Rocket Velocity: γ = 4, v = √15/4c

billy92
Messages
12
Reaction score
0
If a rocket is traveling past the Earth at speed v, and the time interval measured by the astronaut is four times smaller than the time interval measured by an observer on the Earth. What is the value of v

I have attempted the solution and got:

Δt=γΔt'

Where i believe that gamma will be 4.

From this i used:

γ=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

From this i managed to calculate v=\frac{\sqrt{15}}{4}c

However i am slightly concerned about how large this value is for a rocket but cannot seem to see any other solution to this problem. Any suggestions?

Thanks
 
Physics news on Phys.org
Yes, you are right, it takes very large values of velocity to get appreciably large values of gamma. You got the right answer.
 
ghwellsjr said:
Yes, you are right, it takes very large values of velocity to get appreciably large values of gamma. You got the right answer.

Thanks for the quick reply.
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top