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The kinetic energy of Jupiter at the Sun's surface would be of order $GM_{\odot}M_{\rm Jup}/R_{\odot} \sim 4\times 10^{38}$ Joules.

The solar luminosity is $3.83 \times 10^{26}$ J/s.

So in this scenario, the Sun would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale of the convective envelope, which is of order $GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $few $10^{5}$ years.

In this case what will happen is Jupiter will be (quickly) shredded by the tidal field, possibly leaving a substantial core. At an orbital radius of $2 R_{\odot}$, the orbital period will be about 8 hours, the orbital speed about 300 km/s and the orbital angular momentum about $10^{42}$ kg m$^2$ s$^{-1}$. Assuming total destruction, much of the material will form an accretion disc around the Sun, since it must lose some of its angular momentum before it can be accreted.

Some sort of minimal estimate could be to assume the disk is planar and spread evenly between the solar surface and $2R_{\odot}$ and that it gets close to the solar photospheric temperature at $\sim 5000$K. In which case the disk area is $3 \pi R_{\odot}^2$, with an "areal density" of $\sigma \sim M_{\rm Jup}/3\pi R_{\odot}^2$.

In hydrostatic equilibrium, the scale height will be $\sim kT/g m_H$, where $g$ is the gravitational field and $m_H$ the mass of a hydrogen atom. The gravity (of a plane) will be $g \sim 4\pi G \sigma$. Putting in $T \sim 5000$K, we get a scale height of $\sim 0.1 R_{\odot}$.

Given that Earth is in the ecliptic plane and this is where the disk will be, then a large fraction, $> 20$%, of the sunlight reaching the Earth may be blocked. To work out if this is the case, we need to work out an optical depth of the material. For a scale height of $0.1 R_{\odot}$ and a planar geometry, then the density of the material is $\sim 3$ kg/m$^3$. Looking though this corresponds to a column density of $\sim 10^{10}$ kg/m$^{2}$.

For comparison, the solar photospheric density is of order $10^{-12}$ kg/m$^{3}$ and is only the upper 1000 km of the Sun. Given that the definition of the photosphere is where the material becomes optically thick, we can conclude that a tidally shredded Jupiter is optically thick to radiation and indeed the sunlight falling on the Earth would be very significantly reduced - whether or not the amount of radiation impacting the Earth is reduced or increased is a tricky radiative transfer problem, since if the disk were at 5000K and optically thick it would be kicking off a lot of radiation!

How long the accretion disk would remain, I am unsure how to calculate. It depends on the assumed viscosity and temperature structure and how much mass is lost through evaporation/winds. The accreted material will have radiated away a large fraction of its gravitational potential energy, so the energetic effects will be much less severe than scenario 1. However, the Sun will accrete $\sim 10^{42}$ kg m$^2$ s$^{-1}$ of angular momentum, which is comparable to its current angular momentum. The accretion of Jupiter in this way is therefore sufficient to increase the angular momentum of the Sun by a significant amount. In the long term this will have a drastic effect on the magnetic activity of the Sun - increasing it by a factor of a few to an order of magnitude.

The kinetic energy of Jupiter at the Sun's surface would be of order $GM_{\odot}M_\mathrm{Jup}/R_{\odot} \sim 4\times 10^{38}$ joules.

The solar luminosity is $3.83 \times 10^{26}\ \mathrm{J/s}$.

So in this scenario, the Sun would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale of the convective envelope, which is of order $GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $few $10^5$ years.

In this case what will happen is Jupiter will be (quickly) shredded by the tidal field, possibly leaving a substantial core. At an orbital radius of $2 R_{\odot}$, the orbital period will be about 8 hours, the orbital speed about $300\ \mathrm{km/s}$ and the orbital angular momentum about $10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$. Assuming total destruction, much of the material will form an accretion disc around the Sun, since it must lose some of its angular momentum before it can be accreted.

Some sort of minimal estimate could be to assume the disk is planar and spread evenly between the solar surface and $2R_{\odot}$ and that it gets close to the solar photospheric temperature at $\sim 5000\ \mathrm K$. In which case the disk area is $3 \pi R_{\odot}^2$, with an "areal density" of $\sigma \sim M_{\rm Jup}/3\pi R_{\odot}^2$.

In hydrostatic equilibrium, the scale height will be $\sim kT/g m_\mathrm H$, where $g$ is the gravitational field and $m_\mathrm H$ the mass of a hydrogen atom. The gravity (of a plane) will be $g \sim 4\pi G \sigma$. Putting in $T \sim 5000\ \mathrm K$, we get a scale height of $\sim 0.1 R_{\odot}$.

Given that Earth is in the ecliptic plane and this is where the disk will be, then a large fraction, $\gt 20\ \%$, of the sunlight reaching the Earth may be blocked. To work out if this is the case, we need to work out an optical depth of the material. For a scale height of $0.1 R_{\odot}$ and a planar geometry, then the density of the material is $\sim 3\ \mathrm{kg/m^3}$. Looking though this corresponds to a column density of $\sim 10^{10}\ \mathrm{kg/m^2}$.

For comparison, the solar photospheric density is of order $10^{-12}\ \mathrm{kg/m^3}$ and is only the upper $1000\ \mathrm{km}$ of the Sun. Given that the definition of the photosphere is where the material becomes optically thick, we can conclude that a tidally shredded Jupiter is optically thick to radiation and indeed the sunlight falling on the Earth would be very significantly reduced – whether or not the amount of radiation impacting the Earth is reduced or increased is a tricky radiative transfer problem, since if the disk were at $5000\ \mathrm K$ and optically thick it would be kicking off a lot of radiation!

How long the accretion disk would remain, I am unsure how to calculate. It depends on the assumed viscosity and temperature structure and how much mass is lost through evaporation/winds. The accreted material will have radiated away a large fraction of its gravitational potential energy, so the energetic effects will be much less severe than Scenario 1. However, the Sun will accrete $\sim 10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$ of angular momentum, which is comparable to its current angular momentum. The accretion of Jupiter in this way is therefore sufficient to increase the angular momentum of the Sun by a significant amount. In the long term this will have a drastic effect on the magnetic activity of the Sun – increasing it by a factor of a few to an order of magnitude.

The addition of this much energy (if it is allowed to thermalise) would potentially affect the luminosity of the Sun for timescales of tens of thousands of years.

However, I suspect what is more likely is that the kinetic energy would be also be used to do work and lift the convective envelope of the Sun. In other words, the Sun would both increase in luminosity and in radius. If the effects were just limited to the convective envelope (it is unclear whether the plummeting planet could survive falling to even greater depths), then this has a mass of around $0.02 M_{\odot}$ and so could be "lifted" by $\sim 4\times 10^{38} R_{\odot}^2/GM_{\odot}M_{\rm conv} \sim 0.05 R_{\odot}$

So in this scenario, the Sun would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale of the convective envelope, which is of order $GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $few $10^{5}$ years.

The addition of this much energy (if it is allowed to thermalise) would potentially affect the luminosity of the Sun for timescales of tens of thousands of years. The exact effects will depend on where the energy is deposited. Compared with the binding energy of the star, the additional energy is negligible, but if the energy is dissipated in the convection zone then kinetic energy would do work and lift the convective envelope of the Sun. In other words, the Sun would both increase in luminosity and in radius. If the effects were just limited to the convective envelope, then this has a mass of around $0.02 M_{\odot}$ and so could be "lifted" by $\sim 4\times 10^{38} R_{\odot}^2/GM_{\odot}M_{\rm conv} \sim 0.05 R_{\odot}$.

So in this scenario, the Sun would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale of the convective envelope, which is of order $GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $few $10^{5}$ years.

If the planet somehow survived and punched its way to the centre of the Sun, then much less energy would be deposited in the convection zone and the effects would be lessened.

Given that Earth is in the ecliptic plane and this is where the disk will be, then a large fraction, $> 20$%, of the sunlight reaching the Earth may be blocked. To work out if this is the case, we need to work out an optical depth of the material. For a scale height of $0.1 R_{\odot}$ and a planar geometry, then the density of the material is $\sim 3$ kg/m$^3$. Looking though this corresponds to a column density of $\sim 10^{10}$ kg/m$^{-2}$.

Given that Earth is in the ecliptic plane and this is where the disk will be, then a large fraction, $> 20$%, of the sunlight reaching the Earth may be blocked. To work out if this is the case, we need to work out an optical depth of the material. For a scale height of $0.1 R_{\odot}$ and a planar geometry, then the density of the material is $\sim 3$ kg/m$^3$. Looking though this corresponds to a column density of $\sim 10^{10}$ kg/m$^{2}$.