Age of the earth?

1. How old? give a number

So. Let me get this straight, the earth is moving upward at a constant acceleration of 9.8m/s^2? Does this upward movement have a terminal velocity?

The earth continually accelerates upwards at 1g. No, there is no terminal velocity. No, it will never reach the speed of light.

It won't reach the speed of light? why not?

in matter of fact it would reach the speed of light in less than a year:

9.8m/s^2 * 60s/min * 60min/hr* 24hr/day* 365day/year multiplying this out gives us
30904280 m/(sec*yr)

Set this equal to the speed of light with a constant acceleration of 9.8m/s^2 the earth would pass the speed of light in a year. Now that the model of gravity for these two "theories" has been mathematically disproven, please remove from the faq.

please explain what limits the earth from going past C..

Q: "If the Earth's acceleration is constant, wouldn't it be traveling faster than light eventually?"

A: The equations of Special Relativity prevent an object with mass from reaching or passing the speed of light. Due to this restriction, these equations prove that the Earth can accelerate at a constant rate forever in our reference frame and never reach the speed of light. Click here for an in depth explanation.

You guys like landover baptist church? I'm a member. I like being on my knees and praising jesus.

brilliant.

obviously. the acceleration can't be a constant 9.8m/s^2. the earth would be traveling 4.5 billion times the speed of light.

Quote from: owned on July 06, 2011, 10:32:42 PM

brilliant.

Sir, we have a rule about low-content posting in the discussion forums.  If you have something to say, say it.  Otherwise, please don't post.  Consider this a warning.

Yes, I am aware of the speed of light limit from relativity. Yet, the response just links to another thread that goes on a nonsequitor.

So how fast is the earth accelerating currently?

How fast is the earth traveling currently?

What is pushing the earth upwards?

Do you guys really believe this stuff? or are most of you guys just screwing around for the fun factor? mental gymnastics..

Let's say Bob stayed behind on Earth.  He measures the Earth's acceleration to be g = 9.8 m/s^2 -- the same way you might measure the accleration of the automobile going around the curve.  According to special relativity, Alice will measure a different acceleration -- in particular, she will measure it to be g/γ^3, where γ = 1/sqrt(1 - v^2/c^2).  In other words, we have the following differential equation for the velocity v of the Earth:  dv/dt = g/γ^3.

The solution to this equation is v = gt / sqrt( 1 + g^2t^2 / c^2 ), or v = g / sqrt(1/t^2 + g^2/c^2).  We are interested in the infinite time limit of v -- i.e. what happens to v as we wait an arbitrarily long time.  As t --> ∞, v --> g / sqrt(g^2/c^2) = c.

Quote

Let's say Bob stayed behind on Earth.  He measures the Earth's acceleration to be g = 9.8 m/s^2 -- the same way you might measure the accleration of the automobile going around the curve.  According to special relativity, Alice will measure a different acceleration -- in particular, she will measure it to be g/γ^3, where γ = 1/sqrt(1 - v^2/c^2).  In other words, we have the following differential equation for the velocity v of the Earth:  dv/dt = g/γ^3.

The solution to this equation is v = gt / sqrt( 1 + g^2t^2 / c^2 ), or v = g / sqrt(1/t^2 + g^2/c^2).  We are interested in the infinite time limit of v -- i.e. what happens to v as we wait an arbitrarily long time.  As t --> ∞, v --> g / sqrt(g^2/c^2) = c.

uh...dv/dt= g/(1/sqrt(1-v^2/c^2)^3) would be dv/dt= g*(1-v^2/c^2)^3/2....so there would be no speed limit by this equation since there is no asymptotic limit.  so that kinda destroys that argument...

You guys like landover baptist church? I'm a member. I like being on my knees and praising jesus.

Quote from: owned on July 06, 2011, 11:00:49 PM

Quote

Let's say Bob stayed behind on Earth.  He measures the Earth's acceleration to be g = 9.8 m/s^2 -- the same way you might measure the accleration of the automobile going around the curve.  According to special relativity, Alice will measure a different acceleration -- in particular, she will measure it to be g/γ^3, where γ = 1/sqrt(1 - v^2/c^2).  In other words, we have the following differential equation for the velocity v of the Earth:  dv/dt = g/γ^3.

The solution to this equation is v = gt / sqrt( 1 + g^2t^2 / c^2 ), or v = g / sqrt(1/t^2 + g^2/c^2).  We are interested in the infinite time limit of v -- i.e. what happens to v as we wait an arbitrarily long time.  As t --> ∞, v --> g / sqrt(g^2/c^2) = c.

uh...dv/dt= g/(1/sqrt(1-v^2/c^2)^3) would be dv/dt= g*(1-v^2/c^2)^3/2....so there would be no speed limit by this equation since there is no asymptotic limit.  so that kinda destroys that argument...

Are you saying that you're smarter than Einstein?

Quote from: owned on July 06, 2011, 10:29:45 PM

You guys like landover baptist church? I'm a member. I like being on my knees and praising jesus.

Oh my god... Quite a while ago I had an involvement with the guys over there on that forum...
They're even more ridiculous than Tom Bishop and friends. If you act out, they'll ban you and promise eternal damnation on you. It's retarded.

Yeah...you know landover is a joke right? The church doesn't actually exist. I Suspect the same thing here, most of them are faking.

Except Einstein wasn't a math expert, he was a theoretical physicist.  But yes, his math was obviously correct.

Quote from: John Davis on July 08, 2011, 02:30:32 PM

Except Einstein wasn't a math expert, he was a theoretical physicist.  But yes, his math was obviously correct.

Can you be an expert in theoretical physics without being an expert in math?