Calculating Earth's Increased Mass with Velocity Change: A Scientific Analysis

In summary, when the Earth would move 1000 km/s faster than now, its mass would not change, but its velocity would be converted to energy which would then be released during an explosion.
  • #36
I am sorry I don't understand the type of resistance you are referring to. Is it similar to a frictional force??

Space and matter is somehow connected.
Think about a magnetic field, or think about how matter/gravity bends space.

Mass/gravity is increasing due to higher velocity; hence space must bend proportional more due to increasing mass/velocity.

Put the same question above in a different way:
Does it require energy to bend space?
I guess we don’t know ?
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( I am a stranger if my English is not perfect just correct me)
 
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  • #37
it does require energy to bend space. To increase velocity and hence mass, we supply energy which then bends space as it moves. :)
 
  • #38
it does require energy to bend space. To increase velocity and hence mass, we supply energy which then bends space as it moves.
Agree.
Now, - when a body moves something must happen with the space/matter connection.
I mean space bends as a result of the motion of a body right ?

But what is the "space matter connection" ?
And do we have any resistance here?
Think also of the plasma circling around the suns equator, - suddenly every 11 years, > sun storms. Is that due to the space/matter connection?
Does space resist such motion of plasma/matter ?
 
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  • #39
Vin3000
Hmmm
It takes more and more energy to get a smaller and smaller increase in speed.
when its speed is 0.7c, its proper velocity is
[1/sqrrt.(1-v^2/c^2)]v which is [1/sqrrt.(1-0.49)]0.7c = [1/sqrrt.0.51]v = (0.7/0.714)c=0.98c
when speed is 0.6c, its proper velocity is (0.6/0.8)c=0.75c
The speeds do not increase as they are supposed to, it requires extra energy.
 
  • #40
The speeds do not increase as they are supposed to, it requires extra energy.
Agree
 
  • #41
vin300 said:
when its speed is 0.7c, its proper velocity is
[1/sqrrt.(1-v^2/c^2)]v which is [1/sqrrt.(1-0.49)]0.7c = [1/sqrrt.0.51]v = (0.7/0.714)c=0.98c
when speed is 0.6c, its proper velocity is (0.6/0.8)c=0.75c
Can you explain what you're doing here and what you mean by "proper velocity"?

Also folks, please use the quote feature properly so that the link to the original post is maintained and the person quoted is identified.
 
  • #42
Doc Al said:
Can you explain what you're doing here and what you mean by "proper velocity"?

Also folks, please use the quote feature properly so that the link to the original post is maintained and the person quoted is identified.
Proper velocity is the distance measured by the stationary observer on the time elapsed on the moving observer's clock(proper time), which is gamma times v. It can also be interpreted as the velocity of the body, when it becomes equal to c by the formula becomes infinity
 
  • #43
vin300 said:
Proper velocity is the distance measured by the stationary observer on the time elapsed on the moving observer's clock(proper time), which is gamma times v. It can also be interpreted as the velocity of the body, when it becomes equal to c by the formula becomes infinity
Ah, OK. Forgot about that term since I don't use it much. Don't confuse it with the actual speed of the body as measured in some reference frame.
 
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