Sun-Earth Encounter: Acceleration, Roche Limit, Rotation & More

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In summary, the sun's gravity would accelerate the planet on which people live, and would cause them to fall into the sun. If the moon was still orbiting Earth, it would be torn apart by the gravitational pull. If Earth's rotation stopped, too, the dark side of the planet would become uninhabitable.
  • #1
Paige_Turner
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TL;DR Summary
If Earth stopped orbiting the sun and just hung motionless 93M miles away...
  1. What would be the acceleration on the planet from the sun's gravity?
  2. How long would it take to fall into the sun?
  3. What would Roche limit breakup be like for people on the surface?
  4. If the moon was still orbiting us, what would happen to it?
  5. How fast would we be going when we hit?
  6. Would we splash, or just sink into the plasma gas?
  7. If Earth's rotation stopped, too:
    • How long would the dark side remain habitable (with space suits)?
    • If people on the dark side looked up at the sky when Earth hit, what would our atmosphere being blasted away look like?
    • Would they see the sun's surface rise from all horizons and close up as a shrinking dark hole above them?
 
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  • #2
Paige_Turner said:
Summary:: If Earth stopped orbiting the sun and just hung motionless 93M miles away...

  1. What would be the acceleration on the planet from the sun's gravity?
  2. How long would it take to fall into the sun?
  3. What would Roche limit breakup be like for people on the surface?
  4. If the moon was still orbiting us, what would happen to it?
  5. How fast would we be going when we hit?
  6. Would we splash, or just sink into the plasma gas?
  7. If Earth's rotation stopped, too:
    • How long would the dark side remain habitable (with space suits)?
    • If people on the dark side looked up at the sky when Earth hit, what would our atmosphere being blasted away look like?
    • Would they see the sun's surface rise from all horizons and close up as a shrinking dark hole above them?
Consider this as an opportunity to do some physics. Given that the Earth has an approximately circular orbit about the sun with a period of one year, what is the centripetal acceleration of the Earth in that orbit? You already have all the information you need: The radius and orbital period.

Consider using the formula: ##a=\frac{v^2}{r}##
Or, see if you can use the above to derive ##a=\frac{4\pi^2r}{t^2}##
 
  • #3
jbriggs444 said:
what is the centripetal acceleration of the Earth in that orbit?
Wouldn't it be easier to just use the Newton grav formula?All you need is masses and distance. You could almost do it in your head. Once Earth stops orbiting and starts falling, of what use is orbit info like the centrip force calc?
 
  • #4
Paige_Turner said:
Wouldn't it be easier to just use the Newton grav formula?All you need is masses and distance. You could almost do it in your head. Once Earth stops orbiting and starts falling, of what use is orbit info like the centrip force calc?
It is easiest if you use the information you already have: The orbital radius which you've already given and the length of the year which is usually memorized (365 days).

If you have to look up the mass of the Sun and Newton's universal gravitational constant, you're working too hard.

In any case our calculations of the Sun's mass are actually based on the orbital period of planets like the Earth. Our knowledge of ##G## and ##M_\text{sun}## are much less precise than our knowledge of their product. We weigh the sun by measuring ##G## with a device like a Cavendish balance. We determine the product ##GM_\text{sun}## with orbital measurements. Then we divide by ##G## to obtain the mass of the sun.
 
  • #5
Paige_Turner said:
Wouldn't it be easier to just use the Newton grav formula?All you need is masses and distance. You could almost do it in your head. Once Earth stops orbiting and starts falling, of what use is orbit info like the centrip force calc?
And yes, I could calculate maybe 3 of those things, like time to fall from 1 AU. But I was really more interested in the other stuff like what Roche breakup looks like from the inside.

Basically, the film Melancholia was a fascinating premise, but the physics was so horrible that it was hard to watch (like a jupiter-size planet passing so close to Earth that it sucks up our outer atmosphere, but has no other effects). What I'd really like is to see the movie with accurate events.
 
  • #6
Paige_Turner said:
And yes, I could calculate maybe 3 of those things, like time to fall from 1 AU. But I was really more interested in stuff like what Roche breakup looks like from the inside.

Basically, the film Melancholia was a fascinating premise, but the physics was so horrible that it was hard to watch (like a jupiter-size planet passing so close to Earth that it sucks up our outer atmosphere, but has no other effects). What I'd really like is to see the movie with accurate events.
To each their own, I suppose.

This site is supposed to be a learning resource. We try to help each other learn how to do physics. We are not in the "give a man a fish" business. We are in the "teach a man to fish" business.

The visual details of an unsurvivable and impossible apocalypse are not terribly interesting to me. Seems like a lot of work to calculate something of no value. Helping the person who is asking to do their own work seems like a much better use of time.
 
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  • #7
Paige_Turner said:
Summary:: If Earth stopped orbiting the sun and just hung motionless 93M miles away...
Paige_Turner said:
What I'd really like is to see the movie with accurate events.
Accurate events? What actually stops the Earth's orbit so it can "hang motionless"? That's a pretty non-physical premise...
 
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  • #8
jbriggs444 said:
> This site is supposed to be a learning resource.
Umm... hat's why I'm here. That's why we're all here. This isn't amazon or pornhub.

> The visual details of an unsurvivable and impossible apocalypse are not terribly interesting to me.

Thank you for sharing that. I guess I was addressing people who want to answer my questions.

LKT

jbriggs444 said:
We try to help each other learn how to do physics. We are not in the "give a man a fish" business. We are in the "teach a man to fish" business.

The visual details of an unsurvivable and impossible apocalypse are not terribly interesting to me. Seems like a lot of work to calculate something of no value. Helping the person who is asking to do their own work seems like a much better use of time.
 
  • #9
Paige_Turner said:
LKT
??

https://acronyms.thefreedictionary.com/LKT

1626479120037.png
 
  • #10
Paige_Turner said:
Umm... hat's why I'm here. That's why we're all here. This isn't amazon or pornhub.
Asking for answers without displaying a willingness to put in the work. Guess I'm out.
 
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  • #12
Doesn't require calculating, true?

Unless I'm wrong (and I frequently am), an artificial satellite falling straight through the Earth and back to its starting point takes 90 minutes - which is - not coincidentally - the same time it takes to complete an orbit. So the only calc you need to do for a fall from orbit to ground is ... divide by four.

Same with Earth-Sun.

Yeah?I too would like to see the OP make an attempt at their own answers. This isn't Google Central.
 
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  • #13

DaveC426913 said:
Yeah?

Yep.
 
  • #15
Paige_Turner said:
Luxine K. T.
Whelp, that clears things right up. Not.

Thread is done.
 
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  • #16
DaveC426913 said:
Unless I'm wrong (and I frequently am), an artificial satellite falling straight through the Earth and back to its starting point takes 90 minutes - which is - not coincidentally - the same time it takes to complete an orbit.
This is true for a surface satellite--a satellite that, to a good enough approximation, is orbiting right at the surface of the planet (or start). (IIRC, it is strictly true only if the planet or star has constant density.)

AFAIK it is not true for an orbit that is way, way, way above the surface, like the Earth's orbit around the Sun.
 
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1. What is the Roche limit and how does it affect the Sun-Earth encounter?

The Roche limit is the distance at which the gravitational forces of a celestial body, such as the Sun, can overcome the self-gravity of a smaller object, such as Earth. In the case of the Sun-Earth encounter, the Roche limit determines the closest distance that Earth can approach the Sun without being torn apart by the Sun's gravity.

2. How does the Sun's rotation affect the Sun-Earth encounter?

The Sun's rotation plays a significant role in the Sun-Earth encounter as it affects the Sun's magnetic field and solar wind, which can have impacts on Earth's atmosphere and magnetic field. Additionally, the Sun's rotation causes changes in its gravitational pull, which can affect the trajectory of Earth during the encounter.

3. What is the acceleration of Earth during the Sun-Earth encounter?

The acceleration of Earth during the Sun-Earth encounter is determined by the gravitational force of the Sun. This force causes Earth to accelerate towards the Sun, but Earth's own orbital velocity around the Sun helps to balance this acceleration, resulting in a stable orbit.

4. How does the Sun-Earth encounter affect Earth's orbit?

The Sun-Earth encounter can slightly alter Earth's orbit due to the gravitational forces at play. However, Earth's orbit is generally stable and any changes caused by the encounter are typically small and do not significantly impact Earth's position in the solar system.

5. What are some potential effects of a close Sun-Earth encounter?

A close Sun-Earth encounter could potentially result in disruptions to Earth's magnetic field and atmosphere due to changes in the Sun's magnetic field and solar wind. It could also cause changes in Earth's climate and weather patterns. Additionally, if Earth were to pass within the Sun's Roche limit, it could potentially be torn apart by the Sun's gravitational forces.

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