1

Cells, Microbes & Viruses / Re: Does the fittest always survive?

« on: 06/10/2021 16:00:34 »

"Survival of the Fittest", means the "best fit" for the present environment.  So, the newer strain could have just been better adapted to a wider range of conditions, if the petri dish was an environment that suited the original strain, there would no reason for the new strain to have an advantage over it since both could flourish under those conditions.
However, if you changed the conditions so that was more to the liking of one and less to the other, then one strain would dominate.

2

Physics, Astronomy & Cosmology / Re: If the earth stopped orbiting the sun?

« on: 25/09/2021 22:28:54 »

This works out to be about 64 days.

Thanks Janus. Have you taken into account that the earth is starting from a stand still it will take some time to get going?

Yes.  It, like anything else, will start "falling" immediately, and at the same acceleration.  However, that is not to say that the mass of the Earth is completely irrelevant.  It can have an effect on impact time, just in the opposite manner that you seemed to imply.  Increasing the mass of the Earth would decrease the time to impact.  The reason for this is that not only is the Earth falling towards the Sun, but the Sun would fall towards the Earth due to Earth's gravity.  Since the Sun is some 333,000 times more massive than the Earth, the amount it would "fall" towards the Earth is pretty insignificant, and we can safely ignore it.
However, if the Earth had a mass that was a much larger fraction of the Sun's, you would have to account for it. 

3

Physics, Astronomy & Cosmology / Re: If the earth stopped orbiting the sun?

« on: 25/09/2021 22:08:21 »

    Anyway, other articles consider the journey toward the sun to be over in about 65 days but that seems to be because they reckon the tidal forces from suns gravity would rip the planet apart before we even reach the outer edges of the sun.

Tidal forces wouldn't effect the answer significantly.  The Roche limit( as measured in multiples of the primary's radius) is dependent on the relative densities of the bodies.
The Earth is some 4 times denser than than the Sun.  If we treat the Earth as a fluid body, this puts the Roche limit at 1.55 Sun radii from the center of the Sun our  just over 1/2 Sun radius above its surface.
Treated as a rigid body, the Roche limit ends up at ~0.8 the Sun's radius, or under the Sun's surface.  The actual Roche limit for the Earth will likely fall somewhere between.
But even if we put it at 1/2 the Sun's radius away, the Earth is just ~350,000 km from the surface, and will be moving close to its final speed of over 616 km/sec.  In other words, it is only around 10 min from surface impact.

4

Physics, Astronomy & Cosmology / Re: If the earth stopped orbiting the sun?

« on: 25/09/2021 21:34:42 »

If the earth stopped orbiting the sun. This is a hypothetical question and may have an interesting answer. Can anyone do the math and reveal the answer as to the arrival velocity and time period for the earth as it makes contact with the sun. The earth has great mass so I'm sure it will be slow getting started on its journey to the sun. I haven't calculated the speed or time for this event as I have no understanding of the equations involved. I hope someone can give the answer?

The quick and dirty method of computing the fall time is to assume the Earth's orbit is an extremely eccentric ellipse with the Earth's present distance being at perihelion.  This puts the Semi major axis at half the perihelion distance, and you can just calculate half the period of such an orbit.  This works out to be about 64 days.
A more accurate answer which will give you the time to reach the Sun's surface can be found by using the equation at the bottom of this article:
https://en.wikipedia.org/wiki/Equations_for_a_falling_body
Just use the Sun's radius for x and the Earth's orbital radius for r.
Impact speed can be arrived at by taking the difference between the gravitational potential energy at the suns surface and that at Earth orbital distance, and solving for the velocity needed for the Earth to have a equal kinetic energy.
I get ~616.64 km/sec
The fall time from Jupiter distance is ~ 2 years.

5

Physics, Astronomy & Cosmology / Re: How does Time Dilation happen on on the Atomic Scale?

« on: 22/08/2021 18:38:53 »

Hi @Janus
   Your reply looks good.   Just check these numbers:

and fast for 1.8 month.
Since the respective rates are 1/3 and 3, he would see it tick off 3.6 mo while ticking slow, and then another 3.6 mo while ticking fast,

I think that  1.8   should be edited to  1.2,      everything else seems OK   3 x 1.2 = 3.6   etc.

Yep, thanks for catching that. I went back and edited it.

6

Physics, Astronomy & Cosmology / Re: How does Time Dilation happen on on the Atomic Scale?

« on: 21/08/2021 22:59:03 »

So eventually...

Why is Time Dilating then?

If the Earth Twin eats 100 mangoes in 1 Earth Year

The Space Twin would have eaten 100 mangoes in it's 1 Space Year Journey.

If both video recordings are Compared, there would be 100 mangoes in each.

So like the Space Twin's video would be Shorter?
6months?

So then if that SpaceTwin video is played at 0.5x speed(half) then it would resemble Earth Twin's video length?

But then that would show SpaceTwin in Slow Motion Right?

Jeez!
🥴
Mind goes Mangoes understanding all this...wat do u intelligent folks eat?

ps - does anyone feel Utube songs are sloww n play them at 1.25x
Dis has nothin 2 do wit da pond n everythin 2 do wit a DiscoToad!
🤭
(Solly)

Assume The space twin travels at 0.8 c both out and back.
The round trip takes 1 Earth year, meaning he got 0.4 ly from earth when he turned back ( as measured from the Earth)
Our earth observer watching him will see him eat mangoes 1/3 as fact as he himself does (a part of this is due to the time delay caused by the increasing distance between them.)*
After 1/2 year, the space twin reverse direction back towards Earth. however since this occurs when he is 0.4 light years from Earth, the Earth twin doesn't see this occur for another 0.4 years or 0.9 years after the space twin left. During this whole time, he sees the Space twin eating mangoes 1/3 as fast. Since he has eaten 90 mangoes by this time, he will have seen the Space twin eat 30 mangoes. Once he sees the space twin reverse direction, he will see him eating mangoes 3 times faster than himself.**
But by the time he actually sees the space twin eat 30 mangoes and reverse direction, the space twin has already traveled most of the way back to him, and arrives back at Earth just 1/10 of a year later (having been gone a total of 1 year) during this time the Earth time eats another 10 Mangoes, while watching the space twin eat 30 mangoes.
Thus he eats a total of 100 while the space twin eats just 60.

If we look at things from the space twin's perspective, we get this:
The space twin recedes at 0.8c from the Earth, during which time he sees the Earth twin eat mangoes 1/3 as fast. After he has eaten 30 mangoes he reverses direction to head back to Earth ( the fact that he has eaten 30 mangoes by the time he turns around is something that he and the Earth twin must agree on.)
during this time he sees the Earth twin eat 10 Mangoes.
Now here comes a major difference between him and the Earth twin. The Earth twin had to wait to see the effects of the Space twin turning around because that event took place 4/10 of a light year away. But for the space twin, his reversing direction happens right there, so he has no delay. He immediately sees The Earth twin start eating mangoes 3 times faster than himself.  He eats the same number of mangoes for the return trip as on the outbound trip, so he eats 30 more mangoes, for a total of 60. During the return leg he sees the Earth twin eat three times as much as himself, or 90 mangoes, plus the 10 he'd seen him eat during the outbound leg gives 100 for a total.  This is the same end result the Earth twin arrived at.
* what he sees is the combination of this increasing lag and time dilation. If he were to factor out the light delay, he would conclude that the space twin was eating mangoes at  rate 6/10 his rate.
** again a combination of a ever decreasing time lag and time dilation, factoring out the light delay has him concluding that the space twin once again is eating mangoes 6/10 as fast during the return leg.

7

Physics, Astronomy & Cosmology / Re: What Propels Light ?

« on: 05/08/2021 21:49:16 »

Hi again.

Did this one ever get answered?

Thank ewe Halcy. So what is the speed of light in my room ? I'm not in a vacuum !

     Light in air is 1.0003 times slower than light in a vacuum, which slows it all the way down from 299,792,458 meters per second to 299,702,547 meters per second.

[taken from forbes.com]

     I hate air.  I know everyone else likes the stuff but I'm deeply suspicious of it.  Air is a mixture of gases and so it is deeply sinister stuff.  None the less, it's considered a (single) medium by most people.
    On a similar note, light never travels at the speed of light in a vacuum - because there are no vacuums found in the universe.  We believe that there is interstellar gas, neutrinos (CNB) and other photons (CMB) flying around in every part of space. 
     I think most of us accept that matter seems to slow down light but it's also quite likely that any form of energy does.  For example, other photons may interact with and hinder the passage of photons.   NOTE:  I've not seen any hard and reliable references to back that up.  It just seems likely.  I'd be grateful for any good references that support or refute that.

Light does not interact with neutrinos, and no, other photons have no effect on its speed.  As far as interstellar gas goes, we are talking about a few atoms per cubic meter.  Since the spacing between the particles is so much larger than the wavelength of the photons, they only interact by scattering the light which only occurs if a photon strikes a particle.
But, because the particles are so few and spread out, this occurs rarely. Put another way, the mean free path of a photon( The average distance it travels before interacting with one of those particles) in interstellar space is pretty long.

8

Physics, Astronomy & Cosmology / Re: What Propels Light ?

« on: 05/08/2021 15:19:08 »

how does light on the right hand side of a flame know to go one way and light on the other side know to go the other way ?

A sensible sheep will not try to walk through a fire.

But photons are fearless, and will travel in every possible direction.

Some photons do make it through a candle flame from the far side .
- Outside the candle, air is cool and clear, so light from the near side has an unobstructed path to your eye
- But a photon from the far side has to pass through a hot maelstrom of unstable reaction products, which are more likely to absorb it, and then reradiate the energy from a closer point
- Some photons do make it all the way through the candle flame - but they are hard to see against the brightness of the nearer side

PS: Is there such a thing as a sensible sheep?

Thank ewe Evan. I guess this is similar to the way that light takes a gazillion years to finally leave the sun, but (confewesed sheep)...I thought as soon as light interacts with something it gets absorbed or converted into something else. So how does it survive all those collisions in the sun before it finally makes a run for it ?


sensible sheep ? guffaw chortle  !!

The photons that leave the surface are not the same photons that started at the core.  If a photon interacts with an atom, it adds energy to that atom. That atom will end up shedding all or part of that energy (depending on the type of interaction) by emitting another photon. That photon won't necessarily be emitted in the same direction as the original photon.
So while "a" photon eventually leaves the surface, it removed by a chain of many, many, of these types of reactions from the photon that started at the core.

9

Question of the Week / Re: QotW - 21.07.06 - Why is it cold up mountains?

« on: 06/07/2021 15:46:57 »

Air temperature in the Troposphere( the lowest atmospheric layer), is driven by being heated by the ground, which was, in turn heated by the Sun.   Higher altitudes, being further from the ground are heated less by this effect. Now while it is true that air near a mountain is close to the ground of the Mountain, that air is always being mixed with air that comes from the much larger regions where air at that altitude is much further from the ground.
As far as heat rising goes. Yes, warmer air, in general, will rise in colder air, But in doing so, it has to climb against gravity, and it can't do this for free, it has to give up some of its heat energy in doing so. So even warm air rising from the surface will not be as warm at altitude as it was near sea level.

10

Chemistry / Re: Can elements contain multiple isotopes within one sample?

« on: 24/06/2021 14:51:51 »

Any random sample of an element can contain a number of isotopes.  For example, Uranium in its natural form is made up of both U-238 and U-235.  Since only U-235 is fissile, "raw" Uranium has to be enriched (The percentage of U-235 increased) in order for it to be useful for power plants or nuclear weapons(Nuclear weapons need a much higher enrichment than reactor fuel needs)

11

Physics, Astronomy & Cosmology / Re: Another Relativity paradox

« on: 08/06/2021 16:31:31 »

Hi all.

   Does anyone have some knowledge or insight about this "paradox" in the theory of Relativity?

   Imagine a submarine underwater.
Initially:
The submarine is at rest relative to the fluid and has adjusted it's tanks so that it has equal density with the fluid and remains at a depth of 100 metres.  (No thrust required from the engines, it just has neutral buoyancy).
 
Subsequently
    The submarine accelerates rapidly to reach relativistic speeds (let's say 0.9 c) relative to the fluid and then sustains a constant velocity.  This is intended to be a horizontal motion, the fins, bow planes etc. were not set to drive the submarine up or down.

Finally
     As is usual for these sorts of paradoxes,  we have two observers in two different frames of reference.

The submarine commander is at rest inside the submarine.  She should observe length contraction for the fluid in her rest frame and a corresponding increase in density of the fluid.  The submarine retains it's rest characteristics, including density in her frame.

A mermaid is at rest on the ocean floor.  In her rest frame, the density of the fluid has not changed,  however the submarine has undergone length contraction in her frame and it's density has increased.

Question
   Will the submarine rise or sink due to buoyancy?

Background info:   You may like to read the Wikipedia article about Supplee's paradox.
https://en.wikipedia.org/wiki/Supplee%27s_paradox
There is also a similar discussion about a Helium balloon moving through air on another forum.  (I'm not sure I should put links to another forum).

   I do not know the answer.  I can see references to articles in that Wikipedia entry but they seem to demand some application of General Relativity and a complete re-write of the Archimedian principle.  I was wondering if there is a resolution based only on Special Relativity - but I'll take any insight or discussion I can get.

As with most apparent "paradoxes" in Relativity, this one is likely due to only focusing on one aspect of the theory and ignoring the rest.
For example, here, you say that the Water appears more dense due to length contraction from the Sub's frame of reference, and thus should be more buoyant. However, what causes buoyancy?  First you need gravity ( this in of itself means you need GR to properly address the issue).   This in turn causes a difference in the water pressure between the top of bottom of the sub.  It is this difference in pressure ( water pushing up on the bottom more than it pushes down on the top) that produces the net upwards force.
Now you need to consider the effect that water flowing past the sub has on the static pressure of the water. Bernoulli's principle states that a moving fluid's static pressure decreases with fluid velocity.   That suggests that if the sub starts with neutral buoyancy at rest, it will not remain so once it begins to move relative to the water.  Thus if it wants to stay at constant depth, it will need to adjust its bow planes to a new angle.* 
But now, because of length contraction, the Mermaid will measure that angle as being different, changing the effect they produce...

The point is that just considering how length contraction acts on the water or sub as seen by sub or mermaid doesn't give you the full picture as to what is going on.  SR has been proven to be entirely self-consistent. Which means that no such thought experiments can never produce a real paradox, as long as you properly apply it.**

*and this doesn't even go into how Relativity would factor in to calculating the resultant effect.
** Now this, in of itself, does not prove the correctness of SR, just that thought experiment alone is not enough to disprove it.  You need a real life experiment or observation that doesn't match what Relativity predicts for that.

12

Physics, Astronomy & Cosmology / Re: Would altering the moon's weight change its orbit?

« on: 25/05/2021 16:31:58 »

Why is the Moon moving away from the Earth.  Shouldn't it be getting closer, pulled towards the Earth by gravity?

  Earth isn't a perfect sphere, it's oblate  (more like an egg than a good round ball).   The moon is actually causing most of this deformity in the earth (by pulling on it with the force of gravity).   The thing is the earth is spinning on it's own axis quite fast, so these bulges get pulled slightly forward of the line between the earth and the moon.   The moon goes in a prograde orbit around the earth (it moves around the earth in the same direction as the earth spins).  The bulges on the earths surface are then ever so slightly pulling the moon forward a bit,  while at the same time slowing the rotation of the earth.   By giving the moon a bit more of a pull in the direction tangential to it's orbit the moon's speed is increasing.  As that speed increases, the moon drifts out a bit further to an orbit with a greater radius.
    There's some info here:
https://en.wikipedia.org/wiki/Tidal_acceleration

A few corrections here:
While the Earth is oblate (a greater radius at the equator), an egg is prolate( a greater polar radius).
The Earth is oblate due to its spin, giving the equator a radius 21 km greater than the polar radius.
The tidal effect of the Moon, while it does distort this shape some, is far from the dominant effect, as it is in the range of 1 meter.
A complicating factor with the Earth-Moon system is that there are two tides: The Earth tide, and the ocean tide. And these are not in phase with each other.  The ocean tides are effected by the way the water interacts between basins and coastlines.
So ocean tidal bulges tend to be drug along by the Earth's rotation more than Earth tides are.   

13

Physics, Astronomy & Cosmology / Re: Would altering the moon's weight change its orbit?

« on: 24/05/2021 18:39:02 »

Where did all this new angular momentum come from?

No new angular momentum. That's part of the discussion, like the skater's arms or a coalsescing galaxy: change the mass distribution and the rotational speed changes.

We're mining the moon, not mining the Earth. g (the gravitational acceleration at lunar radius) is unaffected by the mining of the moon.

True, but it's another way of changing the weight of the moon, and the one (the OP said "any change...") that would alter R.

The equation for orbital period is T = 2pi sqrt(a^3/G(m1+m2)
(In many practical cases m2 is small enough compared to m2, that you really don't need to include it)
Thus, if we imagine magically transferring some percentage of the Moon's mass to the Earth, The sum m1+m2 doesn't change, so at first glance neither does the orbital period.
However, The barycenter will shift closer to the Earth, which means the Moon would have to travel around a longer path in the same amount of time if it were to maintain a circular orbit.  Ergo, it would need a higher orbital velocity.
But since we magically transferred just the mass, and changed nothing else, our new less massive Moon, will have too low an orbital speed to hold a circular orbit, and will transition to an elliptical one, with a perigee lower than its present distance. And since this in turn decreases "a", the semi-major axis of the orbit, you'll see a decrease in the length of the orbital period.

That being said, there is no way to magically transfer mass from the Moon to Earth, and how it is done would have a large effect.    An efficient means would be by magnetic rail gun mounted on the surface of the Moon. (With one at each pole they will always be pointing in pretty much the right direction.)
You fire your mass opposite to the Moon's orbital velocity (~1 km/sec) to drop it into an Earth atmosphere grazing orbit (it would likely need to be equipped with some type of retro-rocket package)
This would also act like a rocket engine adding orbital velocity to the Moon with each launch.
But, we are talking about pretty insignificant changes. 
The other thing that the OP should be made aware of is that orbits aren't that fragile.  People sometimes get the idea that the slightest deviation will cause an orbiting object to either fall to the planet or fly off into space.  This is far from the case.  In order to get the Moon to fly off you would need to increase its orbital velocity by a good fraction of its present orbital speed of 1 km/sec, and decrease it by even more to put it into an Earth intersecting trajectory.
In reality, anything we do on the Moon would not have any significant effect on it. ( for example, even if you were to transfer the entire population to of the Earth to the Moon, you'd only increase the mass of the Moon by roughly 1/100,000,000,000.)

14

New Theories / Re: What exactly is gravity?

« on: 23/05/2021 15:43:28 »

Let's compare this scenario to waves created by a ship moving in a pond. Scenario presented on your animations shows a situation, where a wave is propagating perpendicuralry to the motion of a boat in a pond, which moves together with the boat in relation to a stationary observer - it propagates perpendicularly in the frame of boat and diagonally in the frame of stationary observer. However wave propagating perpendicuralry to a moving boat in a stationary pond will behave differently - it will propagate perpendicularly to the moving boat in the frame of stationary bystander and diagonally in the frame of the boat.

If the motion of light is not affected by the motion of it's source in the frame of a stationary observer, then it's the second option (with stationary pond), which should be applied to this scenario.

With the pond, the waves travel relative to the water which acts like a medium.  There is no medium for light traveling in a vacuum.  So the waves made by a boat in the pond is not a valid analogy.
But neither does light behave exactly like a bullet fired from a gun according to Newtonian physics.  A bullet fired from a moving train has its speed increased relative to the ground as measured from the ground, light does not. While the observer would measure it traveling at an angle along a diagonal, they would also measure it as moving at c along the diagonal.
You can also examine the train scenario by reversing the roles. Put the source on the ground and have the observer on the "moving" train.   The train observer will measure the light as traveling at a diagonal relative to the train.  In other words, it doesn't matter which one, the source or observer, you assume is moving, the observer observes exactly the same thing, and he can't tell which one is "really" moving. The very idea of who is "really moving" and who is "really stationary" is meaningless.

15

General Science / Re: Binary star: What color is the sky on a planet?

« on: 19/05/2021 16:37:13 »

Blue.  The color of the sky is due to how it scatters light. 
A "red" dwarf doesn't just emit red light but in the the whole visible spectrum. It's spectrum is just more centered towards the red. It would appear to be light orange to our eyes.
A blue dwarf, also doesn't emit just Blue light, but has its spectrum shifted to the blue. It would appear to be very light blue. (though if you were to look directly at either star, they would overload your retina and they would appear white in the sky.)
The mixed light from each star would result in a more even spectrum ( the red star making up for the Blue and vice-versa).
This evens out the spectrum reaching the planet.
The blue part of the spectrum would scatter out giving you a blue sky during the day.
At sunset or sunrise, things could depend on the visual separation between the two stars, and which one sets first.
If the red star sets later, you could see a redder sunset, and if the Blue star sets later a paler sunset. 
But even here, the atmosphere plays a huge role.  More dust/smoke etc. in either situation will result in a more red sunset/sunrise.

16

Physics, Astronomy & Cosmology / Re: How to calculate the orbital speed of the ISS around the centre of the earth?

« on: 03/05/2021 18:20:12 »

If you want to work it out via orbital mechanics, the equation is v^2 = u(2/r-1/a), also known as the vis-viva equation.
u is the gravitational parameter or GM (with M being the mass of the planet).  For the Earth, it is 3.987e14 m^3/s^2
r is the present distance of the orbiting object from the  center of the Earth. 
a is the semi-major axis of the orbit, or average orbital distance from the center of the Earth (this allows you to account for elliptical, non-circular orbits.)
For a circular orbit, r=a, and the equation reduces to V^2 = u/r
With the numbers given in the OP, and assuming a circular orbit, you get 7.652 km/sec.
However, the actual radius of the Earth (at the Equator) is 6378 km.  making this slight correction gives 7.664 km/sec, which when rounded up to two significant digits, gives 7.7 km/sec
The Wiki article on the ISS lists its orbital speed as 7.66 km/sec, which either of the two above answers would round out to.

Also, using the smaller, more accurate radius of 6378 km, and using chiralSPO's method, we get 7.64 km/sec, much closer to to the given value.

17

Physics, Astronomy & Cosmology / Re: Why does acceleration become so difficult at high velocities?

« on: 29/04/2021 14:49:51 »

What is it that it that resists acceleration at higher velocities. If in relativity you are only aware within a frame of reference, why is high velocity acceleration so difficult?

In Relativity, nothing, as far as the accelerating frame is concerned, inhibits further acceleration. So, if you were accelerating at 1g.   You would measure your velocity as changing by 9.8 m/s every sec.
But, this doesn't mean that if you, for example, started at rest with respect to the Earth, that after doing this for 1 year by your clock, you would measure yourself as moving at a bit over 1c relative to the Earth (it ends up being closer to 0.79c)

This is a result of how velocities add up in Relativity.
Imagine you in this accelerating ship, and after accelerating to 0.1c relative to Earth, you drop off a space buoy ( this buoy maintains the same velocity relative to the Earth you had when you released it). You continue to accelerate until you are moving 0.1 c relative to this buoy. This will take the same amount of time by your clock as it did to get to 0.1c relative to the Earth.
If you now measure your speed relative to the Earth, you will find that is not 0.1c+0.1c = 0.2c,  but (0.1c+0.1c)(1+0.1c(0.1c)/c^2) = ~1.98c
If you now drop off another buoy, accelerate until you are moving at 0.1c relative to it, you will measure your velocity as being ~0.292c relative to the Earth.
Keep doing this and the consecutive velocities relative to earth will be:  0.381c, 0.463c, 0.538c, 0.605c, 0.665c...
Each time, the relative velocity to Earth increases by a smaller amount, even though, you noticed nothing changing about your acceleration.
It is not that "something resists your acceleration", it is that the very nature of time and space determine your measurements.

Relativity is a model for time and space and their measurement, It has nothing to do with outside influences inhibiting acceleration.

18

Physics, Astronomy & Cosmology / Re: Why do we not build a light speed ship?

« on: 27/04/2021 17:36:24 »

The KE needed to get 1 kg up to just 90% of c is ~ 1e17 joules

But, that it only part of the problem. Since this would have to be achieved via rocket engine, you need to use the Rocket Equation ( as per Kryptid's post),  and it gets even more difficult.

It isn't enough to work out just how much energy it would take to get the payload up to velocity, but since you are carrying your own fuel, during the earlier parts of the acceleration you have to accelerate the fuel you are going to need for later parts of the acceleration.
As you begin to get nearer to c, relativistic effects become more prevalent and you have to account for them as well.
The Rocket equation that takes this into account gives:
M0/M1 = e^(atanh(v/c) c/Ve)
M0 is the starting  mass of the ship (ship + fuel),
M1 is the final mass ( ship)
v is the final velocity
Ve is the exhaust velocity of your rocket.

An important factor to take away is that the amount of fuel needed to reach a given velocity is heavily dependent on the exhaust velocity of the rocket.

The best exhaust velocity for any engine presently under development is for the DS4G*, with a Ve of 193,000 m/s (compared to the ~4500 m/s for a contemporary chemical rocket)

Using that in the equation above, it works out that for your ship to reach even 4% of c, it would have to use roughly 1/2 the mass of Jupiter in fuel per kilogram of ship mass. To push that up to 5% of c would increase that to 3000 times the mass of the Sun. ( For practical purposes, since v is such a small fraction of c, you don't actually need to use the Relativistic form of the Rocket equation to get a reasonably accurate answer in these examples.)

But let's say that we could push the exhaust speed up to 10% of c. Then it would take over 90 million kg of fuel per kg of ship to reach just 95% of c and 3 times as much to get up to 96% of c. ( and if you want to slow back down afterwards, this jumps to 8.1 x 10 ^15 kg and 7.29 x 10^16 kg respectively.  )

As far as times go:
If you could maintain a 1g acceleration ( as measured in the ship), then:

in 1 yrs ship time and 1.19 yrs Earth time you would achieve 0.77c and have traveled 0.56 ly.
in 2 yrs ship time and 3.75 yrs Earth time you would achieve 0.97c and have traveled 2.90 ly (not yet even as far as the nearest star system to our own)
in 3 yrs ship time, and 11.8 yrs Earth time you would achieve 0.9969c and have traveled 9.7 light years ( There are 7 star systems within this distance; most of the stars are red dwarfs .)

If you want to stop at your destination, you would need to accelerate for half the trip and decelerate for the second half.
Doing so would get you to, for example, Barnard's star ( a red dwarf with ~14% the mass of the Sun) in ~4.25
 yrs ship time and and bit over 9 years Earth time.



* and while the VASIMR has a high exhaust velocity, it doesn't produce a high thrust, and wouldn't be able to achieve an acceleration of 1g.    This, unfortunately, is a common feature with rockets,  You tend to trade high exhaust velocity (thus efficiency) for  thrust.   ION engines for example are very efficient, but have very low thrust, thus low acceleration.  The type of energy sources needed to get around this just aren't achievable/practical at this time.

19

Physics, Astronomy & Cosmology / Re: Did Cassini travel through time?

« on: 21/04/2021 16:59:54 »

If an object's velocity effects it's perceived time, did Cassini move into the future by traveling upwards of 44/km for 13 years? Does time dilation work that way under the speed of light? Was there a clock on Cassini's computer? If so did they have to recalibrate as it traveled?

At that speed, over 13 years, time dilation would only amount to a difference of roughly 4.4 seconds.
That just means that over the trip, Cassini's clock would have measured that much less time than the Earth clock.*

And there really wasn't any reason for Cassini to keep track of its own time over the trip.  It would collect data, and send it back to Earth.  It made no difference that by  Cassini's clock it collected that data at 12:00:04, and according to Earth it was collected at 12:00:00
This is different than, for example, the GPS system, which does require a tight synchronization between satellites and ground in order to maintain accurate results.


*In realitiy it is much more complicated than that. The Earth itself orbits the Sun at some 30 km/sec. Cassini didn't have a constant speed relative to the Sun over its whole trajectory. You also have to account for gravitational time dilation due to differing distance from the Sun...  But even after all this is taken into account, the total time difference accumulation would be too small to matter.

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Physics, Astronomy & Cosmology / Re: How do you measure the distance to a star or galaxy?

« on: 01/04/2021 15:30:35 »

There are a number of methods used, depending the distance range being measured:
https://en.wikipedia.org/wiki/Cosmic_distance_ladder