3 added 584 characters in body edited Oct 24 '15 at 21:11 Rob Jeffries 63.8k55 gold badges137137 silver badges209209 bronze badges Let's assume the white dwarf has a mass of $$0.6 M_{\odot}$$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use Kepler's third law, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($$8.1\times 10^{8}$$ m). The tidal forces this close to a white dwarf are very large. The Roche limit for the total tidal disintegration of a satellite, in synchronous rotation, held together only by its own gravity is roughly $$d = 1.44 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$ where $$R_{WD}$$ is the radius of the white dwarf (similar to the radius of the Earth), $$\rho_{WD}$$ is the average density of the white dwarf (a few times $$10^{9}$$ kg/m$$^3$$) and $$\rho_p$$ the density of the planet (let's assume 5000 kg/m$$^3$$). Thus $$d \simeq 6 \times 10^{8}$$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating. I guess it will be an observational selection effect that such objects will be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were closer they would have already disintegrated and wouldn't be seen! EDIT: On reading the paper - the authors claim that these objects are not tidally disintegrating. In fact they argue that this must be debris from a rocky planet precisely because the density must be large enough to avoid tidal disintegration according to the formula above. However I find the whole discussion rather incoherent. They specifically talk about "disintegrating planetesimals" (note the tense) which are being evaporated in a Parker-type wind due to heating by the radiation from the white dwarf. I cannot see where they explain then how the planetesimals disintegrate. Let's assume the white dwarf has a mass of $$0.6 M_{\odot}$$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use Kepler's third law, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($$8.1\times 10^{8}$$ m). The tidal forces this close to a white dwarf are very large. The Roche limit for the total tidal disintegration of a satellite, in synchronous rotation, held together only by its own gravity is roughly $$d = 1.44 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$ where $$R_{WD}$$ is the radius of the white dwarf (similar to the radius of the Earth), $$\rho_{WD}$$ is the average density of the white dwarf (a few times $$10^{9}$$ kg/m$$^3$$) and $$\rho_p$$ the density of the planet (let's assume 5000 kg/m$$^3$$). Thus $$d \simeq 6 \times 10^{8}$$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating. I guess it will be an observational selection effect that such objects will be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were closer they would have already disintegrated and wouldn't be seen! Let's assume the white dwarf has a mass of $$0.6 M_{\odot}$$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use Kepler's third law, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($$8.1\times 10^{8}$$ m). The tidal forces this close to a white dwarf are very large. The Roche limit for the total tidal disintegration of a satellite, in synchronous rotation, held together only by its own gravity is roughly $$d = 1.44 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$ where $$R_{WD}$$ is the radius of the white dwarf (similar to the radius of the Earth), $$\rho_{WD}$$ is the average density of the white dwarf (a few times $$10^{9}$$ kg/m$$^3$$) and $$\rho_p$$ the density of the planet (let's assume 5000 kg/m$$^3$$). Thus $$d \simeq 6 \times 10^{8}$$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating. I guess it will be an observational selection effect that such objects will be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were closer they would have already disintegrated and wouldn't be seen! EDIT: On reading the paper - the authors claim that these objects are not tidally disintegrating. In fact they argue that this must be debris from a rocky planet precisely because the density must be large enough to avoid tidal disintegration according to the formula above. However I find the whole discussion rather incoherent. They specifically talk about "disintegrating planetesimals" (note the tense) which are being evaporated in a Parker-type wind due to heating by the radiation from the white dwarf. I cannot see where they explain then how the planetesimals disintegrate. 2 added 33 characters in body edited Oct 24 '15 at 19:42 Rob Jeffries 63.8k55 gold badges137137 silver badges209209 bronze badges Let's assume the white dwarf has a mass of $$0.6 M_{\odot}$$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use Kepler's third law, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($$8.1\times 10^{8}$$ m). The tidal forces this close to a white dwarf are very large. The Roche limit for the total tidal disintegration of a satellite, in synchronous rotation, held together only by its own gravity is roughly $$d = 1.26 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$$$d = 1.44 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$ where $$R_{WD}$$ is the radius of the white dwarf (similar to the radius of the Earth), $$\rho_{WD}$$ is the average density of the white dwarf (a few times $$10^{9}$$ kg/m$$^3$$) and $$\rho_p$$ the density of the planet (let's assume 5000 kg/m$$^3$$). Thus $$d \simeq 5 \times 10^{8}$$$$d \simeq 6 \times 10^{8}$$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating. I guess it will be an observational selection effect that such objects will be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were losercloser they would have already disintegrated and wouldn't be seen! Let's assume the white dwarf has a mass of $$0.6 M_{\odot}$$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use Kepler's third law, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($$8.1\times 10^{8}$$ m). The tidal forces this close to a white dwarf are very large. The Roche limit for the tidal disintegration of a satellite held together only by its own gravity is roughly $$d = 1.26 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$ where $$R_{WD}$$ is the radius of the white dwarf (similar to the radius of the Earth), $$\rho_{WD}$$ is the average density of the white dwarf (a few times $$10^{9}$$ kg/m$$^3$$) and $$\rho_p$$ the density of the planet (let's assume 5000 kg/m$$^3$$). Thus $$d \simeq 5 \times 10^{8}$$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating. I guess it will be an observational selection effect that such objects will be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were loser they would have already disintegrated and wouldn't be seen! Let's assume the white dwarf has a mass of $$0.6 M_{\odot}$$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use Kepler's third law, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($$8.1\times 10^{8}$$ m). The tidal forces this close to a white dwarf are very large. The Roche limit for the total tidal disintegration of a satellite, in synchronous rotation, held together only by its own gravity is roughly $$d = 1.44 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$ where $$R_{WD}$$ is the radius of the white dwarf (similar to the radius of the Earth), $$\rho_{WD}$$ is the average density of the white dwarf (a few times $$10^{9}$$ kg/m$$^3$$) and $$\rho_p$$ the density of the planet (let's assume 5000 kg/m$$^3$$). Thus $$d \simeq 6 \times 10^{8}$$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating. I guess it will be an observational selection effect that such objects will be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were closer they would have already disintegrated and wouldn't be seen! 1 answered Oct 24 '15 at 18:57 Rob Jeffries 63.8k55 gold badges137137 silver badges209209 bronze badges Let's assume the white dwarf has a mass of $$0.6 M_{\odot}$$ (there's probably a more accurate value, but most white dwarfs are close to this...). With a period of 4.5 hours we can use Kepler's third law, assuming the planetary mass is negligible compared to the white dwarf, to infer an orbital radius of 0.0054 au ($$8.1\times 10^{8}$$ m). The tidal forces this close to a white dwarf are very large. The Roche limit for the tidal disintegration of a satellite held together only by its own gravity is roughly $$d = 1.26 R_{WD} \left( \frac{\rho_{WD}}{\rho_p}\right)^{1/3},$$ where $$R_{WD}$$ is the radius of the white dwarf (similar to the radius of the Earth), $$\rho_{WD}$$ is the average density of the white dwarf (a few times $$10^{9}$$ kg/m$$^3$$) and $$\rho_p$$ the density of the planet (let's assume 5000 kg/m$$^3$$). Thus $$d \simeq 5 \times 10^{8}$$ m and is very similar to the actual orbital radius of the planet. i.e. It will be tidally disintegrating. I guess it will be an observational selection effect that such objects will be detected at the tidal breakup radius, since if they were further way they would not be disintegrating and would not be detected, and if they were loser they would have already disintegrated and wouldn't be seen!