Ummm, not sure where they got these numbers from but Earth’s escape velocity is not 7000mph and escaping the sun’s gravitational pull (leaving the solar system from Earth) is not 30,000mph. Respectively the numbers are approximately 25,000mph and 94,000mph. You’re welcome.
Also, even if the manhole cover was going at above 12 km/s the trajectory has to be right for that to result in orbit. Most paths it would take would result in it going up and then coming back down again. Similarly, if somehow it did manage more than 50 km/s and wasn’t destroyed in the atmosphere, it might have the velocity to escape the sun’s gravity, but probably wouldn’t be on the right path to do it. Most likely it would fall into the sun.
So, assuming the 125,000 mph (55 km/s) velocity is correct, the most likely outcome is that it was a reverse-meteor, something that burned up going up through the atmosphere, not down. And even if it did have enough speed to get out of the atmosphere, and there was enough of it left, it most likely fell right back down through the atmosphere somewhere else, either burning up on re-entry or hitting the ground (or the water) somewhere else.
Ignoring that it burned up and ignoring losses due to drag if it somehow didn’t. Isn’t the point of escape velocity that it explicitly won’t come back down.iar least not on earth. Your trajectory won’t matter as you have enough velocity to escape the gravity of earth and will orbit the sun. Further if you managed the solar system escape velocity you will end up orbiting the galactic core. Trajectory doesn’t matter if you have escape velocity. Correct trajectory just minimizes the delta v needed to reach that escape velocity.
Exactly. It’s the minimum speed required to get into orbit assuming you get the direction correct. If you launch vertically, you’ll almost certainly come back down, no matter how far out into space you go. The only consideration is that if you go far enough out you might be influenced by the gravity of something else like the moon which could change your trajectory.
Ah sorry, I should have specified that the post not only got the escape velocity wrong as you pointed out they also got the speed of light wrong near the end.
42.1 km/s is the speed required relative to the sun’s surface for objects launching from Earth’s surface. You need to look at the value labelled V_te, which is the speed relative to the minor body the object is launching from. In this case, it is 16.6 km/s.
Ummm, not sure where they got these numbers from but Earth’s escape velocity is not 7000mph and escaping the sun’s gravitational pull (leaving the solar system from Earth) is not 30,000mph. Respectively the numbers are approximately 25,000mph and 94,000mph. You’re welcome.
Gotta love Tumblr. Just massive amounts of disinformation and bullshit all the time.
Also it would have atomized.
yeah, and it is not “research” to check it. They literally teach it in primary school physics.
Not in our freedom schools!
I mean, what for? Knowing that number isn’t a life skill.
That’s 11.2 km/s and 42.1 km/s.
Also, even if the manhole cover was going at above 12 km/s the trajectory has to be right for that to result in orbit. Most paths it would take would result in it going up and then coming back down again. Similarly, if somehow it did manage more than 50 km/s and wasn’t destroyed in the atmosphere, it might have the velocity to escape the sun’s gravity, but probably wouldn’t be on the right path to do it. Most likely it would fall into the sun.
So, assuming the 125,000 mph (55 km/s) velocity is correct, the most likely outcome is that it was a reverse-meteor, something that burned up going up through the atmosphere, not down. And even if it did have enough speed to get out of the atmosphere, and there was enough of it left, it most likely fell right back down through the atmosphere somewhere else, either burning up on re-entry or hitting the ground (or the water) somewhere else.
Ignoring that it burned up and ignoring losses due to drag if it somehow didn’t. Isn’t the point of escape velocity that it explicitly won’t come back down.iar least not on earth. Your trajectory won’t matter as you have enough velocity to escape the gravity of earth and will orbit the sun. Further if you managed the solar system escape velocity you will end up orbiting the galactic core. Trajectory doesn’t matter if you have escape velocity. Correct trajectory just minimizes the delta v needed to reach that escape velocity.
At least that’s all my recollection.
Escape velocity means you could stay in orbit. It doesn’t guarantee anything if you launch at the wrong angle.
Exactly. It’s the minimum speed required to get into orbit assuming you get the direction correct. If you launch vertically, you’ll almost certainly come back down, no matter how far out into space you go. The only consideration is that if you go far enough out you might be influenced by the gravity of something else like the moon which could change your trajectory.
That is not the definition of escape velocity. Escape velocity is the minimum velocity to escape a body’s gravity well entirely. Orbital is much lower
That is not the definition of escape velocity. Escape velocity is the minimum velocity to escape a body’s gravity well entirely. Orbital is much lower
I like how they are implying the speed of light is only 500000mph (as opposed to 671,000,000 mph or 1,080,000,000kph)
?
Ah sorry, I should have specified that the post not only got the escape velocity wrong as you pointed out they also got the speed of light wrong near the end.
I didn’t see that reference but the image is all blurry now.
94000mph is relative to the sun’s surface. Relative to the Earth’s surface, it is around 37000mph, which means they were still wrong.
Not according to Wikipedia. At Earth relative to Sun is 42.1 km/s = 94,175 mph. From the sun’s surface is 617.5 km/s = 1.38 x 10e6 mph.
Source: https://en.m.wikipedia.org/wiki/Escape_velocity
42.1 km/s is the speed required relative to the sun’s surface for objects launching from Earth’s surface. You need to look at the value labelled V_te, which is the speed relative to the minor body the object is launching from. In this case, it is 16.6 km/s.