# Sun and its gravity

1. Jul 2, 2014

### enquirealways

If size of the sun is increased million times ( trillion times or gazillion times), keeping its mass same and then the planets are redistributed outside of it.

Will the planets still revolve around it

Last edited: Jul 2, 2014
2. Jul 2, 2014

### Chronos

The law of gravity doesn't really care much about mass density, just mass. Instead, let's pretend the earth was at a distance of 1,000,000 AU from the sun. Using Kepler's 3rd law, it would theoretically complete an orbit around the sun in one billion years. Of course at such a distance the earth would be about 15.8 light years from the sun. Further complicating matters, there are about 50 other stars within that same distance.

3. Jul 2, 2014

### Bandersnatch

Changing the size to within the radius of an orbit wouldn't affect the orbiting body.
It goes the other way around too: compressing the Sun into a black hole wouldn't affect the orbits of any of the planets.
Look up Shell Theorem for geometrical explanation of why it is so.

If the Sun were blown up to anything higher than ~200 times its current radius, its outer parts would pass the Earth, no longer affecting it gravitationally(again, shell theorem explains this effect), and the force of gravity towards the centre of the orbit would no longer remain the same, causing the Earth to spiral outwards.

Distibuting the Sun's mass over an infinite volume(i.e., "gazillion times" larger) would be equivalent to nullifying any gravitational influence it may have had on Earth, as the mass remaining within the Earth's orbital radius would be negligible, and Earth would just float away rectilinearly at its current orbital speed.

4. Jul 3, 2014

### enquirealways

How can sun affect earth at such a large distance?

It's as if a pole is connecting the two. But it's not.

If we divide the entire distance of 15.8 ly into small parts of day 0.1 light year each, it would take a lot of time for the effect/any change to propogate from one part to another.

So, how is this synchnoricity maintained at such a large distance as far as effect of gravity is concerned.

I think, if experiments are done using rotating magnets, they would never attract iron fillings beyond a certain distance, even if a huge amount of time is involved. I know that nature of these two forces is diffrent but may be some similarity is there.

5. Jul 3, 2014

### enquirealways

When the outer parts of sun pass the earth, why wouldnt they affect the earth.

The parts that pass would have mass, so they should attract the earth in some other/ opposite direction.

6. Jul 3, 2014

### Chalnoth

Gravity is an infinite-range force. It falls off with distance slowly enough that its effect never disappears.

7. Jul 3, 2014

### Bandersnatch

Because the pull from the part that has passed the orbit near the planet(and which is now pulling outwards) is exactly cancelled by the pull of the mass that has passed the orbit on the oposite side(which still pulls in the same direction as always - towards the centre).

It's like if you could get to the centre of the Earth, there'd be immense pressure from all that rocks, but no gravity - everything cancels out, as there is the same amount of mass on each side pulling you.

Here, we've got an in-between case, where you're not at the centre of the mass, so there's still gravity, but some of the mass is already outside and cancels out.

It's not easy to dress in words. Look up Shell Theorem for a technical explanation.

8. Jul 3, 2014

### Chronos

In a spherical object of uniform density, gravity peaks at the surface and tends toward zero as you approach the core. Under shell theorem, the gravity of the 'shell' cancels. Only mass deeper than a particular shell is of any consequence. Earth, however, is not uniformly dense, nor is the sun. Density generally increases as you approach the core. In 1939 Bullen calculated the maximum gravity on earth is located about 2920 km below the surface (re: http://rsnz.natlib.govt.nz/volume/rsnz_69/rsnz_69_02_002150.pdf). Something similar can be expected for the sun.

9. Jul 4, 2014

### Bandersnatch

Chronos, do note that while true, your post might be confusing the matter, as this case is different.
Here, we have a constant distance from the centre, so the force of gravity will fall as soon as solar matter starts to cross the orbit, regardless of density distribution(as long as it's spherically symmetric, that is).

10. Jul 4, 2014

### Chronos

Density distribution is relevant here. Assuming density increases as you approach the core [which is a very reasonable assumption], a gravitational gradient is the natural consequence. There is a differential equation that relates distance to density under such circumstances. The mass of layers external to the particular shell you happen to occupy at any give distance from the center cancel out. If inner layers are more dense than the outer layers, you get a net gravitational attraction towards the center.

11. Jul 4, 2014

### Bandersnatch

That makes no sense for a constant distance R, does it? All the mass within the orbit will(or should, to my understaning) contribute exactly the same amount of force regardless of distribution, so any mass removed beyond the radius will lower the force.

We're not getting closer to the centre. Is there something I'm missing or are we not on the same page here?

12. Jul 4, 2014

### Chronos

What I'm saying is you can have more mass inside the shell than outside. That produces a gravitational gradient.

13. Jul 4, 2014

### Chalnoth

Even if you have no density gradient, you'll still have a gravitational gradient. If I remember correctly, the strength of the force of gravity in the case of constant density is proportional to the distance from the center (and, of course, drops as $1/r^2$ once you go beyond the surface of the spherical body).

14. Jul 4, 2014

### Bandersnatch

Once again, this is not the case discussed. We're not descending below the Sun's envelope, we've got the outer shells expanding past a specific radius(while keeping mass constant).
The factor from the force's of gravity inverse square dependence on distance can't compensate for the loss of mass, as we're never getting closer. Any mass passing the orbit will inevitably reduce the gravitational pull on the planet.

It's the case when in $a=GM/R^2$ only M changes, and it always decreases. No density distribution can increase a when M falls.