Consistent Trajectory for a non-zero rest mass particle?

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The discussion revolves around verifying the consistency of a trajectory defined by the equation dx/dt = cgt/√(1+g²t²) for a non-zero rest mass particle. The main concern is whether the velocity remains less than the speed of light (c) for all time (t). As t approaches infinity, the calculated velocity approaches c, which raises questions about the trajectory's consistency. Participants note a potential typo in the original post, leading to confusion about the limits involved. Ultimately, the trajectory appears inconsistent with the expectations set by relativistic physics.
DeldotB
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Homework Statement


Good day all!
Quick question:
As part of a problem statement, I'm asked to verify if the trajectory: \frac{dx}{dt}=\frac{cgt}{\sqrt{1+g^2t^2}}
Is "consistent".

Homework Equations


None

The Attempt at a Solution



Im not sure what "consistent" means. Does it mean, \frac {dx}{dt} < c for all t? If that's so, I run into a problem because in the limit as t approaches infinity, the velocity = the speed of light (the limit goes to c). Am I approaching this the wrong way? (The trajectory is supposed to be "consistent")
 
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You may want to check the limit of that expression for large ##t##.
 
PeroK: Not sure what you mean...
I get "c" as the limit. Maybe my work is wrong? Lim\, \, t\rightarrow \infty (\frac{cgt}{\sqrt{1+(9.8))^2t^2}})=cg(Lim\, \, t\rightarrow \infty (\frac{t}{\sqrt{1+(9.8))^2t^2}}))=cg(5/49)=c. So as t approaches infinity, the velocity approaches c.
 
That's correct, but inconsistent with the limit of ##cg## you gave in the original post!
 
Ah, I see. A miss-type. Well, nevertheless, this trajectory doesn't seem to be consistent even though my assignment is saying it should be.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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