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50

The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons). You can do the calculation: If you're 10 light-years away from ...


25

Although I agree with all three of the above answers let me present a slightly different perspective on the problem. It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light ...


14

This is not paradoxical and it is not necessary for any physical phenomenon to a priori have to obey any particular law. Some phenomena do have to obey inverse-square laws (such as, particularly, the light intensity from a point source) but they are relatively limited (more on them below). Even worse, gravity and electricity don't even follow this in ...


12

The only stars you can reliably see are ones that are spewing enough photons at your eyeballs to appear stable. Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough. So your question is basically embroiled in observer bias; it ...


9

It's simpler than you (probably) think. In your example of defining speed: this is a change of position $s$ in a time $t$. The units of distance are metres and the units of time are seconds, so the units of velocity are metres per second. So far so good. Now consider acceleration: this is a change of velocity $v$ in a time $t$, so the units of acceleration ...


9

Allow me to channel something akin to the anthropic principle here. You can only see the stars that have a lot of photons reaching your eye. If a star were so far away that photons were reaching your eyes only occasionally then the star would be too dim for you to see in in the first place. Even if you could see the photons, the star would appear to ...


7

There are numerous distance indicators used for within the galaxy. The most common way is by using intrinsic magnitude. By knowing how bright an object would be if we were close, we can determine how far away it is by how dim it is. There are many types of stars where we have a rough idea of how bright they should be due to characteristics of the star: ...


6

Length and distance are not vector quantities (they are scalar quantities), but position and displacement are vector quantities (at least according to common terminological conventions). Here is how all of these are defined. Note that I am restricting the discussion here to vectors in three-dimensional Euclidean space $\mathbb R^3$. Every point in ...


6

A star radiates in all directions. You would still see the star regardless of the number of steps you take to any side, just not the same photons. A laser radiates in only one direction (or in a very small cone). If you took a large enough step to the side (larger than the angular size of the emitted beam) so as to exit this cone, then you would no longer ...


5

This really depend on all sorts of details, but for the idealized situation, the following data will suffice: focal lengths for iPhone cameras. Call that $F$. detector size for iPhone cameras. Call the width of the detector $d$. In addition, call the distance between the camera and the two points $L$. The camera's field of view ($FOV$) (or angle of view ...


5

As has been said, this is probably a very subjective question/answer. Not only that, but the composition of galaxies, and even regions within a galaxy, varies a great deal. Then there is the question of what constitutes as being part of the galaxy as opposed to perhaps a small orbiting dwarf galaxy. The answer you got from the Quora seems to be pretty ...


5

Absolutely not. Even though a laser beam is very narrow, it does spread out, and by the time it reaches the moon it will be a couple of hundred meters wide. The moon is not very reflective (with an albedo it only reflects 1/8th of the light falling on it). Thet reflected light is spread out further by the roughness of the lunar surface so that, by the time ...


5

You have re-discovered the Arrow Paradox from Zeno of Elea (5th century BC). Physics doesn't deal with that. There is obviously movement, and all we do is trying to resemble and predict it with mathematical models. That is how we face the question of understanding nature. You may find a better answer among philosophers than physicists for that question. ...


5

If you assume it is a lamp that is infinitely far away, the lamp size will also have to be infinitely large :) What I think you want to do is let the actual light come from infinity, but place the lamp 1AU (=~150,000,000 km) away, with a radius of 695,500 km (actual values). If this is not possible or impractical in your software for some reason, just ...


4

That's a big question, because there are loads of ways used to measure cosmic distances. have a look at the wikipedia article on the cosmic distance ladder for the gory details. The idea of the ladder is to start with nearby objects like stars. We can measure their distance using a method called parallax. Back in the 90s a satellite called Hipparcos used ...


4

Yes. More specifically, if $d$ is the distance between the planets in their rest frame, then in the astronaut's frame the distance between the planets will be $\frac{d}{\gamma}$ so the travel time as measured from his frame will be \begin{align} t_\mathrm{astro} = \frac{d/\gamma}{v} = \frac{1}{\gamma}\frac{d}{v} \end{align} Notice that the quantity $d/v$ ...


4

As always, a spacetime diagram is crucial for insight. However, this can be dispensed with without one. What am I missing here? The event that Moe's wristwatch reads $1\mu s$ has coordinates, in Joe's frame, of $(2.29\mu s, 618m)$. In other words, according to Joe, Moe's wristwatch reads $1\mu s$ when Joe's clock reads $2.29\mu s$. According to Joe, ...


3

Super novae were known a long time ago. But they were not understood as a the death throes of a star. In spite of the apparent immutability of the heavens, Chinese astronomers were aware that new stars could appear. In 185 AD, they were the first to observe and write about a supernova, now known as the SN 185. The brightest stellar event in ...


3

Looking at this mathematically is one way to do it, but you could also look at it in a more intuitive manner. As you correctly pointed out, the velocity of something ($v$) is the distance travelled ($s$) divided by the time taken ($t$). Acceleration is defined to be how fast your velocity is changing. So in terms of units, you would ask how many metres ...


3

Sound pressure level (SPL) in dB is defined relative to a pressure $p_{ref}=20\mu Pa$ $$L_p=20\log_{10}\left(\frac{p_{rms}}{p_{ref}}\right )$$ 75dB corresponds to acoustic pressure of 0.11 Pa, you can use this online calculator to easily check other SPLs. Acoustic velocity is proportional to acoustic pressure through acoustic impedance $Z=\rho c$ where ...


3

According to the generalized uncertainty principle, the Planck length is, in principle, within a factor of order unity, the shortest measurable length – and no improvement in measurement instruments could change that. (Wikipedia) $$1 \text{ Planck length} = 1.61619926 \times {10}^{-35} m$$


2

One neat trick for middle ranges requires a dynamically bound system whose components have measurable proper motions. There are a reasonable number of globular clusters that qualify. If you project those motions across the sky, they will appear to come together (to some approximation) in two places (one forward and one backward), and the directions of those ...


2

These physical phenomena (gravity, Coulomb force) are forces caused by an object you can consider pointlike. That is, for the inverse square law to hold, the object emits the force uniformly in all directions from one point. That means that at any distance (call it R) from the object, you'll feel the same force as you would anywhere over the surface of a ...


2

As far as we can tell (up to energy scales we've measured so far), spacetime is a nice and smooth manifold. It might happen that the smoothness is approximate and spacetime is discrete at a much more microscopic scale, or it could turn out that spacetime is smooth all the way through. Short answer: We don't know. About the notion of energy quantization: ...


2

I'm not sure why you are using the displacement-time formula if you have the shape of the graph. The distance covered by the particle is given by the area under the graph from point 0. After t = 2s, the distance the particle would have covered is the area under the trapezium, that is: $$\dfrac12 (1+2)(20) = 30$$ The position of the particle becomes $8 m + ...


2

The most intuitive way of understanding acceleration is to understand it in terms of Taylor series expansions $$\sum_{n=0}^{\infty} \dfrac{f^{(n)}(u)}{n!} (t-u)^n$$ A good entry level discussion on how one applies the Taylor series expansion to the question of position-velocity-acceleration can be found in this short paper from this website attributed to ...


2

Yes, lots of people can. This is because there is nothing to calculate. Without a spec for the desired units of the answer it can be as simple as "12000 lunar orbits/day". If you do want to convert this to other units, then you have to provide a clear definition of this unit of length you are calling a "lunar orbit", of course. Without that, nobody can ...


2

A key result from SR that must be kept in mind when making these kinds of comparisons is that, when speaking of elapsed time, there is more than one elapsed time to consider. In a Newtonian universe, there is universal time so the elapsed time according to the astronaut equals the elapsed time according to mission control. In a SR universe, the elapsed ...


2

The cosmological distance to a faraway galaxy only depends on its cosmic redshift $z_\text{cos}$, i.e. the redshift caused by the expansion of the universe; it does not depend on the Doppler redshift $z_\text{dop}$ caused by the galaxy's motion within its local cluster. So we have $$ \frac{a_0}{a(t)} = 1 + z_\text{cos}. $$ Now, what we observe is the total ...



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