Any Astronomy Buffs Here?

[DevilHorse]

Awesome! I would LOVE to own that but I’m reluctant to inquire about the cost - the answer would likely just depress me.

Lots of orrerys available on ebay.
https://www.ebay.com/sch/i.html?_fro...rrery&_sacat=0
Some are really nice (not as nice as the above) but on the order of $1k.

Larry
DevilHorse

[DevilHorse]

Interesting find by the Curiosity rover:
https://twitter.com/latestinspace/st...47149215907842

Anybody hear about the Green Comet that will soon pass by earth?
https://www.indystar.com/story/news/...3/69791991007/

Larry
DevilHorse

A "Where To Find The Green Comet" article:
https://gizmodo.com/how-to-see-green...tltX5xbSx8HP3w

Larry
DevilHorse

[DevilHorse]

On Saturday, an astronomer in the Crimea of all places, discovered (5 days ago - 1/21/2023) an asteroid heading toward earth. The Asteroid is not big (the size of a truck, 10-30 feet) and would burn up in our atmosphere if it entered our atmosphere. It will pass over the southern tip of South America when it visits.

https://www.jpl.nasa.gov/news/nasa-s...arth-this-week

BTW, the speed as it passes will be about 25,000 Miles per hour.

Larry
DevilHorse

[Skydog]

On Saturday, an astronomer in the Crimea of all places, discovered (5 days ago - 1/21/2023) an asteroid heading toward earth. The Asteroid is not big (the size of a truck, 10-30 feet) and would burn up in our atmosphere if it entered our atmosphere. It will pass over the southern tip of South America when it visits.

https://www.jpl.nasa.gov/news/nasa-s...arth-this-week

BTW, the speed as it passes will be about 25,000 Miles per hour.

Larry
DevilHorse

So just fast enough to orbit the earth in an hour’ish.

Speaking of earths circumference I just saw this old nugget: If you created a metal band (no, not the Metallica type!) exactly the circumference of the earth (~40,000 km) and then made it 10 meters longer, how much space would there be between the new band and the earth? Enough for an ant to crawl under? A fox? A man?

Bonus: What if you added 10 meters to a band the circumference of a basketball? How much room would there be between the extended band and the ball?

I know you know these answers Devilhorse but they are surprising to anyone coming across them the first time. They were to me at least.

[camion]

So just fast enough to orbit the earth in an hour’ish.

Speaking of earths circumference I just saw this old nugget: If you created a metal band (no, not the Metallica type!) exactly the circumference of the earth (~40,000 km) and then made it 10 meters longer, how much space would there be between the new band and the earth? Enough for an ant to crawl under? A fox? A man?

Bonus: What if you added 10 meters to a band the circumference of a basketball? How much room would there be between the extended band and the ball?

I know you know these answers Devilhorse but they are surprising to anyone coming across them the first time. They were to me at least.

I consulted this book when computing the answer.

consider-a-spherical-cow.jpg

I got about 1.6 meters of standing (hunched) room.

The change in radius would be the same for a basketball band, but it's awfully hard to stand on a bball.


Adding 10 m to the band would be a minuscule change compared to the size of the earth. Luckily the average human is also minuscule compared to the size of the earth.

[DevilHorse]

So just fast enough to orbit the earth in an hour’ish.

Speaking of earths circumference I just saw this old nugget: If you created a metal band (no, not the Metallica type!) exactly the circumference of the earth (~40,000 km) and then made it 10 meters longer, how much space would there be between the new band and the earth? Enough for an ant to crawl under? A fox? A man?

Bonus: What if you added 10 meters to a band the circumference of a basketball? How much room would there be between the extended band and the ball?

I know you know these answers Devilhorse but they are surprising to anyone coming across them the first time. They were to me at least.

Orbits depend on trajectory.

For comparison to today's asteroid, geosynchronous satellites are positioned over 22,000 miles radius (Earth radius is ~4000 miles, so Geo Sats are 18,000 miles above the surface of the Earth). Their speed needed to stay in orbit is about 6,700 miles per hour. The speed for a satellite to orbit the Earth, at the Earth's surface (assuming no friction from the atmosphere of course) is about 7,000 miles per hour. So this asteroid at 2,200 miles above the surface of the earth would require a speed of about 6,800 miles per hour to keep the orbit from running off the rails. At a speed of 25,000 miles per hour, this asteroid would never be in an Earth bound orbit.

The problem you suggested was in a text for intermediate mechanics as I recall; although it was a string in that book. The answer was the same.

Larry
DevilHorse

[Skydog]

I consulted this book when computing the answer.

consider-a-spherical-cow.jpg

I got about 1.6 meters of standing (hunched) room.

The change in radius would be the same for a basketball band, but it's awfully hard to stand on a bball.


Adding 10 m to the band would be a minuscule change compared to the size of the earth. Luckily the average human is also minuscule compared to the size of the earth.

You got it. I was surprised that adding just 10m to a 40,000m belt would create such a gap. And that the gap size is the same whether you add 10m to 40,000 or 10m to 1m.
Isn't math/science grand?

[Skydog]

Orbits depend on trajectory.

For comparison to today's asteroid, geosynchronous satellites are positioned over 22,000 miles radius (Earth radius is ~4000 miles, so Geo Sats are 18,000 miles above the surface of the Earth). Their speed needed to stay in orbit is about 6,700 miles per hour. The speed for a satellite to orbit the Earth, at the Earth's surface (assuming no friction from the atmosphere of course) is about 7,000 miles per hour. So this asteroid at 2,200 miles above the surface of the earth would require a speed of about 6,800 miles per hour to keep the orbit from running off the rails. At a speed of 25,000 miles per hour, this asteroid would never be in an Earth bound orbit.

The problem you suggested was in a text for intermediate mechanics as I recall; although it was a string in that book. The answer was the same.

Larry
DevilHorse

I shouldn't have used the word orbit. I meant the asteroid speed was such that it could traverse the world’s circumference in ~1hr. But that would be a very low flying “orbit” and would need several “steering rockets” and a massive amount of fuel to keep it from flying away from the surface.

[Skydog]

...

The problem you suggested was in a text for intermediate mechanics as I recall; although it was a string in that book. The answer was the same.

Larry
DevilHorse

Ahhh... so this is the string theory I hear so much about. Easier to understand than I expected!

[camion]

I shouldn't have used the word orbit. I meant the asteroid speed was such that it could traverse the world’s circumference in ~1hr. But that would be a very low flying “orbit” and would need several “steering rockets” and a massive amount of fuel to keep it from flying away from the surface.

This talk of orbits, acceleration, trajectory caused me think of an interesting problem that I don't know the answer to, so I will pose it here.

If the earth rotated fast enough that someone standing on the equator would weigh 0 pounds, what would be the length of the day? To make the arithmetic simple you may assume that the radius of a spherical earth is 4,000 miles (or 6500 km).

[DevilHorse]

This talk of orbits, acceleration, trajectory caused me think of an interesting problem that I don't know the answer to, so I will pose it here.

If the earth rotated fast enough that someone standing on the equator would weigh 0 pounds, what would be the length of the day? To make the arithmetic simple you may assume that the radius of a spherical earth is 4,000 miles (or 6500 km).

I assume you want the answer rather than a full derivation. Permit me a bit of math..

The acceleration of gravity is 32 feet/second^2 (m/s^2); the centrifugal force at the equator is: 0.111 ft/s^2 (that centrifugal force evaporates as you leave the equator towards the poles). That means that gravity is 288 times stronger than the centrifugal force at the equator today.

The math says that the earth would have to spin about 16.9 times faster for the centrifugal force to balance g (the gravitational acceleration). [Acceleration for centrigugal force changes with the square of the velocity; assuming of course that the radius of the earth doesn't change ]. That translates to a rotational period (a new 'day') 5070 seconds (1 hour and 24 minutes).

But this is only for points along the equator. If you were off the equator, things get a little hairier because: 1) radius/velocity on the surface is different 2) gravity is no longer in the same direction as centrifugal force.
To understand a bit how the affect diminishes with latitude, consider that there is 0 centrifugal for to speak of at the north or south pole; so it doesn't matter how fast you spin, you are still bound by gravity.

Other tidbits:

The velocity of a person at the equator today is: 1040 MPH (I assume that folks prefer english units to MKS). The velocity at the surface, to a balanced centrifugal force, is 17800 MPH.

As far as this problem goes:

Speaking of earths circumference I just saw this old nugget: If you created a metal band (no, not the Metallica type!) exactly the circumference of the earth (~40,000 km) and then made it 10 meters longer, how much space would there be between the new band and the earth? Enough for an ant to crawl under? A fox? A man?

Bonus: What if you added 10 meters to a band the circumference of a basketball? How much room would there be between the extended band and the ball?

The answer is very simple to approximate:

C = 2 * Pi *R

a little calculus for differentiation suggests that:

dC = 6.3 * dR # where the d means a change on the variable

so if you change the circumference 10 meters, you change the Radius 10/6.2 meters = 1.59 meters = 5.2 feet (off the earth's ground).

5.2 feet is also the answer to the height above the surface of the basketball.

Larry
DevilHorse

[camion]

I assume you want the answer rather than a full derivation. Permit me a bit of math..

The acceleration of gravity is 32 feet/second^2 (m/s^2); the centrifugal force at the equator is: 0.111 ft/s^2 (that centrifugal force evaporates as you leave the equator towards the poles). That means that gravity is 288 times stronger than the centrifugal force at the equator today.

The math says that the earth would have to spin about 16.9 times faster for the centrifugal force to balance g (the gravitational acceleration). [Acceleration for centrigugal force changes with the square of the velocity; assuming of course that the radius of the earth doesn't change ]. That translates to a rotational period (a new 'day') 5070 seconds (1 hour and 24 minutes).

But this is only for points along the equator. If you were off the equator, things get a little hairier because: 1) radius/velocity on the surface is different 2) gravity is no longer in the same direction as centrifugal force.
To understand a bit how the affect diminishes with latitude, consider that there is 0 centrifugal for to speak of at the north or south pole; so it doesn't matter how fast you spin, you are still bound by gravity.

Other tidbits:

The velocity of a person at the equator today is: 1040 MPH (I assume that folks prefer english units to MKS). The velocity at the surface, to a balanced centrifugal force, is 17800 MPH.

As far as this problem goes:

Speaking of earths circumference I just saw this old nugget: If you created a metal band (no, not the Metallica type!) exactly the circumference of the earth (~40,000 km) and then made it 10 meters longer, how much space would there be between the new band and the earth? Enough for an ant to crawl under? A fox? A man?

Bonus: What if you added 10 meters to a band the circumference of a basketball? How much room would there be between the extended band and the ball?

The answer is very simple to approximate:

C = 2 * Pi *R

a little calculus for differentiation suggests that:

dC = 6.3 * dR # where the d means a change on the variable

so if you change the circumference 10 meters, you change the Radius 10/6.2 meters = 1.59 meters = 5.2 feet (off the earth's ground).

5.2 feet is also the answer to the height above the surface of the basketball.

Larry
DevilHorse

Ah, that brings back memories of the days when I used to do physics, before I slid over into computer work.

I’m actually a bit surprised that the hypothetical day is still over an hour long, but then again when you start squaring things they add up quickly.


Thinking a bit more about it I realize that this is the same thing as orbiting the earth at the equator at an altitude of zero feet. And I recall the announcers back in the day saying it took 90 minutes for the early astronauts to orbit the earth.

[DevilHorse]

On Saturday, an astronomer in the Crimea of all places, discovered (5 days ago - 1/21/2023) an asteroid heading toward earth. The Asteroid is not big (the size of a truck, 10-30 feet) and would burn up in our atmosphere if it entered our atmosphere. It will pass over the southern tip of South America when it visits.

https://www.jpl.nasa.gov/news/nasa-s...arth-this-week

BTW, the speed as it passes will be about 25,000 Miles per hour.

Larry
DevilHorse

Post encounter article:
https://www.cnn.com/2023/01/26/world...scn/index.html

The interesting statement for me:

Before Thursday’s close pass, the asteroid had a circular orbit that took about 359 days to complete around the sun. Now, scientists estimate that the asteroid’s orbit is elongated, extending that single orbit of the sun to 425 days.

In 425 days, 2023 BU should cross roughly in the same spot in space (according to Newtonian mechanics), but (obviously) Earth will be approximately 60 days further along in its' orbit.

Larry
DevilHorse

[DevilHorse]

I found this article about newly discovered moons around Jupiter (not via the JWST BTW):
https://www.cnn.com/2023/02/06/world...f-4807bc0ddd3c

The article also discusses the latest count of moons for the other 3 gas giants.

Good to know that Jupiter is back on top with the most moons; Saturn held the crown for a while.

I haven't yet seen a good discussion of the difference between "a moon" and planet orbiting space rocks. Rocks don't get 'rounded' (much) due to gravity. I'm sure there are plenty of rocks and other things in the rings of the various planets that span the range from large moon to space dust.

Larry
DevilHorse

[Skydog]

I found this article about newly discovered moons around Jupiter (not via the JWST BTW):
https://www.cnn.com/2023/02/06/world...f-4807bc0ddd3c

The article also discusses the latest count of moons for the other 3 gas giants.

Good to know that Jupiter is back on top with the most moons; Saturn held the crown for a while.

I haven't yet seen a good discussion of the difference between "a moon" and planet orbiting space rocks. Rocks don't get 'rounded' (much) due to gravity. I'm sure there are plenty of rocks and other things in the rings of the various planets that span the range from large moon to space dust.

Larry
DevilHorse

To be a moon it has to be round(ish). I know that because moons are by definition romantic and misshapen rocks aren't romantic.

YW. If you have any more scientific questions, don't hesitate to ask!

[DevilHorse]

To be a moon it has to be round(ish). I know that because moons are by definition romantic and misshapen rocks aren't romantic.

YW. If you have any more scientific questions, don't hesitate to ask!

Thank you for that detailed clarification.

I submit for your approval, a competing romantic rock formation:
heartRock1.jpg

Happy Valentines Day (in a week) to all Stargazers.

Larry
DevilHorse

[OldPhiKap]

I haven't yet seen a good discussion of the difference between "a moon" and planet orbiting space rocks.

As long as you can tell the difference between a moon and a space station, I’ll fly with you anytime.

[BLPOG]

As long as you can tell the difference between a moon and a space station, I’ll fly with you anytime.

Frankly, I was disappointed with the lack of "that's no moon" memes related to the recent balloon sighting, but maybe that's a reflection of me avoiding garbage areas of the internet rather than those garbage areas failing to produce the cultural references I want to see.

[DevilHorse]

There was some buzz on the internet about a 1 meter asteroid, recently noticed, that was going to crash into the atmosphere over Western Europe (SE UK) on late Saturday/Sunday morning-ish. A 1 meter asteroid will vaporize/explode and was not predicted to hit the ground.
Here it was:
https://twitter.com/i/status/1624978164597268480

Larry
DevilHorse

[DevilHorse]

Here is the latest in Asteroid Naming:
https://cruxnow.com/vatican/2023/03/...alendar-reform

I suppose an asteroid would have to NOT crash into a planet/moon to receive a name; posthumous (this is an anthropomorphic use) asteroid naming wouldn't make much sense.

Larry
DevilHorse