apologies if this is a silly question, but I'm assuming this problem involves time dilation? We learned about it in the last lesson but I'm still unfamiliar with how to apply it within problems.

 

[Chestermiller]

Would Δx in this case be the 5 hours? As it mentions in the problem, the planets are 5 hours apart in distance and I'm on the shuttle measuring by my watch. I'm sorry if I'm missing any obvious parts of the problem...we only recently began learning about these types of problems and they make almost no sense to me.

I think that they meant to say that Δx is 5 light-hours. Have you learned the following equation yet?
$$(ΔS)^2=(Δt)^2-(Δx)^2=(Δt')^2-(Δx')^2$$
If so, what do the parameters in this equation represent physically with regard to your problem?

Chet

 

[relativelnr00]

I think that they meant to say that Δx is 5 light-hours. Have you learned the following equation yet?
$$(ΔS)^2=(Δt)^2-(Δx)^2=(Δt')^2-(Δx')^2$$
If so, what do the parameters in this equation represent physically with regard to your problem?

Chet

This looks new to me, so I may not have actually learned it yet.

(ΔS)2=(Δt)2−(Δx)2=(Δt′)2−(Δx′)2

So the values I currently have are:

x = 5 hours
t = 2.5 hours

The spacetime interval would be measured from within the shuttle, as both events of departing and leaving occur within the observer's frame of reference.

Thus, the equation so far would be:

(ΔS)^2=(2.5)^2−(5)^2=(Δt′)2−(Δx′)2

I think..?

Thanks a lot for your help so far Chet. I really appreciate it!

 

[Chestermiller]

This looks new to me, so I may not have actually learned it yet.

(ΔS)2=(Δt)2−(Δx)2=(Δt′)2−(Δx′)2

So the values I currently have are:

x = 5 hours
t = 2.5 hours

The spacetime interval would be measured from within the shuttle, as both events of departing and leaving occur within the observer's frame of reference.

Thus, the equation so far would be:

(ΔS)^2=(2.5)^2−(5)^2=(Δt′)2−(Δx′)2

I think..?

Thanks a lot for you help so far Chet. I really appreciate it!

This is not correct yet. Here's a hint: Δt' = 2.5 hours (the time interval measured on the shuttle's clock), Δx = 5 light-hours (the distance as reckoned from the Earth frame of reference). Δt is the travel time as reckoned from the Earth frame of reference; this is what you are trying to solve for.

Chet

 

[relativelnr00]

This is not correct yet. Here's a hint: Δt' = 2.5 hours (the time interval measured on the shuttle's clock), Δx = 5 light-hours (the distance as reckoned from the Earth frame of reference). Δt is the travel time as reckoned from the Earth frame of reference; this is what you are trying to solve for.

Chet

So then:

(ΔS)2=(Δt)2−(Δx)2=(Δt′)2−(Δx′)2 would in fact be:

(ΔS)2=(Δt)^2−(5)^2=(Δ2.5)^2−(Δx′)^2

Removing (ΔS)2 means:

(Δt)^2−(5)^2=(Δ2.5)^2−(Δx′)^2

Thus:

(Δt)^2+(Δx′)^2=(Δ2.5)^2+(5)^2

Thus (Δt)^2+(Δx′)^2=31.25

But it seems too easy this way if (Δt)^2 is in fact (2.5)^2...

 

[Chestermiller]

You need to substitute a value for Δx'. Has the shuttle's position changed within it's own frame of reference, or is it at rest within its own frame of reference. What does that tell you about Δx'. Using this value for Δx', what do you get for Δt?

Chet