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Like a ball falls on the earth due to gravity, if suppose we are able to place a ball close enough to the sun. Will the ball fall into the sun due to its gravity or will it be pushed away due to its energy?

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    $\begingroup$ What energy would push the ball away? $\endgroup$ – HDE 226868 Jan 24 '16 at 21:11
  • $\begingroup$ there is no steam :) so may be the nuclear energy released on the surface of sun due to hydrogen helium fusion $\endgroup$ – Anupam Rekha Jan 24 '16 at 21:15
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    $\begingroup$ Are you talking about radiation pressure from photons? $\endgroup$ – HDE 226868 Jan 24 '16 at 21:17
  • $\begingroup$ I think radiation pressure pushes very small particles only. $\endgroup$ – Anupam Rekha Jan 24 '16 at 21:24
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    $\begingroup$ That's true; I'm just not sure what other forces you think would be at work to push it away. $\endgroup$ – HDE 226868 Jan 24 '16 at 21:25
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Short answer is yes. The only way radiation pressure would be a factor was if the ball was in orbit around the Sun, this would mean that the slight pressure it feels would cause it to drift away slowly (assuming its made of the most heat resistant material on Earth!). But in your question you said "placed" near the sun not in orbit around it, so the strength of the sun's gravity would overcome all other forces and pull it in.

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    $\begingroup$ Haha oops, thanks @Joan.bdm. I got so carried away with my explanation I forgot the exact wording of the question. $\endgroup$ – Dean Jan 25 '16 at 11:38
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The answer depends on the size and mass of the ball. It also depends on its ability to reflect light (albedo $A$), but let's forget that for a moment.

Pressure vs. gravity

Solar pressure decreases with $R^2$ (the inverse square law). At Earth, which is located at a distance of $1\,\mathrm{AU}$ from the Sun, we receive an irradiance $S_0 = 1361\,\mathrm{W}\,\mathrm{m}^{-2}$. Since the momentum of a photon of energy $E$ is $p = E/c$, the pressure at a distance $R$ from the Sun is $$ P = \frac{S_0}{c(R/\mathrm{AU})^2}. $$ If the ball's radius is $r$, this pressure will exert a force $$ F_\gamma = \pi r^2 P = \frac{\pi r^2 S_0}{c (R/\mathrm{AU})^2}\qquad(\mathrm{away\,from\,the\,Sun}). $$ Meanwhile, if the ball's mass is $m$, the gravitational force exerted by the Sun on the ball is $$ F_g = \frac{G M_\odot m}{R^2},\qquad(\mathrm{toward\,the\,Sun}) $$ where $G$ is the gravitational constant and $M_\odot$ is the mass of the Sun.

Threshold for falling

The threshold for the ball falling into the Sun is found by equating the two oppositely directed forces: $$ \frac{\pi r^2 S_0}{c (R/\mathrm{AU})^2} = \frac{G M_\odot m}{R^2}. $$ The first thing to notice is that $R$ cancels out; the reason being that flux density and gravity both follow the inverse square law. Second, re-arranging terms we see that the ball will fall if its mass per area is greater than this threshold (if I have calculated correctly): $$ \frac{m}{r^2} \gtrsim \frac{\pi S_0}{c G M_\odot } \mathrm{AU}^2 \simeq 2.4\times10^{-4}\,\mathrm{g}\,\mathrm{cm}^{-2}. $$ Now you can plug in your favorite numbers. You will find that for most macroscopic objects, such as a football, a rock, and even a sand grain, it will fall. On the other hand, small "balls" such as dust grains and atoms, will tend to be pushed away from the Sun. An example of a macroscopic object that won't fall is a solar sail which seeks to maximize area per mass.

For a given density, say $\rho = 2.5\,\mathrm{g}\,\mathrm{cm}^{-3}$ which is characteristic of rocky materials, you can also calculate the maximum size before it falls toward the Sun: $$ r_\mathrm{max} = \frac{3}{4\pi\rho} \,(2.4\times10^{-4}\,\mathrm{g}\,\mathrm{cm}^{-2}) \sim 0.1\mathrm{-}1\,\mu\mathrm{m}. $$

Albedo

The above calculations hold of all the radiation is absorbed by the object. If a fraction of it is reflected, the radiation will transfer more momentum to the object, and for a prefectly reflecting object, the radiation force will be roughly twice the amount above (the exact factor depends on the geometry of the object).

Extinction curves

It should be noted, though, that treating small particles as rigid spheres with a geometric cross section becomes imprecise when their size is comparable to wavelength of the light; rather, their absorption/scattering cross section should be treated quantum mechanically. In practice, extinction curves — i.e. the cross section as a function of the wavelength of the light for a given dust particle size distribution — are measured observationally by comparing the light from unobscured star with similar stars behind dust clouds, and subsequently fitted with various functional forms.

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  • $\begingroup$ Very well calculated answer @pela but I have one issue. If you calculate the mass per area ratio to be 2.4*10^-4 (which I calculate to be roughly the density of carbon dioxide in 3D) then how could you say that a solar sail would not fall? Surely it would be made from more dense material than a gas like carbon dioxide? $\endgroup$ – Dean Jan 25 '16 at 14:51
  • $\begingroup$ @Dean: I'm not sure what you mean with "the density of carbon dioxide in 3D". Density is mass per volume, while the calculated factor is mass per area. That is, the goal is to make the sail as thin as possible. To be honest, I don't know much about solar sails, but from the Wiki article's section on materials I see that with nanotube mesh weaves, you can reach 1e-5 g/cm² (this is called the "sail loading factor"), while lithium sails — the lightest metal — go down to 2e-6 g/cm². This is 20 kg for a 1 km² sail. Pretty damn light! $\endgroup$ – pela Jan 25 '16 at 19:28
  • $\begingroup$ Also, launching a space craft from Earth with a solar sail would put it in an initial orbit around the Sun, in which case, as you describe in your answer, it wouldn't fall toward the Sun even without the sail. $\endgroup$ – pela Jan 26 '16 at 9:54
  • $\begingroup$ I added an extra dimension to your mass to area calculation to be able to compare it to a list of materials with known density. This would in theory just occupy the extra dimension facing away from the sun so not to affect your calculations. In regards to what you said about launching the sail from earth, that's what I way trying to get at when asking you why the solar sail would not fall. I thought it was important to highlight he fact that there's a huge difference in energy required to keep a free falling particle away from the sun than it is for one in orbit. $\endgroup$ – Dean Jan 26 '16 at 11:10
  • $\begingroup$ nevertheless your solar sail idea does look promising if they can get the densities theh claim! $\endgroup$ – Dean Jan 26 '16 at 11:11

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