Time and distance at constant C: A sieries of questions for Umbran or other physicists.

[Scott DeWar]

I think it was down to 10 mm to the right of the decimal, but I will have to re work it tomorrow.

 

[Umbran]

Well, we have 35 km/sec. Let's look at what the stuff to the right of the decimal would be...

35.abcdef

a = 100s of meters
b = 10s of meters
c = meters
d = tenths of meters (decimeters)
e = hundredths of meters (centimeters)
f = thousandths of meters (millimeters)


I gave gamma = 1.0000678. I need at least 6 significant digits in the speed to get 1.00006, so I need to know the speed down to decimeters per second.
To get 1.0000678 I'd need to know down to the millimeter per second.

 

[freyar]

Another way to work out the spread of values in the "gamma factor" (ie, how many digits your answer is good to) would be to calculate based on the range of input speeds. So if we know the speed is between 34.5 and 35.5 km/s, we know gamma is between 1.0000000066.... and 1.0000000070... . What's interesting is that some of the calculator programs I regularly use know enough about significant digits to just say that the answer is "1" unless you really force them to give more digits.

 

[Scott DeWar]

I will get to this when I am done with all of the game threads I am in, be right back professors!

 

[Scott DeWar]

Ok, I am going to attempt some math in this post. Please, no comments until I post -END-

this is take two on my incorrect math.


[1 / (1 - 35035^2 / 300,000^2)^ -2] =
1 / (1,227,451,225 / 90,000,000,000) ^-2 =
1/ (.01363834694)^2 =
1 / 0.1167833333 =
8.56286570597

Meteor 2015 TB 145 Velocity =

78,830 MPH = [78,830 * 1.6] = 126,128 KPH; 126,128/3600 = 35,035 Km/S or 35.035555556 [repeat the 5]

Time dilation formula

1/sqrt(1-v^2/c^2)

[1/(1-(35.035555556^2/300,000,000^2)^-2]=
[1/(1-1227.4901531/9*10^16)^ -2]=

I will continue later, but take note that the numbers I came up with for meters/second has a repeating 5 at the mm level. I also added the 000 to the velocity of C.

-old text-
So, as an exorcise of thought, I am going to say that a ship has taken a mining crew to dig for . . . we will say platinum . . . . yup, that sounds good, to the meteor 2015 TB 145 and it was sent to be timed with the rock's passing by earth at 1.3 lunar distances on 31 Halloween.

So, given the above formula on time dilation is the final number I have 8.56286570597, what is the final answer's unit of measure? Seconds per day of decrease?

IE: 100 days on the rock and you will be 856.2865 seconds younger then those people on earth?

 

[MarkB]

One thing to bear in mind is that the article only provides a simple figure for the asteroid's speed, not an actual velocity. Is it moving at 78,000 mph relative to the solar system in general, or is that its relative speed of approach to Earth? Bear in mind that Earth is travelling around the sun at 70,000 mph.

 

[Scott DeWar]

I am going to take the "relative to earth" choice here just for the sake of a math test. this is all it is. just a test for me.

 

[Staffan]

78,830 MPH = [78,830 * 1.6] = 126,128 KPH; 126,128/3600 = 35,035 Km/S or 35.035555556 [repeat the 5]

[stuff]

I will continue later, but take note that the numbers I came up with for meters/second has a repeating 5 at the mm level. I also added the 000 to the velocity of C.

Except the "source" number isn't the speed in meters per second. It's the speed in MPH, which has four (or possibly five, but zeroes at the end usually doesn't count) significant digits. More properly, you'd say that the speed is 35,040 m/s. Those trailing fives are "false" accuracy.

Hmm, come to think of it, you actually only have two significant digits, because you use 1.6 as the conversion factor between mph and km/h. Plug in 1.609 instead and you get up to four again. That gets you to 35,230 m/s... which actually works out well as an illustration of why significant digits are a thing.

 

[Scott DeWar]

**BANG BANG BANG** That is the sound of my head hitting the table top.

 

[Umbran]

Except the "source" number isn't the speed in meters per second. It's the speed in MPH, which has four (or possibly five, but zeroes at the end usually doesn't count) significant digits.

I am going to be picky, because this is science, and we are talking about precision.

The article says it has a speed of "35 kilometers per second (78,000 mph)". These are, actually, both given with two significant digits. In the absence of other graphic elements telling us which are significant (a bar above or below the rightmost significant zero is common), you generally assume that all three trailing zeroes are not significant. The other indicator is that, typically, shifts in units of measure do not introduce or remove significant figures - so the km/sec and mph figures should have the same number of significant figures.

https://en.wikipedia.org/wiki/Significant_figures
Concise rules

• All non-zero digits are significant
• Zeros between non-zero digits are significant.
• Leading zeros are never significant.
• In a number with a decimal point, trailing zeros, those to the right of the first non-zero digit, are significant.
• In a number without a decimal point, trailing zeros may or may not be significant. More information through additional graphical symbols or explicit information on errors is needed to clarify the significance of trailing zeros.

By the way, if you are going to do a calculation in metric units, and they *hand* you a metric measurement (km/sec), it doesn't make a whole lot of sense to use the Imperial measure, and then convert it to metric. 35 km/s = 35,000 m/s, and off you go!