Relative Motion: Collision of Two Spaceships on Earth

AI Thread Summary
Two spaceships, moving towards each other at 0.8c and 0.6c, collide after 1 second according to an observer on Earth, with a distance of 4.2 x 10^8 m between them. The velocities of the spaceships relative to each other are incorrectly calculated as 1.4c, which violates the principles of relativity. The correct approach requires using the relativistic velocity addition formula to determine their speeds in each other's frames. The collision time remains consistent at 1 second in the Earth frame, but must be recalculated for the frames of the individual spaceships. The discussion highlights the importance of considering relativistic effects in high-speed scenarios.
olga11
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Homework Statement



An observer on Earth sees two spaceships moving in opposite directions and finally they collide. At t=0 the observer on Earth says that the spaceship 1 which moves to the right with Ua=0.8c is at the point A and the spaceship 2 which moves to the left with Ub=0.6c is at the point B. The distance AB=L=4,2.10^8 m.
When do the two spaceships collide to the Earth frame of reference?
What is the velocity of the spaceship 2 to the frame reference of the spaceship 1?
What is the velocity of the spaceship 1 to the frame reference of the spaceship 2?
When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

Homework Equations





The Attempt at a Solution



Let D be the point of the collision and AD=x, so DB=L-x
The velocity is constant.
spaceship 1: AD=x=0,8c.t
spaceship 1: DB=L-x=0,6c.t
By addition
L=AD+DB
L=(0,8c+0,6c).t
t=4,2.10^8/(1,4.3.10^8)
t=1sec
Is it ok so far?

A hint to go on, please?
 
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The velocity of the spaceship 2 to the frame reference of the spaceship 1 is Uba=Ua+Ub=1.4c
The velocity of the spaceship 1 to the frame reference of the spaceship 2 is also Uab=Ua+Ub=1.4c

When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

t=AB/Uab=4,2.10^8/(1,4.3.10^8)=1 sec

Something is wrong. It cannot be the same.
 
You don't consider the fact that the speeds are comparable to the speed of light.
 
1. Homework Statement

An observer on Earth sees two spaceships moving in opposite directions and finally they collide. At t=0 the observer on Earth says that the spaceship 1 which moves to the right with Ua=0.8c is at the point A and the spaceship 2 which moves to the left with Ub=0.6c is at the point B. The distance AB=L=4,2.10^8 m.
When do the two spaceships collide to the Earth frame of reference?
What is the velocity of the spaceship 2 to the frame reference of the spaceship 1?
What is the velocity of the spaceship 1 to the frame reference of the spaceship 2?
When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

2. Homework Equations



3. The Attempt at a Solution

Let D be the point of the collision and AD=x, so DB=L-x
The velocity is constant.
spaceship 1: AD=x=0,8c.t
spaceship 1: DB=L-x=0,6c.t
By addition
L=AD+DB
L=(0,8c+0,6c).t
t=4,2.10^8/(1,4.3.10^8)
t=1sec
Is it ok so far?
 
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