Traveling Speed of light..

If someone could travel at the speed of, say, 0.99c ... can the time to reach
the moon be calculated by t = distance / velocity ??? or would space between
the moon and earth contract and some other formula be needed.

Can anyone explain this please.


------------------------

I would think traveling at the speed of light to the moon would result in a
very hard, deadly landing.

BTJustice

"Mike" wrote in message
s.com...


If someone could travel at the speed of, say, 0.99c ... can the time to
reach
the moon be calculated by t = distance / velocity ??? or would space
between
the moon and earth contract and some other formula be needed.

Can anyone explain this please.


------------------------

I would think traveling at the speed of light to the moon would result in a
very hard, deadly landing.

BTJustice

"Mike" wrote in message
s.com...


If someone could travel at the speed of, say, 0.99c ... can the time to
reach
the moon be calculated by t = distance / velocity ??? or would space
between
the moon and earth contract and some other formula be needed.

Can anyone explain this please.


------------------------

"Mike" wrote in message
s.com...


If someone could travel at the speed of, say, 0.99c ... can the time to
reach
the moon be calculated by t = distance / velocity ??? or would space
between
the moon and earth contract and some other formula be needed.

Can anyone explain this please.


------------------------


The key to the answer is the fact that you, as the observer, defined the
"speed" to be "0.99c". Since that is the observed speed - plain ol'
Newtonian calculations would suffice to determine the time between the
"leaving earth" event and the "arriving at the moon" event. The actual
values you come up with will depend on whether you are measuring things
while traveling or measuring it while watching. Both will be correct.

-ralph

"Mike" wrote in message
s.com...


If someone could travel at the speed of, say, 0.99c ... can the time to
reach
the moon be calculated by t = distance / velocity ??? or would space
between
the moon and earth contract and some other formula be needed.

Can anyone explain this please.


------------------------


The key to the answer is the fact that you, as the observer, defined the
"speed" to be "0.99c". Since that is the observed speed - plain ol'
Newtonian calculations would suffice to determine the time between the
"leaving earth" event and the "arriving at the moon" event. The actual
values you come up with will depend on whether you are measuring things
while traveling or measuring it while watching. Both will be correct.

-ralph

On Mon, 06 Oct 2003 06:52:08 GMT, "rdc" wrote:

The key to the answer is the fact that you, as the observer, defined the
"speed" to be "0.99c". Since that is the observed speed - plain ol'
Newtonian calculations would suffice to determine the time between the
"leaving earth" event and the "arriving at the moon" event. The actual
values you come up with will depend on whether you are measuring things
while traveling or measuring it while watching. Both will be correct.

Let's say I am measuring while traveling. I am traveling 0.99c. How long would
it take to reach the moon according to my measurements?

On Mon, 06 Oct 2003 06:52:08 GMT, "rdc" wrote:

The key to the answer is the fact that you, as the observer, defined the
"speed" to be "0.99c". Since that is the observed speed - plain ol'
Newtonian calculations would suffice to determine the time between the
"leaving earth" event and the "arriving at the moon" event. The actual
values you come up with will depend on whether you are measuring things
while traveling or measuring it while watching. Both will be correct.

Let's say I am measuring while traveling. I am traveling 0.99c. How long would
it take to reach the moon according to my measurements?

Mike wrote:

Let's say I am measuring while traveling. I am traveling 0.99c. How long would
it take to reach the moon according to my measurements?

At 0.99c you would experience time-dilation according to the Lorentz
factor gamma = (1 - 0.99^2)^(-1/2) = 7.1. So while your trip would
take about 3.84*10^8 m / 2.97*10^8 m/s = 1.3 seconds as measured by
an observer stationary WRT the earth-moon system, a stopwatch aboard
your ship would show the trip to take only 1.3 s / 7.1 = 0.18 second.
Another way to describe the situation would be to say that from your
point of view the earth-moon distance would seem shorter by a factor
of 7.1, yielding the same reduction of the travel time you experience.

--
Odysseus

Mike wrote:

Let's say I am measuring while traveling. I am traveling 0.99c. How long would
it take to reach the moon according to my measurements?

At 0.99c you would experience time-dilation according to the Lorentz
factor gamma = (1 - 0.99^2)^(-1/2) = 7.1. So while your trip would
take about 3.84*10^8 m / 2.97*10^8 m/s = 1.3 seconds as measured by
an observer stationary WRT the earth-moon system, a stopwatch aboard
your ship would show the trip to take only 1.3 s / 7.1 = 0.18 second.
Another way to describe the situation would be to say that from your
point of view the earth-moon distance would seem shorter by a factor
of 7.1, yielding the same reduction of the travel time you experience.

--
Odysseus

Mike Accelerators now prove a space ship traveling at 94% of "C" has
dead astronauts aboard. Bert