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# Quickest way to get to The Super Earth, will this work?

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Forgive me, but, what do you mean by "Utilize all of the mass in the universe?" Can you explain that thoroughly, I think that would do me more good than Wikipedia's definitions.

For a massive object to reach c, it must have infinite energy. Since mass and energy are related, and since we are speaking of infinite energy, infinite mass is implied.

In mathematical terms, when we give c a value of 1, the equation $E=mc^2$ becomes $E=m$. Thus, infinite m = infinite E. This paradox is resolved because only a particle with a rest mass of zero is permitted to travel at c. Thus, $E=mc^2$ becomes, $0=0*1^2$, which equals zero. ;)

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Why not, if ten times less thrust can propel the craft at 1/10 the speed of light, why can't ten times more propel it AT the speed of light??

It doesn't make sense.:)

You'll get no argument from me on that point, it doesn't make sense from a common sense stand point but relativistic travel seldom makes sense from our point of view. On way to look at it is this, the closer you to the speed of light the more massive your vehicle becomes. As you pour more energy into accelerating your craft more of that energy goes into making your craft more massive than it does into making it go faster until eventually almost none goes into speed and it all goes into mass. Personally I think your vehicle would collapse into a black hole of it's own making if it got massive enough but any way you look at it nothing made of mass can achieve "C" Not even something as small as a sub atomic particle.

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Damn, I type too slow:doh:

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So if my craft had thousands of G's of thrust propelling it, it would be so heavy that the thrust wouldn't propel it ten times faster than if it had hundreds of G's of thrust?

Speed can create drag!

That actually makes a lot of sense.:)

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So if my craft had thousands of G's of thrust propelling it, it would be so heavy that the thrust wouldn't propel it ten times faster than if it had hundreds of G's of thrust?

Speed can create drag!

That actually makes a lot of sense.:)

Actually it's worse than that, eventually no matter how many gees of thrust you had you would get no acceleration at all.

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Actually it's worse than that, eventually no matter how many gees of thrust you had you would get no acceleration at all.

Yes, because it would weigh so much that the thrust wouldn't be able to push it, and eventually the weight will push against the thrust, and thats when time travel starts.

I just figured out time travel, I never understood it.

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Yes, because it would weigh so much that the thrust wouldn't be able to push it, and eventually the weight will push against the thrust, and thats when time travel starts.

I just figured out time travel, I never understood it.

Actually you'll never get to that point, it's forever barred from our experience due to the speed of light being faster than anything can go. Of course there are certain other theories but they aren't as widely accepted.....:)

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I actually already brought this up back on Page 2 of this thread, but the energy needed to get any mass up to a any given velocity is found by.

$E = mc^2 \left ( \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}-1 \right )$

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I actually already brought this up back on Page 2 of this thread, but the energy needed to get any mass up to a any given velocity is found by.

$E = mc^2 left ( sqrt{frac{1}{1-frac{v^2}{c^2}}}-1 right )$

How do I deciver that, what does it mean.

I don't know formula's yet.:(

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Why not, if ten times less thrust can propel the craft at 1/10 the speed of light, why can't ten times more propel it AT the speed of light??
There’ve already been some good answers, so I’ll just try throwing in a bit of practical math.

The main thing you need to know to use special relativity in rocket (or laser-powered lightsail, or whatever) calculations is the Lorentz factor:

$\gamma = \left ( 1 - \left (\frac{v}{c} \right )^2 \right)^{- \frac12}$.

It’s used a lot. In the case of mass dilation, as a body increases in relative speed, its mass increases according to the formula $M = M_0 \cdot \gamma$.

Since acceleration is proportional to force divided by mass ($A=\frac{F}{M}$), The force needed to accelerate a body by the same amount at different speeds looks like this:

Speed (c)                          Force
.01                                 1.00
.05                                 1.00
.1                                  1.01
.25                                 1.03
.5                                  1.15
.75                                 1.51
.9                                  2.29
.99                                 7.09
.999                               22.37
.9999                              70.71
.99999                            223.61
.999999                           707.11
.9999999                         2236.07
.99999999                        7071.07
.999999999                      22360.68
.9999999999                     70710.68
.99999999999                   223606.80

The closer you get to 1 c, the greater the force required.

It’s important to understand that this isn’t just theory. Although we’ve never accelerated a spacecraft to much more than .00005 c, very small bodies – electrons, protons, and sometimes more massive atomic nuclei – are routinely accelerated to high fractions of c in particle accelerators, requiring more and more force as the speed approaches 1 c.

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How do I deciver that, what does it mean.

I don't know formula's yet.:)

Formulas may look complicated, but they're easy once you get the hang.

E = mc^2 left ( sqrt{frac{1}{1-frac{v^2}{c^2}}}-1 right )"

The E on the left refers to Energy, measured in Joules.

The m to the right of the = refers to Mass, the mass of the spaceship in kilograms

The C squared means the speed of light in meters per second, squared. The creates a very, very large number

The big crooked thingie is the symbol for square-root. So, we will take the square root of everything under it.

Under the square root thingie we have a fraction within a fraction, and the inner-most part is V squared divided by C squared.

V is the velocity you want the spaceship to travel at.

C is the speed of light again.

Does this help?

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How do I deciver that, what does it mean.

I don't know formula's yet.:(

Formulas may look complicated, but they're easy once you get the hang.

$E = mc^2 \left ( \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}-1 \right )$

The E on the left refers to Energy, measured in Joules.

The m to the right of the = refers to Mass, the mass of the spaceship in kilograms

The C squared means the speed of light in meters per second, squared. The creates a very, very large number

The big crooked thingie is the symbol for square-root. So, we will take the square root of everything under it.

Under the square root thingie we have a fraction within a fraction, and the inner-most part is V squared divided by C squared.

V is the velocity you want the spaceship to travel at.

C is the speed of light again.

Does this help?

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• 2 weeks later...

Formulas may look complicated, but they're easy once you get the hang.

$E = mc^2 left ( sqrt{frac{1}{1-frac{v^2}{c^2}}}-1 right )$

The E on the left refers to Energy, measured in Joules.

The m to the right of the = refers to Mass, the mass of the spaceship in kilograms

The C squared means the speed of light in meters per second, squared. The creates a very, very large number

The big crooked thingie is the symbol for square-root. So, we will take the square root of everything under it.

Under the square root thingie we have a fraction within a fraction, and the inner-most part is V squared divided by C squared.

V is the velocity you want the spaceship to travel at.

C is the speed of light again.

Does this help?

I under stand it completely, well, the E=MC2/Finding out how much speed a space ship will travel part because of you.

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