Playing Kerbal Space Program has led me to this channel. Building a comm relay network, like everything in KSP, is a sharp learning curve. Your vids are very informative, thank you.
Wow. I've been looking all over for instructions how to calculate the orbital period and this quality content has provided me what I've been looking for. Thank you so much for making this as easy for first timer tackling this subject.
Ahhhh so that is why the "t" is capitalized. I picked up a physics final exam from a professor at my local college to see if i could pass physics 1. And the exam is pretty straightforward but there are a few things that i havent seen. The "T" being one of those things. And i remember from reading an old book on vectors that sometimes people will capitalize or italicize or bold a vector instead of using an arrow to indicate. Cleared that up for me. Thank you!
Excellent video. Thanks for the help. I really like the way you show the equations being derived, as this give a lot more understanding for the students.
Hey, I know you probably won't see this but thank you so much. I have been programming a satellite system calculator, and need the orbital period. I found one video on how to do that, but it uses a standard gravitational parameter which would be really hard to program in, and I couldn't figure out period. After watching this though, I know not only how to program it, but what it all means. Thank You.
@@stepbystepscience Oval orbits, like the orbit of Pluto around the sun. For example: A rocket orbits earth in a circular orbit at 100 km height, with a velocity of 7484 m/s. Then if the rocket accelerates (velocity prograde) with 1000 m/s for example, what will the orbital path be? And how long would it take for 1 orbit?
What if orbits are not circular, and have a noted difference between apogee and perigee. The velocity will change, but does the formula for orbital period change?
The equations for elliptical orbit are different. Here is the link to a website that discusses the differences: web.njit.edu/~gary/320/Lecture6.html#:~:text=Orbital%20Velocity,be%20a%20very%20useful%20equation.&text=%3D%20%5B2pa%20%2F%20P(1,%5D%20(e%20sin%20q).
perhaps, you should also show the difference between elliptical orbit and the circular orbit theory; which one applies in what given situation; I first tried this given formulae (applicable to circular orbit theory), but failed when I used it to solve orbital period in the case of 'low Earth orbiting satellites'. However, it was solved correctly when 'elliptical orbit' theory using Kepler's third law on his laws of planetary motion was applied. Thanks.
م is it the same equation used to find orbital period when it's circular orbit only or I can use it when I have ellipse orbit (with perigee and apogee)
Thank you for the video! But right at the crucial moment, I'm lost -- @8:53, where did those figures for gravity and mass come from?? I'm studying this tutorial to try to find the mass of Saturn given Titan's orbital period, mass, and orbit radius. The problem I have is that even after all of these equation conversions and simplifications I still don't have enough information. Even if I rearrange your equation (at 8:53) to find M instead of T, I don't have a figure for G :( I'm a layman but even I figure I can't use 9.81 unless the object is 'on' earth?! I also suspect that my lecturer doesn't want me Googling the gravity on Saturn. Can anybody hellllp? I'm a maths illiterate BioMed student who's accidentally found herself in a physics course and my brain has left the building.
I'm very confused, I'm watching videos like this because I'm back into KSP and love it but also want to learn the math, and I'm confused because both this video and another of an actual university professor, both have equations that seem to get the wrong orbital period. I know I don't have any mods installed to give me precisely the exact period, but I have timed it and my equation comes out right (within error) while the equation here at 6:45 is wildly wrong. So in my spreadsheet instead I have been using: P=((2*pi*a^3)/GM)^(1/2) P being the period, a being the semi-major axis, and this equation has been working for me, but the equation here in the video and of the other video lecture I saw, both come out to the wrong values. Does this equation in the video only work on circular orbits? (did both mine and yours in the spreadsheet assuming a circular orbit with 0 eccentricity, yea seems to check out. So really it should be "can find the period given a radius only if your orbit has an eccentricity of 0) EDIT: Also now even more confused. Maybe just because my math skills are rusty from years out of school, but I'm trying to figure out how these two equations are related since one doesn't seem to reduce to the other so I set one as equal to the other and cancelled out all terms until I get pi=2*pi^2 and thought "oh this works out because pi is for a circle and the orbits will neatly divide into each other, like how going 360 and then another object going 720, they both land at the same spot. But nope, they look like they don't fit together whatsoever...
Should not the force of gravity of the Earth on m2 mass have opposite direction and be representing the centripetal force that compels the smaller mass m2 to deviate from the tangential force of its circular motion? In such situation m2 is subject to two opposite equal forces : the centrifugal force represented by m2*v^2/r and the gravitational force represented by m*M*G/r^2 ? In my opinion it is the attractive gravitational force acting from Earth on m2 that represents the centripetal force and its direction is always towards the centre of the earth while in the drawing it seems having the opposite direction while any body who moves in circular motion is subject to a centrifugal force.
If you are talking about the last problem then it is the orbital period, the time it takes for the satellite to go once around the Earth. Does that help?
ohhhhh i understand now,thank you, thank you, thank you...i love your videos and i have so many questions but first i am gonna watch all your videos...
Very excellent explanation but looking at those big sums I think I will get senseless. Sir plz tell any short trick for how to solve these big calculations in paper pen as in my school under cbse board doesn't allow us to use calculators 😥😢
I've tried this for hours now breaking down every little detail how are u getting 43,058s cause 4x3.14^2x26552000^3 = 7.4x10^23 The u divide that by mu 4x10^14 then u square that which gives u 4.3x10^18 how is that 43,058s
I am not sure what you are doing wrong but (7.4x10^23)/(4.0x10^14) = 1,85x10^9, and then you have to take the square root of that and you get about 43,000 seconds. Does that help?
How is this done, if the orbis is NOT circular? You should at least point out that this ONLY works for circular orbits. That is a special case of orbit, as you know.
@aa bb General. Relativity... Basically the spacetime is distorted by energy in it. Straight lines on a curved surface are called geodesics and those are calculated by Einsteins equation for the 4D spacetime. All objects follow them to the largest distortions of the spacetime, we call it gravity. The geodesics are what create orbits, so they work, and in fact explain, eliptical orbits. I left some stuff out, but that a breief overview.
Can I conclude that the mass m2 is irrelevant? What if mass m2 is very large ( let's say =1/2m1) . Does't it have also a strong gravitational force acting on m1? In your sample there seems to be only one gravitational force. Another question: why is the orbit of the Earth around the Sun not perfectly circular? Is that because of an imbalnace caused by the orbiting moon, hence changing the centre of gravity of their combined mass and hence a subtiel change in r?
No matter HOW many times I try, i CANNOT get that answer with my calculator. I need a video for "dummies" about how to type all this into my TI-83plus. I have to be missing a parenthesis somewhere or something.... :(
Are you using the exponent key on your calculator to put in number with scientific notation. Otherwise your calculator might do the wrong order of operations.
- can you solve this problem for me ? - mass of the earth is 81 times the mass of the moon and the distance between th earth and moon is 60 times th radius of the earth.If R is th radius of th earth ,then th distance between th moon and the point on the line joining th moon and earth where the gravitational force becomes zero is .
Playing Kerbal Space Program has led me to this channel. Building a comm relay network, like everything in KSP, is a sharp learning curve. Your vids are very informative, thank you.
Same, watched this video because I'm planning on setting up a geostationary satellite network around kerbin
Same had some difficulties in building a geostationary network
same haha
This whole channel has been my teacher (and savior) this entire virtual semester. Thank you!!!!!!!!
Such a nice compliment, thank you so much!
Wow. I've been looking all over for instructions how to calculate the orbital period and this quality content has provided me what I've been looking for. Thank you so much for making this as easy for first timer tackling this subject.
Great to hear! Glad it was helpful and thanks for posting your comment.
Thank you so much for this series! It's extremely clear and simple to understand :)
You're very welcome! Glad you like the series.
Thanks for fixing my physics mark. Wished my teacher taught as well as u did
Any time, so glad to be helpful!
Excellent videos, crystal clear both in terms of the theory and the practical excercises!!! Thanks a lot
You're very welcome! Glad you like them.
Ahhhh so that is why the "t" is capitalized.
I picked up a physics final exam from a professor at my local college to see if i could pass physics 1. And the exam is pretty straightforward but there are a few things that i havent seen. The "T" being one of those things. And i remember from reading an old book on vectors that sometimes people will capitalize or italicize or bold a vector instead of using an arrow to indicate.
Cleared that up for me. Thank you!
Glad to have cleared something up!
I would say the T is capitalizes because it is a particular type of time, orbital period (T), not just regular time (t) which gets the lower case t.
Excellent video. Thanks for the help. I really like the way you show the equations being derived, as this give a lot more understanding for the students.
You're very welcome! Glad you like my step by step approach.
Hey, I know you probably won't see this but thank you so much. I have been programming a satellite system calculator, and need the orbital period. I found one video on how to do that, but it uses a standard gravitational parameter which would be really hard to program in, and I couldn't figure out period. After watching this though, I know not only how to program it, but what it all means. Thank You.
I'm very happy to hear that my video was so helpful to you. Thanks for the nice comment.
Well demonstrated!
Good to hear, thanks
My brain exploded after trying to figure out what any of this means
That was quick and easy. Thank you very much
Step by Step....thanks for the comment.
Great video! Please make a video about not circular orbits.
What exactly do you mean?
@@stepbystepscience Oval orbits, like the orbit of Pluto around the sun.
For example: A rocket orbits earth in a circular orbit at 100 km height, with a velocity of 7484 m/s. Then if the rocket accelerates (velocity prograde) with 1000 m/s for example, what will the orbital path be? And how long would it take for 1 orbit?
@@youtubetv1987 you use the semi-major axis of the ellipse instead of the circle radius.
@@MeTube3 oribital period would still be the same.
What if orbits are not circular, and have a noted difference between apogee and perigee. The velocity will change, but does the formula for orbital period change?
The equations for elliptical orbit are different. Here is the link to a website that discusses the differences:
web.njit.edu/~gary/320/Lecture6.html#:~:text=Orbital%20Velocity,be%20a%20very%20useful%20equation.&text=%3D%20%5B2pa%20%2F%20P(1,%5D%20(e%20sin%20q).
Thank you very much.
perhaps, you should also show the difference between elliptical orbit and the circular orbit theory; which one applies in what given situation; I first tried this given formulae (applicable to circular orbit theory), but failed when I used it to solve orbital period in the case of 'low Earth orbiting satellites'. However, it was solved correctly when 'elliptical orbit' theory using Kepler's third law on his laws of planetary motion was applied. Thanks.
Yes, I need to do videos on Kepler's laws and elliptical orbits. Thanks for the comment.
it helped me a lot! thank you :))
You're welcome!
Awesome explanation. 😀
Thanks for the comment.
Awesome video
Thanks!
م is it the same equation used to find orbital period when it's circular orbit only or I can use it when I have ellipse orbit (with perigee and apogee)
This is the equation you use for a circular orbit.
You are amazing
Thank you so much!
Perfect, Thanks!
Welcome!
What happens to distance when mass is doubled and period remain the same?
Then the distance remains the same. Mass does not effect velocity.
Thanks well done..
Most welcome
if this video was posted 3 days ago, i would have passed my exam.. xD.. thank you anyways sir, it really helps 😀
Shush
Thank you for the video! But right at the crucial moment, I'm lost -- @8:53, where did those figures for gravity and mass come from?? I'm studying this tutorial to try to find the mass of Saturn given Titan's orbital period, mass, and orbit radius. The problem I have is that even after all of these equation conversions and simplifications I still don't have enough information. Even if I rearrange your equation (at 8:53) to find M instead of T, I don't have a figure for G :( I'm a layman but even I figure I can't use 9.81 unless the object is 'on' earth?! I also suspect that my lecturer doesn't want me Googling the gravity on Saturn. Can anybody hellllp? I'm a maths illiterate BioMed student who's accidentally found herself in a physics course and my brain has left the building.
Gravitational Constant (G) and the radius of Saturn
I'm very confused, I'm watching videos like this because I'm back into KSP and love it but also want to learn the math, and I'm confused because both this video and another of an actual university professor, both have equations that seem to get the wrong orbital period. I know I don't have any mods installed to give me precisely the exact period, but I have timed it and my equation comes out right (within error) while the equation here at 6:45 is wildly wrong. So in my spreadsheet instead I have been using:
P=((2*pi*a^3)/GM)^(1/2)
P being the period, a being the semi-major axis, and this equation has been working for me, but the equation here in the video and of the other video lecture I saw, both come out to the wrong values. Does this equation in the video only work on circular orbits?
(did both mine and yours in the spreadsheet assuming a circular orbit with 0 eccentricity, yea seems to check out. So really it should be "can find the period given a radius only if your orbit has an eccentricity of 0)
EDIT: Also now even more confused. Maybe just because my math skills are rusty from years out of school, but I'm trying to figure out how these two equations are related since one doesn't seem to reduce to the other so I set one as equal to the other and cancelled out all terms until I get pi=2*pi^2 and thought "oh this works out because pi is for a circle and the orbits will neatly divide into each other, like how going 360 and then another object going 720, they both land at the same spot. But nope, they look like they don't fit together whatsoever...
Would the formula also work in spacial proportions, like a planet around a star?
Yes
@@stepbystepscience would this formula also be the similar to work out the local time on a planet?
Should not the force of gravity of the Earth on m2 mass have opposite direction and be representing the centripetal force that compels the smaller mass m2 to deviate from the tangential force of its circular motion? In such situation m2 is subject to two opposite equal forces : the centrifugal force represented by m2*v^2/r and the gravitational force represented by m*M*G/r^2 ? In my opinion it is the attractive gravitational force acting from Earth on m2 that represents the centripetal force and its direction is always towards the centre of the earth while in the drawing it seems having the opposite direction while any body who moves in circular motion is subject to a centrifugal force.
how would i manipulate this to find mass?
That would be M1 = (4p² . r³) / ( G . T²) with p as pie, because I couldn't find it on my keyboard.
What if mass is are equal which M do you pick or do the 2 objects not orbit each other
How would we solve for T if given the distance and mass only? Not the radius
That is all you need, because G is the gravitational constant.
How do I type that into the calculator? I'm not getting that answer
Do it one piece at a time. Top, bottom, divide then take the square root.
Could you have square when you had velocity^2?
whait,please,you said 12 hours,is that before object fall to the earth or before he lose some orbital height?please answer...
If you are talking about the last problem then it is the orbital period, the time it takes for the satellite to go once around the Earth. Does that help?
ohhhhh i understand now,thank you, thank you, thank you...i love your videos and i have so many questions but first i am gonna watch all your videos...
@@LOVACKOJINEUBIJA1 you are very welcome, thanks for watching and commenting.
Could someone please explain to me how to put Nm^2/Kg^2 into a calculator? Sorry if that's a stupid question I am a bit confused
Good question and a common one, you do not "put" the units into the calculator, you just put the number into the calculator. Do you know what I mean?
@@stepbystepscience Yes I understand. Thank you!
Very excellent explanation but looking at those big sums I think I will get senseless.
Sir plz tell any short trick for how to solve these big calculations in paper pen as in my school under cbse board doesn't allow us to use calculators 😥😢
But that is the way it works.
@@stepbystepscience sir that I have understood but if I can't do these big calculations then everything will go in vain .
Mate u have to do it I too am facing the problem
He'll o Sir Good Evening from India
Greetings from Germany
nice video
Thanks for the comment.
What calculator do I use to calculate the final equation for T? every calculator I tried just gets an error
The error message might mean you are not entering the numbers correctly.
I've tried this for hours now breaking down every little detail how are u getting 43,058s cause 4x3.14^2x26552000^3 = 7.4x10^23
The u divide that by mu 4x10^14 then u square that which gives u 4.3x10^18 how is that 43,058s
I am not sure what you are doing wrong but (7.4x10^23)/(4.0x10^14) = 1,85x10^9, and then you have to take the square root of that and you get about 43,000 seconds. Does that help?
YESSSS, helped me! thank you!!@@stepbystepscience
saved my life
Wonderful, glad to hear it!
thank you!
You are welcome, thanks for commenting.
Is time in years?
Seconds
Why is the pie there and what am i going to solve it with
Circular orbit, circumference is 2 x pi x r.
How did you calculate for minutes and seconds!?
Divide the 43,058 by 60 for minutes, and again by 60 for hours. For days, divide by 24 again.
Thank you
How is this done, if the orbis is NOT circular?
You should at least point out that this ONLY works for circular orbits. That is a special case of orbit, as you know.
Relativity
@aa bb General. Relativity...
Basically the spacetime is distorted by energy in it. Straight lines on a curved surface are called geodesics and those are calculated by Einsteins equation for the 4D spacetime. All objects follow them to the largest distortions of the spacetime, we call it gravity. The geodesics are what create orbits, so they work, and in fact explain, eliptical orbits. I left some stuff out, but that a breief overview.
You're making me regret not having an avid physic teacher
The answer I'm getting is 1.36231 Help! What do I do?
Madison Feehan 😅😅😅
You probably assumed a height of 20 kilometeres, but the satellite is 20 thousand kilometers away from earth surface. So 26552000 m.
thx
you are welcome.
This is starting to hurt my head but I think I get it I don’t think I would be doing this my first year of high school
You got this!
Oh thank goodness 😭
happy to hear that!
what about the orbital period of an asteroid
Samething or do you mean an elliptical orbit?
Can I conclude that the mass m2 is irrelevant? What if mass m2 is very large ( let's say =1/2m1) . Does't it have also a strong gravitational force acting on m1? In your sample there seems to be only one gravitational force.
Another question: why is the orbit of the Earth around the Sun not perfectly circular? Is that because of an imbalnace caused by the orbiting moon, hence changing the centre of gravity of their combined mass and hence a subtiel change in r?
Nice :D
I'm completely lost where your value for G comes from.
It is a constant
No matter HOW many times I try, i CANNOT get that answer with my calculator. I need a video for "dummies" about how to type all this into my TI-83plus. I have to be missing a parenthesis somewhere or something.... :(
Are you using the exponent key on your calculator to put in number with scientific notation. Otherwise your calculator might do the wrong order of operations.
- can you solve this problem for me ?
- mass of the earth is 81 times the mass of the moon and the distance between th earth and moon is 60 times th radius of the earth.If R is th radius of th earth ,then th distance between th moon and the point on the line joining th moon and earth where the gravitational force becomes zero is .
freecodecamp sent me here :/
Who's here from freeCodeCamp's JavaScript Intermediate Algorithm Scripting?
Thanks!
No problem!