Your videos on this and the other gravitational fields videos are some of the most helpful physics videos I’ve ever seen, the points were explained perfectly with no filler and helped me understand something in twenty minutes that I’ve been struggling to wrap my head around all year.
This is one of the best channels I have discovered recently. Your explanations and animations are really good. But I was really surprised to see that you have just 400 subs. But, keep making such videos, your channel will surely grow.
Thanks for watching glad to hear it. I’ll admit I just used Keynote with the “magic move” transitions and move animations to get things moving around lol
@@TheBrainFillerone last question: How long it took you to edit this video? Ps: once i was making similar animation video about the variation of gravity due to depth and height but sadly, my project got deleted and I got demotivated, but now you motivated me again to make one again😄 (though i must say managing animation is cumbersome, at least in powerpoint)
Sure, it’s been a while now but I’d say it probably took me a lot longer than it should have. Once you’re comfortable with the pipeline it should theoretically only take a couple hours to get the literal effects made (probably took me longer cause I spent a while thinking about how i wanted it to look as i was doing it)
Did you know the gravitational time dilation on a planets surface is determined by the escape velocity? Its like space is falling down into plant and creating a special relativity flow across the object. What physical model of spacetime would support this equivalence?
Hello, you said the velocity would be 0 at the end of the trip. I have a question regarding this. In an empty space with no resistance at all, if the spaceship turns off the engines once it escapes Earth's gravitational influence, by Newton's first law it will displace infinitely through space with velocity equal to escape velocity (Ve). In this situation where the final velocity is Ve, the final Kinetic Energy will be 1/2(m.Ve)^2. In this case, how can the final mechanical energy be 0?
Ah so the mistake is that you think the spaceship could ever “escape Earth’s gravitational influence”. When I said that the velocity would be 0 at the “end” of the trip (this would happen infinitely far into the future), I was saying that the Earth’s gravitational attraction would keep slowing down the ship as it got further and further away (nothing to do with air resistance or anything like that)
@@TheBrainFiller Thank you for responding so quickly. So if I understand correctly, what you mean is no matter how far the spaceship goes, it will always feel earth's gravitational influence. It will get weaker with increasing distance, but never 0. Is that correct?
Totally, gravity follows an inverse square law using newtons equation so if you look at a graph of 1/r^2 you’ll see it never reaches 0 (though it does tend to 0 as r goes to infinity)
@@TheBrainFiller I've known the inverse square law for 18 years now and I'm wondering why this didn't occur to me instinctively that it can never resolve to 0. Today I feel ridiculously dumb but I learnt a lot. Thank you very very much!
Doesn't this formula suggest that you can escape if youre going the speed of light? So then wouldn't the equation be R < 2GM/c^2 because if R is equal you can escape at the speed of light, where as light can't escape from a black hole.
Perhaps yes based on this analysis but the full GR for a non-rotating black hole shows that the coordinate speed of light goes to 0 *at* the schwarzschild radius.
1:40 You said the KE will be 0 by the end of the trip. But if we launched the object with its escape velocity and assuming there is no friction, shouldnt it maintain its escape velocity even at infinity and therefore should have a non zero kinetic energy?
No because the gravitational field is a applying a “drag” force the whole trip. As a simple example, when you throw an apple in the air the reason it comes to a stop and then falls back down has very little to do friction and a lot more to do with Earth’s gravity I suppose maybe the disconnect is that you’re forgetting that earths gravitational field extends out to infinity which admittedly as a concept is pretty weird when I think about it but that’s the model that works
True escape velocity formula ev = c × sqrt[ 1-exp(-2GM / (Rc²)) ], which is allways less than c, read: All Black Holes Are Escapable, Even from Their Black Surfaces
Sir what If I suppose it is going to the moon... And there is a point that the sum of the gravitional forces which affect over it from the earth and the moon is zero
Yeah you absolutely could it’s a fun question where the key insight is that the distance between the earth and the moon is d then the distance between the earth and the rocket is x and the distance between the rocket and the moon is d-x. Then just solve for x in your force equation
@@TheBrainFiller I did it sir and I got from everything I need like earth mass and the moon and the distance and I got the answer... I just wanna thank you for your replying 🌹🌹
Why is the swarzchild radius calculated using kinetic energy formula m*v^2/2 rather than E^2 = (m*c^2)^2 + (p*c)^2 ? I thought E = m*v^2/2 was only valid for speeds much lower than the speed of light.
You’re totally right. It’s actually a bit of a coincidence that classical Newtonian physics gives the right answer here but technically you’d need General Relativity (but you still end up with the same formula)
It’s essentially a matter of definitions. We define the potential energy to be zero at infinity because that’s where gravity/electric fields no longer influence the object and then we want a change in potential energy to tell us how much work is needed to achieve that configuration. These two requirements force the potential energy to be negative
Straight to the point and not 10+ minutes like others
Well done
Thanks I appreciate that and I’m glad it helped
Your videos on this and the other gravitational fields videos are some of the most helpful physics videos I’ve ever seen, the points were explained perfectly with no filler and helped me understand something in twenty minutes that I’ve been struggling to wrap my head around all year.
Appreciate it, I’m definitely most happy with the way those turned out and my double slit one.
This is one of the best channels I have discovered recently. Your explanations and animations are really good. But I was really surprised to see that you have just 400 subs. But, keep making such videos, your channel will surely grow.
That’s very kind, thanks for watching!
Well done, I like the black hole idea, Thanks very much.
THANK YOU!! The explanation was so good
Your videos are so helpful and simple. They help so much for my a level revision, new favourite physics channel :)
Thanks so much! That’s very kind
Underrated Channel. Keep up the great work buddy!
That’s very kind, thank you!
You just found yourself a new follower. Great channel
Awesome thanks for watching
Truly Helpful.
nice work, glad i found your channel
Thanks for watching
Wow great video, concept crystal cleared!!
Which software did you used make all those beautifull animations?
Thanks for watching glad to hear it. I’ll admit I just used Keynote with the “magic move” transitions and move animations to get things moving around lol
@@TheBrainFiller oh that's cool, is there anything similar to magic move in powerpoint?
Yeah I think in PowerPoint it’s called morph
@@TheBrainFillerone last question: How long it took you to edit this video?
Ps: once i was making similar animation video about the variation of gravity due to depth and height but sadly, my project got deleted and I got demotivated, but now you motivated me again to make one again😄 (though i must say managing animation is cumbersome, at least in powerpoint)
Sure, it’s been a while now but I’d say it probably took me a lot longer than it should have. Once you’re comfortable with the pipeline it should theoretically only take a couple hours to get the literal effects made (probably took me longer cause I spent a while thinking about how i wanted it to look as i was doing it)
Good.. Work..
Did you know the gravitational time dilation on a planets surface is determined by the escape velocity? Its like space is falling down into plant and creating a special relativity flow across the object. What physical model of spacetime would support this equivalence?
Thank you again!
My pleasure, thanks for sticking around and watching more than one video!
best explanation ever
Awesome, thanks for watching
Anouther thought now you have a small black hole the size of 9mm will it eventually get bigger.
Great explication
Thanks for watching!
Hello, you said the velocity would be 0 at the end of the trip. I have a question regarding this.
In an empty space with no resistance at all, if the spaceship turns off the engines once it escapes Earth's gravitational influence, by Newton's first law it will displace infinitely through space with velocity equal to escape velocity (Ve). In this situation where the final velocity is Ve, the final Kinetic Energy will be 1/2(m.Ve)^2. In this case, how can the final mechanical energy be 0?
Ah so the mistake is that you think the spaceship could ever “escape Earth’s gravitational influence”. When I said that the velocity would be 0 at the “end” of the trip (this would happen infinitely far into the future), I was saying that the Earth’s gravitational attraction would keep slowing down the ship as it got further and further away (nothing to do with air resistance or anything like that)
@@TheBrainFiller Thank you for responding so quickly. So if I understand correctly, what you mean is no matter how far the spaceship goes, it will always feel earth's gravitational influence. It will get weaker with increasing distance, but never 0. Is that correct?
Totally, gravity follows an inverse square law using newtons equation so if you look at a graph of 1/r^2 you’ll see it never reaches 0 (though it does tend to 0 as r goes to infinity)
@@TheBrainFiller I've known the inverse square law for 18 years now and I'm wondering why this didn't occur to me instinctively that it can never resolve to 0. Today I feel ridiculously dumb but I learnt a lot. Thank you very very much!
Nice!🖤
Doesn't this formula suggest that you can escape if youre going the speed of light? So then wouldn't the equation be R < 2GM/c^2 because if R is equal you can escape at the speed of light, where as light can't escape from a black hole.
Perhaps yes based on this analysis but the full GR for a non-rotating black hole shows that the coordinate speed of light goes to 0 *at* the schwarzschild radius.
1:40 You said the KE will be 0 by the end of the trip. But if we launched the object with its escape velocity and assuming there is no friction, shouldnt it maintain its escape velocity even at infinity and therefore should have a non zero kinetic energy?
No because the gravitational field is a applying a “drag” force the whole trip. As a simple example, when you throw an apple in the air the reason it comes to a stop and then falls back down has very little to do friction and a lot more to do with Earth’s gravity
I suppose maybe the disconnect is that you’re forgetting that earths gravitational field extends out to infinity which admittedly as a concept is pretty weird when I think about it but that’s the model that works
i just camed my pants great video man
Lol thanks for the support
Does it mean that, conversely, this radius would be where you achieve the speed of light had you fall into it?
Yes is does.
Your videos are great! Keep it up ☺️
Thanks a lot! Do you happen to have any suggestions lol?
Gravity inside the black hole follows Newton's shell theorem. Gravity in the center of a black hole is zero.
True escape velocity formula ev = c × sqrt[ 1-exp(-2GM / (Rc²)) ], which is allways less than c, read:
All Black Holes Are Escapable, Even from Their Black Surfaces
Sir what If I suppose it is going to the moon... And there is a point that the sum of the gravitional forces which affect over it from the earth and the moon is zero
Yeah you absolutely could it’s a fun question where the key insight is that the distance between the earth and the moon is d then the distance between the earth and the rocket is x and the distance between the rocket and the moon is d-x. Then just solve for x in your force equation
@@TheBrainFiller I did it sir and I got from everything I need like earth mass and the moon and the distance and I got the answer... I just wanna thank you for your replying 🌹🌹
Use integration
Why is the swarzchild radius calculated using kinetic energy formula m*v^2/2 rather than E^2 = (m*c^2)^2 + (p*c)^2 ? I thought E = m*v^2/2 was only valid for speeds much lower than the speed of light.
You’re totally right. It’s actually a bit of a coincidence that classical Newtonian physics gives the right answer here but technically you’d need General Relativity (but you still end up with the same formula)
@@TheBrainFiller no u don't end up with same answer the shwardschild radius is different in that case
Oh is it different by a constant or something? Cause really by dimensional analysis you’re gonna have to end up with a very similar formula
@@TheBrainFiller yes sir u r right
How can energy be negative?
It’s essentially a matter of definitions. We define the potential energy to be zero at infinity because that’s where gravity/electric fields no longer influence the object and then we want a change in potential energy to tell us how much work is needed to achieve that configuration. These two requirements force the potential energy to be negative
I thought this guy was an AmeriCan
Lol yeah i have a pretty strong American accent…for someone who isn’t American
@@TheBrainFiller Nice man!
nice!
Thanks!
Schwarzschild*
I just saw + - without a parenthesis on a supposed 'science video'. I just stop watching it.
Lol fair we all have our peeves. Thanks for slogging through as far as you did 😂
Schwarzschild is pronounced more like shwarz-shild
Lol yeah I just anglicised it