Perhaps, not the best question asked, but still. If massive objects distort space around themselves causing other objects to orbit, how come these orbiting objects do not fall into the "hole" with time?

forajoho

### Posted Answers

How come planets don't fall into stars as they orbit them? Einstein said a "space-time" curvature is caused because of the gigantic mass of the stars....why don't the planets orbiting these stars just fall right into them?

Does it have something to do with the same reason why electrons never fall into the nucleus that they revolve around? Why does this not happen either?

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physixguruJun17-08, 04:01 AM

It is exactly the same analogy to the electron and nucleus theory.

The electron does NOT orbit the nucleus. Just look at the lowest energy orbital : the s-orbital. It is a sphere around the nucleus. So , prior to any kind of measurement, the electron is basically everywhere around the nucleus.

Same goes for any other orbital (ofcourse the have another shape)

QM proves us that the kinetic energy is higher when being closer to the nucleus, but the potential energy is lower (more negative). The sum of these two really yields a stable equilibrium throughout all energy levels.

According to classical law of thermodynamics any charged particle moving in a curved path should loose its energy continuously and fall into the nucleus but this does not happened as two different force acting which are known as centripetal and centrifugal forces and this was explained by Neils Bohr.

Planets are traveling at the speed necessary to keep from falling deeper into the gravity well. To demonstrate, take a marble and spin it around a large bowl. As long as the marble has the right velocity, it won't fall into the bowl but will continue to revolve around the bowl.

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jimmypoopinsJun17-08, 04:32 AM

It is exactly the same analogy to the electron and nucleus theory.

The electron does NOT orbit the nucleus. Just look at the lowest energy orbital : the s-orbital. It is a sphere around the nucleus. So , prior to any kind of measurement, the electron is basically everywhere around the nucleus.

Same goes for any other orbital (ofcourse the have another shape)

isn't this not correct?

i'm pretty sure the electon orbits the nucleus in an orbital (and spherical) pattern.

the earth and other planets orbit the sun in a semi-circular pattern. i'm pretty sure you can learn the laws involved in some fairly basic classical physics related laws.

i'm a math major and haven't studied the equations in a couple of years but it shouldn't be too hard to figure out what exactly is going on.

Planets are traveling at the speed necessary to keep from falling deeper into the gravity well.

that, however, is correct, similar to how satellites orbit the earth.

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jablonsky27Jun17-08, 04:44 AM

isn't this not correct?

i'm pretty sure the electon orbits the nucleus in an orbital (and spherical) pattern.

the earth and other planets orbit the sun in a semi-circular pattern. i'm pretty sure you can learn the laws involved in some fairly basic classical physics related laws.

i'm a math major and haven't studied the equations in a couple of years but it shouldn't be too hard to figure out what exactly is going on.

no, electrons dont orbit the nucleus. as physixguru already said, only the s-orbital is spherical. other orbitals(p, d, f) have different shapes. p-orbital for example is 'dumbell' shaped. to derive these you need to have knowledge of quantum mechanics.

the planets revolve around the star in elliptical orbits. these can be derived from classical physics related laws.

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WallaceJun17-08, 04:57 AM

Electrons can be described as orbiting around the nucleus in simple models, such as the Bohr model, that do a reasonable job of fitting to the data. As jablonksy pointed out though, in modern quantum mechanics (that does an even better job) this is not how electron orbitals are described.

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jtbellJun17-08, 06:35 AM

How come planets don't fall into stars as they orbit them?

The Earth (for example) continuously falls towards the sun. It is also moving "sideways" with respect to the sun, so it continuously misses the Sun!

Imagine that you're suspended stationary far away from the sun, say at the distance of the Earth. If you simply drop an object from that location, it falls in a straight line and eventually hits the Sun. If you throw the object sideways, it doesn't fall in a straight line, but rather in a curved path. If you don't throw it very hard, it still hits the Sun, but not directly underneath you. If you throw it hard enough, it misses the "edge" of the Sun, whips around, comes back up in an elongated elliptical orbit, and whacks you in the back of the head! :eek: (assuming there's no "air resistance" from gases near the sun, of course)

Does it have something to do with the same reason why electrons never fall into the nucleus that they revolve around? Why does this not happen either?

We have an FAQ about this in the General Physics forum.

Why Donâ€™t Electrons Crash Into The Nucleus In Atoms? (http://www.physicsforums.com/showpost.php?p=862093&postcount=2)

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stevebd1Jun17-08, 07:02 AM

The centripetal acceleration of Earth equals the gravitational pull of Sun at this radius. Basically-

\frac{v^2}{r}=\frac{Gm}{r^2}

where v is the angular velocity of the Earth in metres/second, r is the Earths mean orbit radius in metres and m is the mass of the Sun in kg. Both quantites are expressed in m/s^2

The same equation can be used to explain why the moon doesn't fall into the Earth and why the ISS stays in orbit.

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jablonsky27Jun17-08, 07:05 AM

where v is the angular velocity of the Earth, r is the Earths mean orbit radius and m is the mass of the Sun.

isnt v the linear velocity?

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stevebd1Jun17-08, 07:07 AM

If it's rotating about a point, I'm under the impression it's angular velocity-

http://en.wikipedia.org/wiki/Angular_velocity

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jablonsky27Jun17-08, 07:15 AM

its rotating about a point alright, but in the equation you posted, v is the tangential velocity that the earth has at any given point in its orbit. thats what i meant when i said linear velocity.

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KurdtJun17-08, 07:21 AM

v is the tangental velocity. jimmyp, if you're interested in electron orbitals look up spherical harmonics which describe the shape of orbitals in modern QM for quantum numbers l and m. They are of course solutions to the TISE in spherical polar coords.

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stevebd1Jun17-08, 07:37 AM

I stand corrected, the proper term for v is tangental velocity but the quantities and equations still stand.

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stevebd1Jun17-08, 08:11 AM

Based on the above equation, it seems like the following would apply when looking at the elliptical orbit of planets-

Due to the unlikely nature of a perfectly circular orbit, Earth is sometimes closer to the Sun. Because of conservation of angular momentum, J = vmr where J is the angular momentum (in Nms), m is the mass of the Earth (in kg), r is the mean orbit radius (in m), as the Earth gets closer to the Sun, J & m don't change but because r reduces, v increases and as a result, Earth's centripetal acceleration overcomes the gravitational pull of the Sun at this reduced radius and it manages to overshoot it's mean orbit radius as it continues it's orbit around the Sun. Again, because of the conservation of angular momentum, at this increased radius, J & m don't change but the increased r means a reduced v and the centripetal acceleration is overcome by the Suns gravity, pulling the Earth back into a reduced orbit and the process starts again.

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physixguruJun18-08, 03:37 AM

'v' is the tangential velocity.NO two opinions about it.

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Alan McDougalJun18-08, 09:09 AM

Planets do not fall into stars are due to space bending near the star and the law of momentum.

Eventually most planetary orbits will decay and said planet descends into the star the star will never lets its children escape.

Of course before the planet falls into most stars like our sun, it will go red giant and consume the poor planet in an orgy of fiery death

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junglistJun20-08, 03:38 AM

Eventually most planetary orbits will decay and said planet descends into the star the star will never lets its children escape.

is this true? ive read about the moon's orbital radius increasing by 4mm/year so is it likely that this would also be the case for planets orbiting stars?

does the mass of the sun decrease as it converts matter to energy? would this effect be greater than that of phenomenon that cause orbital decay (drag, etc)?

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WallaceJun20-08, 03:42 AM

is this true? ive read about the moon's orbital radius increasing by 4mm/year so is it likely that this would also be the case for planets orbiting stars?

In general planetary orbits do not decay. In principle the orbit of a planet around a star gives of gravitational waves which will slowly decay the orbit but the key there is slowly. Any star would be looong burnt out before this effect would even be measurable.

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junglistJun20-08, 03:49 AM

In general planetary orbits do not decay. In principle the orbit of a planet around a star gives of gravitational waves which will slowly decay the orbit but the key there is slowly. Any star would be looong burnt out before this effect would even be measurable.

are these gravitational waves due to surface irregularities?

if so, i understand you. :)

is this effect more or less significant than drag as the planet passes through dust and other material thats floating around?

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KurdtJun20-08, 05:19 AM

does the mass of the sun decrease as it converts matter to energy? would this effect be greater than that of phenomenon that cause orbital decay (drag, etc)?

The mass of the sun does decrease, but it is only a tiny fraction of the suns total mass and so this effect is not really measurable either.

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junglistJun20-08, 06:46 AM

Now I'm worried that I'll end up asking questions on this forum for eternity...

...thanks to wallace and kurdt with your informative and hasty answers.

this place is like crack for the inquisitive mind. :D

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WallaceJun20-08, 06:54 AM

are these gravitational waves due to surface irregularities?

if so, i understand you. :)

is this effect more or less significant than drag as the planet passes through dust and other material thats floating around?

Muuuch less significant than pretty much anything you care to mention. We would never have a hope of measuring this effect even if we had billions of years in which to watch the accumulated effect.

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Alan McDougalJun20-08, 07:38 PM

Guys,

While the sun mass does decrease over billions of years, its volume increases.

The sun will, therefore, in about 5 billion years become a red giant and engulf the earth.

So the original question "why does a planet not fall into the sun:? is a valid one

Regards

Alan

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jablonsky27Jun22-08, 11:55 PM

hi alan,

the OP says,

How come planets don't fall into stars as they orbit them? Einstein said a "space-time" curvature is caused because of the gigantic mass of the stars....why don't the planets orbiting these stars just fall right into them?

the poster definitely did not ask what you said. besides, if a star engulfs a planet, its not the same as the planet *falling* into the star.

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D HJun23-08, 12:45 AM

I am addressing several posts here.

The centripetal acceleration of Earth equals the gravitational pull of Sun at this radius. Basically-

\frac{v^2}{r}=\frac{Gm}{r^2}

That equation is valid for circular orbits of bodies with negligible mass about some massive body. A more general expression is the vis-viva equation,

v^2 = GM\left(\frac 2 r - \frac 1 a\right)

where M is the mass of the central mass, v is the magnitude of the velocity vector of the orbital body with respect to the central body, r is the distance between the two bodies, and a is the semi-major axis of the orbit.

ive read about the moon's orbital radius increasing by 4mm/year so is it likely that this would also be the case for planets orbiting stars?

You may have also read that the Moon is slowing the Earth's rotation rate. This means the Earth's angular momentum is decreasing. Angular momentum is a conserved quantity. The Moon's orbital angular momentum increases in lockstep with the decrease in the Earth's rotational angular momentum. The same mechanism is responsible for both phenomena: the tides that the Moon raises in the Earth's oceans (and in the Earth itself. The Earth is not quite solid, so the Earth itself has a tidal bulge.)

I will answer the second part of your question next.

does the mass of the sun decrease as it converts matter to energy? would this effect be greater than that of phenomenon that cause orbital decay (drag, etc)?

There is very little drag from the solar wind at the Earth's orbital distance from the Sun. Gravitational waves are also very, very weak. The Sun is losing mass as it converts mass into energy. It also loses mass to a lesser extent because some mass escapes the Sun in the form of the solar wind. The Earth's orbit does expand because the Sun is losing mass and also because a tidal transfer of solar rotational angular momentum to Earth orbital angular momentum. The former effect is tiny; the latter affect is extremely tiny. While the gravitational force is an inverse square law, tidal effects are an inverse cube law. The tidal transfer of angular momentum from the Sun to the Earth is a very small effect.

are these gravitational waves due to surface irregularities?

They result from the mere act of orbiting. They are also very small. The gravity waves resulting from the Earth orbiting about the Sun have all of 313 watts of power, or four 75 watt light bulbs. The wikipedia article on gravitational waves is a good start. It is, however, just a start. Like everything on wiki, beware. That silly stuff about a mountain on a neutron star is just that -- silly, and maybe even misleading.

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junglistJun23-08, 01:23 AM

Thanks D H.

I had a listen to an old audiotape from Feynman doing a lecture on the 'theory of gravitation' at Caltech or somesuch in the 60's. It helped me understand the further complexities of planetary orbits as he explained some of their properties resulting from proximity of the other planets.

Its only when you understand just how extremely well the planets orbital paths are in equilibrium that you can realise that almost any constant/consistent perturbations we can think of will form part of that equilibrium equation (or will be naturally damped to insignificance).

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