Thanks you so much! Sorry for the bad manners. Last question i would have is whether in this case you need to convert the units from lys to meters and years to seconds? i would assume so.
Solving for u:
L0 *Sqrt(1-(u^2/c^2)) / u = t0
First square both sides giving us L0^2 / u^2 * (1-(u^2/c^2)) = t0^2
next we distribute the L0^2 / u^2 so we get
L0^2 / u^2 - L0^2/c^2 = t0^2 (u^2 gets canceled out in the second term)
so solving for u we get
u^2 = L0^2 / (t0^2 + (L0^2/c^2))...
My apologies this question has given me a bad mood. Steps:
Assuming we are in ship's inertial frame as the ship goes toward regulus, the time we see for the ship to make the trip should be equal to L1 (the contracted distance between the ship and regulus) / u (the velocity of the ship) aka...
I'm not trying to guess i am taking the course and i understand where the time dilation equation and length contraction equation come from, maybe it is as simple as saying 77lys / 40 years produces a speed over c so therefore it is not possible but i thought that maybe time dilation would have...
Looking again i see that i did it wrong the problem I am having is dealing with the 1-u^2/c^2 i don't know how to separate the u out. I was defining t1 as 40 as its the number of years he would need to reach into be alive and L0 i defined as 77 light years.
Assuming i sub t0 for L0/u then the only variable unknown is u but I am left with t1 = L0/u (1/(sqrt(1- (u^2/c^2))) and i don't know how to solve this for u. (Sorry I am on mobile so i can't add this to post)
Homework Statement
A visit to Regulus is on my bucket list. However, it is 77 light years away. Assuming I will live only another 40 years, can I make it to Regulus? How fast would I have to travel (at constant speed) to get there in 40 years?
Homework Equations
t1 = t0...
so would perhaps the conservation of kinetic energy formula work?
or actually its the conservation of angular momentum so (moment of inertia)(angular speed) intial = (moment of inertia)(angular speed) final?
Homework Statement
In a playground there is a small merry-go-round of radius 1.20 m and mass 220 kg. The radius of gyration is 91.0 cm. A child of mass 44.0 kg runs at a speed of 3.00 m/s tangent to the rim of the merry-go-round when it is at rest and then jumps on. Neglect friction between the...