# Tag Info

57

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 ...

40

It's all about the angles made by the object when light from it enters the eye. Consider this crude doodle of an eye looking at two identically sized trees. The light entering the eye from the nearer tree makes a broader angle at the eye, and the further tree makes a sharper angle. The brain interprets this as the further tree seeming to be smaller. ...

36

Suppose you're walking past a nearby tree: As you pass the tree the angle of the line joining you and the tree changes. From your perspective it looks as if you are standing still and the tree has moved backwards. However the Moon is so far away that, as you walk, the angle of the line joining you to the Moon doesn't change by any significant amount: ...

33

The moon is a lot further away than the horizon or nearby buildings, trees, hills or mountains. As you pass these nearby objects, from your perspective you see them apparently move backwards relative to your motion. This is a matter of simple trigonometry. This is because, the angle between your bicycle and (for example) a tree 100 metres away changes ...

30

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 ...

30

45 degree angle for a projectile gives you the maximum distance in a vacuum, but air resistance, as pointed out, changes that a little. With air resistance slowing the ball, you need to to throw a tick under 45 degrees for maximum distance. Also Also, since you throw from above the shoulder, not from the ground, the ball is usually thrown a foot or ...

25

It is because light travels in more-or-less straight rays. Let's assume for simplicity that your eye is like a pinhole camera; it has a pinhole in front and a screen at the back. Then an image forms by the light rays that pass through the pinhole. (from https://commons.wikimedia.org/wiki/File:Pinhole-camera.png) Consider two points on an object, such as ...

24

There are a variety of methods used to measure distance, each one building on the one before and forming a cosmic distance ladder. The first, which is actually only usable inside the solar system, is basic Radar and LIDAR. LIDAR is really only used to measure distance to the moon. This is done by flashing a bright laser through a big telescope (such as ...

23

I am a baseball fan (and a physicst), and your coach is misleading you a little. First, in the absence of air resistance, a 45-degree launch will get the ball there with miminum energy expenditure. But not, as the other answers suggest, minimum time. And time matters. A lot. :-) Your coach should also be telling you to plan your longer throws such ...

21

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 ...

18

It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. ...

13

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 ...

12

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 ...

11

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 blink....

10

Why does my sports coach tell me that when I'm fielding I should throw the baseball 'flat' to get the maximum distance? I thought from physics that you get the most distance from throwing at a 45 degree angle? The second question first, this is true if you are a robot throwing a ball on the Moon (no atmosphere) that releases the ball at the same speed ...

9

An angle is just a measure of the relationship between the arc length of a segment of a circle and the radius of said circle. An object (let's say a pencil) of length $30~\text{cm}$ will always be that length, but the distance the pencil is from your eye can change. The definition of an angle is the arc length, $s$, over the radius, $r$ Even though the ...

8

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: ...

7

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 distances using a method called parallax. Back in the 90s a satellite called Hipparcos used ...

7

Apparent size is not measured as an ordinary size, in meters. It is actually an angle, so it is measured in degrees or radians. See this picture: The left blog is the eye. Look like as the object moves further, the angle becomes smaller. That is what is called perspective. Sometimes people try to compare apparent size and real size, but that makes no ...

7

The intensity goes as $r^{-2}$, not $r^{-3}$. One meter is 100 times further away than 1 cm, thus 1/10,000th the intensity. Side Note 1 How often are you within 10 cm (~3.94 inches) of a halogen bulb? The article states that 15 minutes under constant exposure at 10 cm can elicit erythema. However, who sits under a halogen bulb 10 cm away for 15 minutes ...

6

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. ...

6

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 ...

6

Triangulation. The Earth is not stationary, it moves in a 150 million km (1 AU) radius orbit around the Sun. If you measure the apparent position of a star at different points in that orbit a near enough object will appear to be displaced by a measurable amount, this displacement is called parallax, which is typically measured across a 1 AU baseline. A ...

6

This is one of my favorite questions in astronomy. It's extremely clever really. The Sun emits a certain colour of light that can be analysed. This colour is obviously completely white other than certain frequencies. For example, an object emitting a turquoise light would be emitting every colour frequency other than parts of red perhaps (there are other ...

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 three-...

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 ...

6

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 $\... 6 The behaviour you are describing is a consequence of the virial theorem. Without going into the gory details this tells us that if some interacting system of many objects has an average total potential energy of$<U>$then its average total kinetic energy$<T>$is related to$<U>\$ by: $$<T> = \tfrac{1}{2} <U>$$ The proof of ...

5

The field line picture known from school might be helpful with that: The surface area of the surrounding sphere (and not it's volume) determines the density of the lines sourced by a point charge, corresponding to the field strength.

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