What are the realistic problems of a planet orbiting too close to its sun?

Question

Assume we have a planet similar to earth in size and materials. But we put it in orbit of a mass much greater then our sun so that the year becomes much shorter like 10 days (or put it much closer to get the same result on constant mass ignoring other issues).

Now since the earth remains the same but forces increase so does the force difference on earth surface depending on being a daytime or nighttime. Basically objects on surface will be heavier on night side then on day side due to being closer to sun (though I would expect other forces next to gravity to take effect here).

So I read that this effect is relatively small on earth. But increasing the gravitational acceleration of the sun might increase it significantly. At that point my question is how would this affect the planet? Would we be looking at something like having constant earthquakes basically having small wave of earth itself circling the surface? What about atmosphere pressure changes and winds? Are these types of issues plausible or would the planet have to be so much closer (relatively) that other issues would be way more troubling?

Answer 1

From the title of your question, the most obvious answer is that the temperature of the planet would be very high due to solar radiation. Based on the rest of your question however, it seems that you are more interested in the gravitational effects on the planet due to tidal forces, so I will focus on those effects.

TL;DR Unless the planet is very close to the star the effects will be small. If it is very close, there can be exreme effects due to tidal heating and tidal locking. Eventually the planet could fall into the star.

Dynamic tidal forces

From the accepted answer to the question you linked, we see that for a planet of radius $r$ orbiting at distance $R$ around a star of mass $M$, the correction to apparent gravitational acceleration due to the tidal force and centripetal acceleration is $$ \Delta a = GM \left( \frac{r}{R(R-r)^2} \right). $$ If the planet has mass $m$, the average surface gravitational acceleration is $$ g = \frac{Gm}{r^2}, $$ so the ratio of the correction due to the star to the averge surface gravity is $$ \alpha = \frac{\Delta a}{g} \\ = \frac{M}{m} \left( \frac{r^3}{R(R-r)^2} \right) $$ This can be simplified if we think in terms of the ratio of the radius of the planet to the radius of the orbit $\rho = R/r$, and the ratio of the masses of the star and the planet $\mu = M/m$. This lets us rewrite the equation: $$ \alpha = \frac{\mu}{\rho(\rho - 1)^2} $$

Now let's consider some numbers. For the Earth-Sun system, $\mu = 3.3\times10^{5}$ and $\rho = 2.3\times10^4$, so $\alpha = 2.5\times10^{-8}$ as shown in the linked answer. As you mention in your question, this is a very small effect.

If you want to decrease the orbital period of the Earth to 10 days, we get (from Kepler's third law) that the radius of the orbit has to shrink by a factor of about 11. This means that $\rho$ shrinks by about 11, and $\alpha = 3.4\times10^{-5}$, or about 3 thousandths of a percent. In other words, the tidal forces would still be unnoticible in everyday cicumstances.

If you wanted to continue moving the Earth closer to the sun until the tidal effect was 10% of the normal surface gravity, you would need to have $\rho \approx 150$, which would make the radius of the Earth's orbit about 1 million km, which is a little less than twice the radius of the sun. In other words, for the direct effects of the tidal force to be large, the planet must be extremely close to the star.**

This doesn't mean that there would not be extreme effects on ocean tides, continuous earthquakes, and other geophysical events however. I'm not a geophysicist, but my intuition is the following:

• The ocean tides would be extreme, if there were any liquid water at all. It's possible that all of the water would boil away because:
  • The tidal forces on the earth would result in significant tidal heating, leading to:
    • Large amounts of volcanic activity.
    • Large amounts of seismic activity.
    • Non-trivial contributions to increased surface temperature.

Tidal locking

Note that the energy dissipated by tidal heating comes from the energy stored in the planet's rotation. This means that the planet's rotation will slow down as it loses kinetic energy to tidal heating. Eventually, the planet will become tidally locked. With all the interesting worldbuilding implications that go along with that.

Similarly, there will be tidal heating of the central star. This will have an exremely small effect on the central star, but it can have a large effect on the orbiting planet. The energy for tidally heating the star comes from the orbit of the planet. This means that the planet's orbit will slowly decay until either:

• The star becomes tidally locked to the planet, or
• The planet falls into the sun.

In summary:

• Short term effects are likely to be major, i.e. the planet will survive, but it would likely be uninhabitable due to tidal heating.
• Over the long term, the planet would become tidally locked.
• Over the very long term, the orbit would decay and the planet would end up even closer to the star.

Finally, if the object that the planet is orbiting is any normal star, all of these effects will be small compared to the increased solar radiation.

Answer 2

So... A few things might happen. One, the planet might get within the Roche limit and get torn apart. Two, it's so hot as to be unlivable. (See:mercury) Finally, tidal forces make things unstable.

Part 1: Gaining Radii

So let's look at some numbers. The sun has a mass of ≈1.989e30 kg. But your sun is much bigger. Maybe... 10 times as much mass? We'll call this $M_s$. Next, the orbital period. You said 10 days? Great, that's $P$. Now, mass of our planet. We'll just use the mass of Earth, $m_e$ Ok, let's do some math. Kepler's 3rd law can be phrased as

$$\frac{P_n^2} {a_n^3} = \frac {4\pi^2} {G (M_s+m_e)}$$

where $G$ is the gravitational constant and $a_n$ is the semi-major axis of the planet's orbit.

Ok, let's look at $G_{s}$:

$$\frac {(10\;days)^2}{a^3} = \frac {4\pi^2} {G(1.989*10^{31}\;kg+5.972*10^{24}\;kg)}$$

$a$ come out to ≈29 million km. Mercury, by comparison, is at 57.91 million km. Note that this is merely the length of the semi-major axis of the planet's orbit. For simplicity, though, we'll count this as the average distance of the planet from the sun, or $r$.

Part 2: Effective Temperature

Given $r$ from part 1 and the radii of the planet and the star, we can determine the temperature of the planet at its surface. From Wikipedia's entry on effective temperature, we can see that the equation for absorbed power is $$P_{abs} = \frac {LA_{abs}(1-a)}{4\pi D^2}$$ where $L$ is the luminosity of the star (We'll use the value for our sun, 3.846e24 W), $a$ is the albedo (shininess, to be blunt) of the planet, ranging from 1 to 0 (Earth's is about .3, so we'll use that.), $D$ is the radius of the sun (again, using our sun for an example, it's 695,700 km), and $A_{abs}$ is the area of the planet that absorbs light. Keeping things simple, this is 1/2 the surface area of the planet, or $2\pi r^2$. We have all the things we need to calculate the power the planet absorbs.

But wait, there's more! The stefan-boltzmann law can be used to figure out the temperature of the planet. We know that the law can be phrased

$$P_{rad} = A_{rad}\epsilon\sigma T^4$$

Rearranging leads to the temperature.

Answer 3

Consider Io

https://fineartamerica.com/featured/a-scene-on-jupiters-moon-io-the-most-ron-miller.html

Io is a good model for your scenario because you are interested in the gravity but do not mention the radiation that a planet would encounter so close to a star.

Io has volcanoes galore. This is because of the tidal forces from Jupiter's gravity flexing the crust and heating it up.

Io also is so close to Jupiter that it traverses its magnetic field and this produces effects that I do not really understand.

Io orbits within a belt of intense radiation known as the Io plasma torus. The plasma in this doughnut-shaped ring of ionized sulfur, oxygen, sodium, and chlorine originates when neutral atoms in the "cloud" surrounding Io are ionized and carried along by the Jovian magnetosphere.[54] Unlike the particles in the neutral cloud, these particles co-rotate with Jupiter's magnetosphere, revolving around Jupiter at 74 km/s. Like the rest of Jupiter's magnetic field, the plasma torus is tilted with respect to Jupiter's equator (and Io's orbital plane), so that Io is at times below and at other times above the core of the plasma torus.

I would think that a volcanically body in comparable proximity to a star (with even greater magnetic fields) would do the same. Could this charged plasma cloud protect the close orbiter from some of the star's particle radiation in the manner Earth's magnetic field protects us from radiation?

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ADDENDUM My contribution to this question might be relevant here too:

Which astronomical or cosmological event would explain periodical low-grav-effect on earth-like planet?

I actually did the math to show that an earth sized planet swooping close to Jupiter would experience reduced gravity on the side facing Jupiter. I think one would experience increased gravity on the far side but there seems to be some dispute about that.

The thing about swooping in close to a star (as opposed to a planet) is that stars are radiant and I think at such proximity this would tend to char anyone on the light gravity side thinking they could dunk the ball. And probably strip away any atmosphere, if you thought you would hide on the shady side.