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> It collapses into a small and dense white dwarf, which cools over trillions of years. Its intense gravity can rip apart any surviving planets that stray too c

by ericol 8y ago

> It collapses into a small and dense white dwarf, which cools over trillions of years. Its intense gravity can rip apart any surviving planets that stray too close How is the gravity from the white dwarf different from the previous state (red giants) of the star? Doen't they have basically the same mass? (Actually less, as they burn)

jvanderbot 8y ago
"stray" implies they get closer, eventually coming close enough for tidal effects to get nasty. Everything I understand agrees with a white dwarf not ripping apart planets whose orbits are not distirbed from their previous ones. IANA astronomer/astrophysicist.
dexen 8y ago
To expand on @jvanderbot excellent reply, there's "Roche Limit" [1] - region of space close to massive bodies where gravitation gradient will break up the orbiting body. Given that white dwarfs remain in the cool-down phase for trillions of years, any orbiting body's orbit will slowly decay and eventually it ends up inside the Roche limit. [1] https://en.wikipedia.org/wiki/Roche_limit https://en.wikipedia.org/wiki/Roche_limit
Rooster61 8y ago
It's a question of density and tidal forces. The red giant has that mass spread out across a much larger space, so the object that is "straying" has a much more uniform gradient of gravitational attraction across its mass than the white dwarf. An object can only be close to a very small part of the red giant at a time, while an object getting close to a white dwarf is subjected to much more of the gravity well all at once. That makes the gravitational gradient much more pronounced across the object, which means tidal forces are much stronger and more likely to rip the object apart. A good way to think about it is by looking at where the Roche limit exists for a mass of that size. Since the red giant is much larger but has roughly the same mass as the white dwarf, the Roche limit probably exists somewhere behind the surface, while the same orbit would extend far past the surface of the white dwarf. Let's say we have an orbital body around a giant star whose mass dictates that its Roche limit is 0.1AU, and then lets say that the star's diameter is 0.3AU. The object could never get close enough to the star to break apart. Now, that same star collapses while losing a negligible amount of mass, and now is a white dwarf with a diameter of 0.01AU. The object could now approach close enough that it gets pulled apart by tidal forces since its unimpeded by the stars surface all the way down to 0.1AU. https://en.wikipedia.org/wiki/Roche_limit https://en.wikipedia.org/wiki/Roche_limit EDIT: After reading, I realized that I unnecessarily introduced a non-astronomical term in gravitational gradient when I could have just said tidal force. It's the same thing. My apologies to any sticklers out there :)
blacksmith_tb 8y ago
I am also curious about the cool-off period of white dwarf stars being measured in 'trillions of years' - given that the universe is only 13.8 billion years old, I assume we are just modeling that?
- mlindner 8y ago
  Yes. These will eventually cool into so-called "black dwarfs" that emit very little light. https://en.wikipedia.org/wiki/Black_dwarf https://en.wikipedia.org/wiki/Black_dwarf They cool so slowly because they are so dense it's very hard for energy to escape.
- xref 8y ago
  Right, and the model predicts that no “black dwarfs” exist in the universe yet because it is too young for a white dwarf to have ever cooled off.

by ericol 8y ago

jvanderbot 8y ago

"stray" implies they get closer, eventually coming close enough for tidal effects to get nasty. Everything I understand agrees with a white dwarf not ripping apart planets whose orbits are not distirbed from their previous ones. IANA astronomer/astrophysicist.

dexen 8y ago

To expand on @jvanderbot excellent reply, there's "Roche Limit" [1] - region of space close to massive bodies where gravitation gradient will break up the orbiting body. Given that white dwarfs remain in the cool-down phase for trillions of years, any orbiting body's orbit will slowly decay and eventually it ends up inside the Roche limit. [1] https://en.wikipedia.org/wiki/Roche_limit https://en.wikipedia.org/wiki/Roche_limit

Rooster61 8y ago

It's a question of density and tidal forces. The red giant has that mass spread out across a much larger space, so the object that is "straying" has a much more uniform gradient of gravitational attraction across its mass than the white dwarf. An object can only be close to a very small part of the red giant at a time, while an object getting close to a white dwarf is subjected to much more of the gravity well all at once. That makes the gravitational gradient much more pronounced across the object, which means tidal forces are much stronger and more likely to rip the object apart. A good way to think about it is by looking at where the Roche limit exists for a mass of that size. Since the red giant is much larger but has roughly the same mass as the white dwarf, the Roche limit probably exists somewhere behind the surface, while the same orbit would extend far past the surface of the white dwarf. Let's say we have an orbital body around a giant star whose mass dictates that its Roche limit is 0.1AU, and then lets say that the star's diameter is 0.3AU. The object could never get close enough to the star to break apart. Now, that same star collapses while losing a negligible amount of mass, and now is a white dwarf with a diameter of 0.01AU. The object could now approach close enough that it gets pulled apart by tidal forces since its unimpeded by the stars surface all the way down to 0.1AU. https://en.wikipedia.org/wiki/Roche_limit https://en.wikipedia.org/wiki/Roche_limit EDIT: After reading, I realized that I unnecessarily introduced a non-astronomical term in gravitational gradient when I could have just said tidal force. It's the same thing. My apologies to any sticklers out there :)

blacksmith_tb 8y ago

I am also curious about the cool-off period of white dwarf stars being measured in 'trillions of years' - given that the universe is only 13.8 billion years old, I assume we are just modeling that?

mlindner 8y ago
Yes. These will eventually cool into so-called "black dwarfs" that emit very little light. https://en.wikipedia.org/wiki/Black_dwarf https://en.wikipedia.org/wiki/Black_dwarf They cool so slowly because they are so dense it's very hard for energy to escape.
xref 8y ago
Right, and the model predicts that no “black dwarfs” exist in the universe yet because it is too young for a white dwarf to have ever cooled off.