Interstellar Medium Shielding - Revision history

Tshhmon at 03:02, 6 November 2026

2026-11-06T03:02:59Z

← Older revision		Revision as of 19:02, 5 November 2026
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	=Credit=		=Credit=
	To Tshhmon for writing the article		To Tshhmon for writing the article
	* To Luke Campbell for helping with particle erosion calculation, editing, ~~and~~ physics ~~and~~ math checking and correction		* To Luke Campbell for helping with particle erosion calculation, editing, physics, math checking and correction
	* To Rocketman1999 for helping with dust erosion distribution		* To Rocketman1999 for helping with dust erosion distribution
	* To Kerr for the relativistic flux equation and derivation		* To Kerr for the relativistic flux equation and derivation

	[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Physics & Math & Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]		[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Physics & Math & Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]

Tshhmon at 02:58, 6 November 2026

2026-11-06T02:58:43Z

← Older revision		Revision as of 18:58, 5 November 2026
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	</div>		</div>

	=== ~~Relativistic corrections~~ ===		=== Formula with relativistically corrected energy ===

	At small fractions of the speed of light, relativity has a small effect and the results can be approximated by the methods described above. At 0.1 c, for example, non-relativistic formulas are still fairly close. But as speeds get to larger and larger fractions of light speed, relativity becomes more and more important.		At small fractions of the speed of light, relativity has a small effect and the results can be approximated by the methods described above. At 0.1 c, for example, non-relativistic formulas are still fairly close. But as speeds get to larger and larger fractions of light speed, relativity becomes more and more important.

	~~To include~~ the effects of relativity, first find the parameters β and γ. β is the fraction of light speed of the motion – if you are moving at 0.1 c your β is 0.1. Once you have β you can find γ = 1/(1 - β<sup>2</sup>).		Although we have no idea regarding the intermediate regime between "hypervelocity" and ultra-relativistic, we can nonetheless make a modest improvement to the model by considering the effects of relativity on kinetic energy. To include this, first find the parameters β and γ. β is the fraction of light speed of the motion – if you are moving at 0.1 c your β is 0.1. Once you have β you can find γ = 1/(1 - β<sup>2</sup>).

	With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to		With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to

Tshhmon at 02:46, 6 November 2026

2026-11-06T02:46:44Z

← Older revision		Revision as of 18:46, 5 November 2026
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	''Note: For collisions, there are various regimes which govern the response of the material after being hit by an impactor. At low velocities (below many kilometers per second), the regime is hydrodynamic. For "hypervelocity" impacts, these matters are governed by the crater regime - in which the impactor leaves a crater in the material. On the extreme end -- the ultra-relativistic regime, impactors are so penetrating that they end up being more like extremely large mega-bunches of particles leaving cones of primarily radiation and thermal damage.		''Note: For collisions, there are various regimes which govern the response of the material after being hit by an impactor. At low velocities (below many kilometers per second), the regime is hydrodynamic. For "hypervelocity" impacts, these matters are governed by the crater regime - in which the impactor leaves a crater in the material. On the extreme end -- the ultra-relativistic regime, impactors are so penetrating that they end up being more like extremely large mega-bunches of particles leaving cones of primarily radiation and thermal damage.

	''In this article, velocities between 1% to a hair under c, are considered. However, we have entirely no idea of what an intermediate case--in which the spaceship's velocity is considerably beyond thousands of km/s, but still slow enough that corrections from general relativity are low.''		''In this article, velocities between 1% to a hair under c, are considered. However, we have entirely no idea of what an intermediate case is like--a case in which the spaceship's velocity is considerably beyond thousands of km/s, but still slow enough that corrections from general relativity are low.''

	We shall assume that the hypervelocity crater regime still holds for the most part, bearing in mind that up to some high fraction of c, or more generously, when gamma is a large multiple of 1, it will cease to be even slightly accurate.		We shall assume that the hypervelocity crater regime still holds for the most part, bearing in mind that up to some high fraction of c, or more generously, when gamma is a large multiple of 1, it will cease to be even slightly accurate.

Tshhmon at 02:45, 6 November 2026

2026-11-06T02:45:16Z

Lwcamp at 19:02, 7 March 2026

2026-03-07T19:02:19Z

← Older revision		Revision as of 11:02, 7 March 2026
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	* To Kerr for the relativistic flux equation and derivation		* To Kerr for the relativistic flux equation and derivation

	[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]		[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Physics & Math & Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]

Lwcamp: /* Calculating the Temperature of the Forward Shield */

2026-02-25T05:32:08Z

Calculating the Temperature of the Forward Shield

← Older revision		Revision as of 21:32, 24 February 2026
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	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
	<tr>		<tr>
	<td rowspan="2" nowrap="nowrap" align="right"> <i>T</i> = <span style="font-size: 250%;">(</span>		<td rowspan="2" nowrap="nowrap" align="right"> <i>T</i> =
			<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">(</span>
	<td nowrap="nowrap">   φ		<td nowrap="nowrap">   φ
	<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>		<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>

Lwcamp: /* Calculating Collision Effects */

2026-02-19T06:05:30Z

Calculating Collision Effects

← Older revision		Revision as of 22:05, 18 February 2026
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	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
	<tr>		<tr>
	<td rowspan="2" nowrap="nowrap" align="right"> <i>r</i> = <span style="font-size: 250%;">(</span>		<td rowspan="2" nowrap="nowrap" align="right"> <i>r</i> =
			<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">(</span>
	<td nowrap="nowrap">   3 <i>V</i>		<td nowrap="nowrap">   3 <i>V</i>
	<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>		<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>

Lwcamp: /* Calculating Collision Effects */

2026-02-19T00:28:03Z

Calculating Collision Effects

← Older revision		Revision as of 16:28, 18 February 2026
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	The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.		The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.
	If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. An individual dust grain will blast out a ~~spherical~~ crater with a radius		If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. An individual dust grain will blast out a hemispherical crater with a radius
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
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	<td nowrap="nowrap"> <sup>1/3</sup>		<td nowrap="nowrap"> <sup>1/3</sup>
	<tr>		<tr>
	<td style="border-top:solid 1px black;"> 4 π		<td style="border-top:solid 1px black;"> 2 π
	<td>		<td>
	</table>		</table>
	</div>		</div>
	In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 2.8 mm in radius while the 0.9 c grain will make a crater 1.8 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.		In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 3.5 mm in radius while the 0.9 c grain will make a crater 2.2 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.

	So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that		So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that

Lwcamp: /* Relativistic corrections */

2026-02-19T00:22:34Z

Relativistic corrections

← Older revision		Revision as of 16:22, 18 February 2026
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	With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to		With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to

	~~At relativistic speeds, this is modified to~~
	<div align=center>		<div align=center>
	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >

Lwcamp: /* Calculating Collision Effects */

2026-02-19T00:21:41Z

Calculating Collision Effects

← Older revision		Revision as of 16:21, 18 February 2026
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	where <i>K<sub>h</sub></i> is the cavity strength of the material. The cavity strength can be found from commonly reported material properties using <i>K<sub>h</sub></i> = (2/3) <i>K<sub>c</sub></i> × (1 + ln[2 <i>G</i>/<i>K<sub>h</sub></i>]), for compressive strength <i>K<sub>c</sub></i> and shear modulus <i>G</i>. However for approximate uses, <i>K<sub>h</sub></i> tends to be about 3 to 4 times <i>K<sub>c</sub></i>.		where <i>K<sub>h</sub></i> is the cavity strength of the material. The cavity strength can be found from commonly reported material properties using <i>K<sub>h</sub></i> = (2/3) <i>K<sub>c</sub></i> × (1 + ln[2 <i>G</i>/<i>K<sub>h</sub></i>]), for compressive strength <i>K<sub>c</sub></i> and shear modulus <i>G</i>. However for approximate uses, <i>K<sub>h</sub></i> tends to be about 3 to 4 times <i>K<sub>c</sub></i>.

	An individual dust grain will blast out a spherical crater with a radius		The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.
			If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. An individual dust grain will blast out a spherical crater with a radius
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
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	</table>		</table>
	</div>		</div>
	~~The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.~~		In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 2.8 mm in radius while the 0.9 c grain will make a crater 1.8 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.
	If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 2.8 mm in radius while the 0.9 c grain will make a crater 1.8 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.

	So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that		So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that
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	</table>		</table>
	</div>		</div>

			=== Relativistic corrections ===

			At small fractions of the speed of light, relativity has a small effect and the results can be approximated by the methods described above. At 0.1 c, for example, non-relativistic formulas are still fairly close. But as speeds get to larger and larger fractions of light speed, relativity becomes more and more important.

			To include the effects of relativity, first find the parameters β and γ. β is the fraction of light speed of the motion – if you are moving at 0.1 c your β is 0.1. Once you have β you can find γ = 1/(1 - β<sup>2</sup>).

			With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to

	At relativistic speeds, this is modified to		At relativistic speeds, this is modified to
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	</table>		</table>
	</div>		</div>
	where γ = 1/(1 - β<sup>2</sup>) and β = <i>v</i>/c. At relativistic speeds the kinetic energy of an individual dust grain is likewise modified to <i>E</i> = m (γ - 1) c<sup>2</sup>.

	==Prospects for Shield Materials==		==Prospects for Shield Materials==

← Older revision		Revision as of 18:45, 5 November 2026
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	<br/> Reference for the Daedalus report figure.</ref>		<br/> Reference for the Daedalus report figure.</ref>

	According to <ref>[https://arxiv.org/abs/1608.05284 The interaction of relativistic spacecrafts with the interstellar medium]</ref>, at a certain velocity regime (5% to 20% of the speed of light), impacting particles may have additional erosive effects by leaving "ion tracks". These tracks are essentially trails of damaged material left in the wake of the ion, which has penetrated deeply into the material. However, there are numerous issues with the paper, according to M. Karlusic's comment <ref>https://arxiv.org/abs/1701.04319</ref>, and the ion track effect may not even apply if the shield is made out of conductive metals.		According to a paper<ref>[https://arxiv.org/abs/1608.05284 The interaction of relativistic spacecrafts with the interstellar medium]</ref>, at a certain velocity regime (5% to 20% of the speed of light), impacting particles may have additional erosive effects by leaving "ion tracks". These tracks are essentially trails of damaged material left in the wake of the ion, which has penetrated deeply into the material. However, there are numerous issues with the paper, according to M. Karlusic's comment <ref>https://arxiv.org/abs/1701.04319</ref>, and the ion track effect may not even apply if the shield is made out of conductive metals.

	Another paper, <ref>[https://ui.adsabs.harvard.edu/abs/2021ApJ...908..248D/abstract Damage to Relativistic Interstellar Spacecraft by ISM Impact Gas Accumulation]</ref>, argues that hydrogen and helium atoms at relativistic velocity implant themselves in the material, becoming slowly diffusing gas atoms. These then cause damage through bubble formation, blistering and exfoliation.		Another paper<ref>[https://ui.adsabs.harvard.edu/abs/2021ApJ...908..248D/abstract Damage to Relativistic Interstellar Spacecraft by ISM Impact Gas Accumulation]</ref>, argues that hydrogen and helium atoms at relativistic velocity implant themselves in the material, becoming slowly diffusing gas atoms. These then cause damage through bubble formation, blistering and exfoliation.

	However, even in both cases, the erosion is limited to on the order of a millimeter depth every 4 light years travelled (as these papers generally cover the case of a Breakthrough Starshot spacecraft journeying to Proxima Centauri, the closest star). For relatively large starships, these concerns may not matter much. Extrapolating from the rate, for a meter of material to be eroded the starship would need to travel 4,000 light years.		However, even in both cases, the erosion is limited to on the order of a millimeter depth every 4 light years travelled (as these papers generally cover the case of a Breakthrough Starshot spacecraft journeying to Proxima Centauri, the closest star). For relatively large starships, these concerns may not matter much. Extrapolating from the rate, for a meter of material to be eroded the starship would need to travel 4,000 light years.
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	Interstellar dust grain density ranges from a few hundred to a few thousand grains per cubic kilometer<ref>https://openstax.org/books/astronomy/pages/20-1-the-interstellar-medium </br> Reference for interstellar dust grain density.</ref>. For the rest of this section, we'll assume a density of 500 grains/km<sup>3</sup>. This translates to 5×10<sup>-7</sup> grains per cubic meter. Note that the interstellar dust cloud which Earth is moving through, has an order of magnitude higher density: 10<sup>-6</sup> dust grains per cubic meter<ref>https://ntrs.nasa.gov/citations/20050215611</br> Reference for the local interstellar dust density.</ref>.		Interstellar dust grain density ranges from a few hundred to a few thousand grains per cubic kilometer<ref>https://openstax.org/books/astronomy/pages/20-1-the-interstellar-medium </br> Reference for interstellar dust grain density.</ref>. For the rest of this section, we'll assume a density of 500 grains/km<sup>3</sup>. This translates to 5×10<sup>-7</sup> grains per cubic meter. Note that the interstellar dust cloud which Earth is moving through, has an order of magnitude higher density: 10<sup>-6</sup> dust grains per cubic meter<ref>https://ntrs.nasa.gov/citations/20050215611</br> Reference for the local interstellar dust density.</ref>.

	A light year is 9.454 quadrillion meters long, so under this theoretical model there would be ≈ 5×10<sup>9</sup> dust grains per square meter per light year, colliding with ~~your~~ shield.		A light year is 9.454 quadrillion meters long, so under this theoretical model there would be ≈ 5×10<sup>9</sup> dust grains per square meter per light year, colliding with the shield.

	As mentioned above, dust makes up about 1% of the mass of the ISM. Under the assumption of one atom per cubic centimeter and with the distribution of atom masses given above, this gives a mass of dust swept out of about ρ<sub>d</sub> = 2×10<sup>-7</sup> kg/m^2/ly. The dust density can change considerably between different parts of the galaxy but for our example we will use this value of ρ<sub>d</sub>.		As mentioned above, dust makes up about 1% of the mass of the ISM. Under the assumption of one atom per cubic centimeter and with the distribution of atom masses given above, this gives a mass of dust swept out of about ρ<sub>d</sub> = 2×10<sup>-7</sup> kg/m^2/ly. The dust density can change considerably between different parts of the galaxy but for our example we will use this value of ρ<sub>d</sub>.

	==Calculating Collision Effects==		==Calculating Collision Effects==
	''Note: For collisions, there are various regimes which govern the response of the material after being hit by an impactor. At low velocities (below many kilometers per second), the regime is hydrodynamic. For "hypervelocity" impacts, these matters are ~~entirely~~ governed by the crater regime - in which the impactor leaves a crater in the material. On the extreme end -- the ultra-relativistic regime, impactors are so penetrating that they end up being more like ~~big,~~ mega-bunches of particles leaving cones of primarily radiation and thermal damage.		''Note: For collisions, there are various regimes which govern the response of the material after being hit by an impactor. At low velocities (below many kilometers per second), the regime is hydrodynamic. For "hypervelocity" impacts, these matters are governed by the crater regime - in which the impactor leaves a crater in the material. On the extreme end -- the ultra-relativistic regime, impactors are so penetrating that they end up being more like extremely large mega-bunches of particles leaving cones of primarily radiation and thermal damage.

	''In ~~the purview of~~ this article, velocities ~~tend to fall~~ between 1% to a hair under c. However, we have entirely no idea of what ~~the~~ intermediate case ~~is like~~ -- ~~when you~~'~~re already many~~ thousands of km/s, but still ~~below the relativistic regime? We can only guess -- so bear in mind the following is pure and utter conjecture~~.''		''In this article, velocities between 1% to a hair under c, are considered. However, we have entirely no idea of what an intermediate case--in which the spaceship's velocity is considerably beyond thousands of km/s, but still slow enough that corrections from general relativity are low.''

	''Now, dust grain impacts might end up as a hybrid between the hypervelocity crater and the ultra-relativistic "cone". With increasing velocity, the dust grains will penetrate deeper and deeper, along with secondary showers and exotic effects at such high kinetic energies. The original, roughly hemispherical/parabolic crater shape might change to resemble more that of a cone with increasing velocity. This concludes our speculation, and we shall ~~move on to the calculation.''~~		We shall assume that the hypervelocity crater regime still holds for the most part, bearing in mind that up to some high fraction of c, or more generously, when gamma is a large multiple of 1, it will cease to be even slightly accurate.

	~~Let's~~ assume that the hypervelocity crater regime still holds for the most part, bearing in mind that up to some high fraction of c, or more generously, when gamma is a large multiple of 1, it will cease to be even slightly accurate.

	We can use the method of calculating crater volume that was presented in the article on [[Beam-Target_Interactions\|beam-target interactions]]: The volume eroded <i>V</i> is related to the energy deposited in the impact event <i>E</i> by		We can use the method of calculating crater volume that was presented in the article on [[Beam-Target_Interactions\|beam-target interactions]]: The volume eroded <i>V</i> is related to the energy deposited in the impact event <i>E</i> by