I suspect we'd have bigger problems than simply losing the moon.
The moon orbits around the Earth with a mean orbital velocity of 1.022 km/s and has a mass off roughly 0.07342 x 10^24 kg. This is a big object. In order for the Earth to lose the moon for good, it would have to, at minimum, achieve escape velocity from the Earth's orbit. Escape velocity for the moon, defined as where kinetic energy is equal to the gravitational potential energy of the orbiting object, is given be the formula
vescmoon=√2μd
where d is the distance between the moon and the earth's centers of gravity and μ
is the standard gravitational parameter of the Earth. At perigee, this distance is roughly 365e3 km, resulting in an escape velocity of 1.478 km/s [1]. Let's put some boundary values on an object that would accelerate the moon to that velocity.
The easiest way for a celestial object to accelerate the moon is for it to hit dead-center towards it's center of mass in the same direction as its velocity vector precisely at the moon's perigee, when velocity is maximized. The moon's velocity at perigee is 1.076 km/s.
Given conservation of linear momentum, relativistic effects on the momentum of very fast objects and the upper bound on velocity set by the speed of light in a vacuum, a good estimate of a low-end mass that could possibly achieve an orbital escape of the moon would be a relativistic projectile at around 0.9 times the speed of light that is somehow made of a tarball that gloms onto the moon immediately. We have a change in momentum needed for the moon [2], and the inelastic projectile's velocity (~270e3 km/s), so we can solve for the (rest) mass that this would require. From our reference frame, the relativistic object would have an apparent momentum of
p=γm0v=m0v√1−v2c2
consistent with the apparent increase in mass for objects moving at relativistic speeds. For an object moving at 0.9c the parameter γ is 1√0.19
, which equals about 2.3. The rest mass of the object moving at this speed is enormous, 2.509e17 kg or roughly 251 trillion tons [3].
This works out to a relativistic ball of silly putty roughly the size of Uranus' moon Rosalind. I'm talking about a celestial body with a mean radius of 35 km that would tower into the stratosphere if somehow translated to sea level.
That being said, it's a tiny fraction of the moon's mass.
Even so, the kinetic energy that would have to be imparted to the moon by this tarball would be massive. Some of this energy would be expressed by the increase in its linear velocity, but the rest would have to be dissipated as heat or radiation. We are talking about a seriously large amount of energy here. The relativistic tarball that would hit the moon would have had a kinetic energy of 2.918e34 J [4] and would have transferred only an amount of kinetic energy amounting to 3.769e28 J [5] to the moon in order to get it to escape velocity. These numbers are too large for metric prefixes and show just how ridiculously large the amount of energy stored by a relativistic asteroid would be.
By looking up estimates of the constant pressure heat capacity of the moon, I was able to figure out an estimate of just how hot this would make our friendly celestial neighbor. Yes, yes, I know the heat capacity will change, and I'm making a lumped capacitance assumption here - remember that this is a thought experiment. When I plugged all of this in, this resulted in the mind-boggling temperature change of roughly 570,000,000 Kelvins [6].
For a brief moment, the moon would become over 100 times hotter than the center of the sun. The entire moon. An incredible burst of radiation at all wavelengths would bathe the half of the earth in fire and gamma rays and the moon (and our tarball) would instantaneously become a soup of rapidly disintegrating elementary particles that are blown apart by their own radiation pressure. In the brief moments in which the Earth and Moon reach thermal equilibrium before disintegrating, the heat transfer to the Earth would obey some form of the Stefan-Boltzmann law for heat transfer between ideal blackbody radiators, assuming that at the frequency of gamma rays being emitted the relative emissivity of each body is basically moot:
Q=σ(AsurfacemoonT4moon−AsurfaceearthT4earth)
where for the purposes of this estimate the surface areas of the Earth and Moon exposed to each other are one-half of their respective surface areas each (approximately 205 million square kilometers and 19 million square kilometers, respectively).
For a brief moment, one half of the Earth would be bathed in 1.14e41 Watts of radiation [7] with intensity peaking at a wavelength of 5.08 picometers [8], well into the gamma ray range.
I honestly struggled with a way to contextualize this much energy transfer. It is 300 billion times the normal radiative emissions of the sun. The raw power emitted would be equivalent to carpet-nuking the Earth with 0.86 MOLES of the oversized Soviet Tsar Bomba every second [9] or roughly enough to drop 1250 on every square millimeter of Earth. The energy emitted in the first second would be an appreciable fraction of the mass-energy of the Earth and would be 100 billion times greater than the kinetic energy of the Earth's orbit around the Sun.
In any case, the amount of energy that the moon would absorb as part of that collision that didn't contribute to the velocity increase exceeds the total gravitational binding energy of the moon by several orders of magnitude - five, in fact. It is also more energy than the gravitational binding energy of the Earth (which is roughly 2.2×1032J
). I expect the surface of the earth would have begun boiling before the first of the ionized particles from what was once the moon hit the surface.
Uh, before we back away slowly, we have to remember that this heating effect, where kinetic energy dissipates, will hold not merely for our relativistic projectile but for any perfectly inelastic collision. A perfectly elastic collision would require a smaller projectile but still release an enormous amount of energy (the minimum required kinetic energy as above) to be transferred directly to the moon. Real collisions will be somewhere between these two extremes (I am not an astrophysicist, so I don't even know where to look) but even the minimum amount of energy transferred to the moon is cataclysmic.
Then let's step back and realize that given the extreme unlikelihood of a relativistic projectile of that nature, we will have to have a much more massive projectile. Cutting the speed of the projectile to 0.1c from 0.9c, for example, due to relativistic effects of momentum, would require a celestial body of rest mass 27 times larger, although the increase in mass required is less extreme as you get farther and farther away from appreciable fractions of the speed of light. Nevertheless, extrapolating this relationship we quickly get masses for the projectiles such that they can be described as rogue planets. If we further realize that the energy is not transferred directly to the center of mass of the moon, it is inevitable that what we will see will result in, minimally, a massive ejection of mass from the surface of the moon into decaying Earth orbits, as well as a violently energetic explosion that will make what that vegan guy in Scott Pilgrim did to the moon look like a love tap.
However when Todd vegan-bombed the moon, twice, without collateral damage, he had his vegan powers to help him. We can hardly expect a planetoid intruder in our solar system to have slogged through vegan academy.
Like I said, we'll have other problems, and I'm not even sure the vegans will be able to save us.
[8] Taking the wavenumber for of the peak of the Planck's law distribution at the ludicrous temperature we are using on the moon by taking the approximation λ=hc4.965kBT at ((4.965 boltzmann constant 570e6 K)/hc)^(-1)
[9] "Total annihilation" radius of the Tsar bomba was estimated at approximately 35 km, surface of the earth estimated at 200 million square miles and the yield of the Tsar bomba at 220 petajoules. 1.14e41 watts/(220 petawatts) - Wolfram|Alpha
Source for thermal calculation values: Moon Fact Sheet
, Szurgot M. "On the heat capacity of asteroids, satellites and terrestrial planets." In 43rd Lundar and Planetary Science Conference (2012)
Thanks to Mark Eichenlaub, chief of the physics police, for correcting my grandiose thought experiments for the third time, making me remember that integrating Planck's Law over all frequencies and exposed solid angles results in the Stefan-Boltzmann Law, and giving me still more excuses to actually get cracking on my copy of the Feynman Lectures.
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OMG, this Quora answer is the pinnacle of social-media wisdom:
Profile photo for Josh Velson Josh Velson alternatehistory.com contributor, and avid alternate history readerUpvoted by Eurico Chagas , MS Civil Engineering & Physics, University of Illinois (1978) and Vegard Stornes Farstad , M.Sc. Physics & Education, Norwegian University of Science and Technology (2002)Author has 3K answers and 10.6M answer viewsUpdated 6y Originally Answered: Recently an astroid passed so close to Earth that it went inside the moon's orbit. What would happen to Earth if one hit the moon and knocked it off its orbit and we lost it for good?
I suspect we'd have bigger problems than simply losing the moon.
The moon orbits around the Earth with a mean orbital velocity of 1.022 km/s and has a mass off roughly 0.07342 x 10^24 kg. This is a big object. In order for the Earth to lose the moon for good, it would have to, at minimum, achieve escape velocity from the Earth's orbit. Escape velocity for the moon, defined as where kinetic energy is equal to the gravitational potential energy of the orbiting object, is given be the formula vescmoon=√2μd where d is the distance between the moon and the earth's centers of gravity and μ
is the standard gravitational parameter of the Earth. At perigee, this distance is roughly 365e3 km, resulting in an escape velocity of 1.478 km/s [1]. Let's put some boundary values on an object that would accelerate the moon to that velocity.
The easiest way for a celestial object to accelerate the moon is for it to hit dead-center towards it's center of mass in the same direction as its velocity vector precisely at the moon's perigee, when velocity is maximized. The moon's velocity at perigee is 1.076 km/s.
Given conservation of linear momentum, relativistic effects on the momentum of very fast objects and the upper bound on velocity set by the speed of light in a vacuum, a good estimate of a low-end mass that could possibly achieve an orbital escape of the moon would be a relativistic projectile at around 0.9 times the speed of light that is somehow made of a tarball that gloms onto the moon immediately. We have a change in momentum needed for the moon [2], and the inelastic projectile's velocity (~270e3 km/s), so we can solve for the (rest) mass that this would require. From our reference frame, the relativistic object would have an apparent momentum of p=γm0v=m0v√1−v2c2
consistent with the apparent increase in mass for objects moving at relativistic speeds. For an object moving at 0.9c the parameter γ is 1√0.19
, which equals about 2.3. The rest mass of the object moving at this speed is enormous, 2.509e17 kg or roughly 251 trillion tons [3].
This works out to a relativistic ball of silly putty roughly the size of Uranus' moon Rosalind. I'm talking about a celestial body with a mean radius of 35 km that would tower into the stratosphere if somehow translated to sea level.
That being said, it's a tiny fraction of the moon's mass.
Even so, the kinetic energy that would have to be imparted to the moon by this tarball would be massive. Some of this energy would be expressed by the increase in its linear velocity, but the rest would have to be dissipated as heat or radiation. We are talking about a seriously large amount of energy here. The relativistic tarball that would hit the moon would have had a kinetic energy of 2.918e34 J [4] and would have transferred only an amount of kinetic energy amounting to 3.769e28 J [5] to the moon in order to get it to escape velocity. These numbers are too large for metric prefixes and show just how ridiculously large the amount of energy stored by a relativistic asteroid would be.
By looking up estimates of the constant pressure heat capacity of the moon, I was able to figure out an estimate of just how hot this would make our friendly celestial neighbor. Yes, yes, I know the heat capacity will change, and I'm making a lumped capacitance assumption here - remember that this is a thought experiment. When I plugged all of this in, this resulted in the mind-boggling temperature change of roughly 570,000,000 Kelvins [6].
For a brief moment, the moon would become over 100 times hotter than the center of the sun. The entire moon. An incredible burst of radiation at all wavelengths would bathe the half of the earth in fire and gamma rays and the moon (and our tarball) would instantaneously become a soup of rapidly disintegrating elementary particles that are blown apart by their own radiation pressure. In the brief moments in which the Earth and Moon reach thermal equilibrium before disintegrating, the heat transfer to the Earth would obey some form of the Stefan-Boltzmann law for heat transfer between ideal blackbody radiators, assuming that at the frequency of gamma rays being emitted the relative emissivity of each body is basically moot: Q=σ(AsurfacemoonT4moon−AsurfaceearthT4earth)
where for the purposes of this estimate the surface areas of the Earth and Moon exposed to each other are one-half of their respective surface areas each (approximately 205 million square kilometers and 19 million square kilometers, respectively).
For a brief moment, one half of the Earth would be bathed in 1.14e41 Watts of radiation [7] with intensity peaking at a wavelength of 5.08 picometers [8], well into the gamma ray range.
I honestly struggled with a way to contextualize this much energy transfer. It is 300 billion times the normal radiative emissions of the sun. The raw power emitted would be equivalent to carpet-nuking the Earth with 0.86 MOLES of the oversized Soviet Tsar Bomba every second [9] or roughly enough to drop 1250 on every square millimeter of Earth. The energy emitted in the first second would be an appreciable fraction of the mass-energy of the Earth and would be 100 billion times greater than the kinetic energy of the Earth's orbit around the Sun.
In any case, the amount of energy that the moon would absorb as part of that collision that didn't contribute to the velocity increase exceeds the total gravitational binding energy of the moon by several orders of magnitude - five, in fact. It is also more energy than the gravitational binding energy of the Earth (which is roughly 2.2×1032J
). I expect the surface of the earth would have begun boiling before the first of the ionized particles from what was once the moon hit the surface.
Uh, before we back away slowly, we have to remember that this heating effect, where kinetic energy dissipates, will hold not merely for our relativistic projectile but for any perfectly inelastic collision. A perfectly elastic collision would require a smaller projectile but still release an enormous amount of energy (the minimum required kinetic energy as above) to be transferred directly to the moon. Real collisions will be somewhere between these two extremes (I am not an astrophysicist, so I don't even know where to look) but even the minimum amount of energy transferred to the moon is cataclysmic.
Then let's step back and realize that given the extreme unlikelihood of a relativistic projectile of that nature, we will have to have a much more massive projectile. Cutting the speed of the projectile to 0.1c from 0.9c, for example, due to relativistic effects of momentum, would require a celestial body of rest mass 27 times larger, although the increase in mass required is less extreme as you get farther and farther away from appreciable fractions of the speed of light. Nevertheless, extrapolating this relationship we quickly get masses for the projectiles such that they can be described as rogue planets. If we further realize that the energy is not transferred directly to the center of mass of the moon, it is inevitable that what we will see will result in, minimally, a massive ejection of mass from the surface of the moon into decaying Earth orbits, as well as a violently energetic explosion that will make what that vegan guy in Scott Pilgrim did to the moon look like a love tap.
However when Todd vegan-bombed the moon, twice, without collateral damage, he had his vegan powers to help him. We can hardly expect a planetoid intruder in our solar system to have slogged through vegan academy.
Like I said, we'll have other problems, and I'm not even sure the vegans will be able to save us.
[1] 398600.4418 km^3/s^2*2/(365e3 km))^0.5 - Wolfram|Alpha
[2] Page on wolframalpha.com
[3] Page on wolframalpha.com
[4] (1/sqrt(1-(0.9c)^2/c^2)-1)2.509e14 tonnes * c^2
[5] 0.5(2.509e14 tonnes + 0.07342e24 kg)((1.478 km/s)^2-(1.076 km/s)^2)
[6] 2.918e34 J /(698 J/(kg K) * (0.07342e24 kg+2.509e17 kg)
[7] (5.670400e-8 J/(sm^2K^4))(19e6 km^2(5.7e8 K)^4-205e6 km^2(300 K)^4)
[8] Taking the wavenumber for of the peak of the Planck's law distribution at the ludicrous temperature we are using on the moon by taking the approximation λ=hc4.965kBT at ((4.965 boltzmann constant 570e6 K)/hc)^(-1)
[9] "Total annihilation" radius of the Tsar bomba was estimated at approximately 35 km, surface of the earth estimated at 200 million square miles and the yield of the Tsar bomba at 220 petajoules. 1.14e41 watts/(220 petawatts) - Wolfram|Alpha
Source for thermal calculation values: Moon Fact Sheet , Szurgot M. "On the heat capacity of asteroids, satellites and terrestrial planets." In 43rd Lundar and Planetary Science Conference (2012)
Thanks to Mark Eichenlaub, chief of the physics police, for correcting my grandiose thought experiments for the third time, making me remember that integrating Planck's Law over all frequencies and exposed solid angles results in the Stefan-Boltzmann Law, and giving me still more excuses to actually get cracking on my copy of the Feynman Lectures. 106.3K viewsView upvotes View 5 shares