I have accidentally done this in KSP when i figured out that i didnt have enough delta v to return to kerbin normally like they do in the videos but then i remembered where the most efficient place to burn was
@@mita_pita He's referring to the game Kerbal Space Program (KSP), in which a real-life space program is simulated. Unlike the reality, rockets are programmed and controlled by humanoids called Kerbals, not humans. So that is
Yes, a spacecraft can lift off from the earth and ascend to a specific orbital altitude directly. It gets a little more complicated if you are trying to meet up with something already in orbit (i.e. space station) when precise timing is required and you might not be able to launch at precisely the right time. The spacecraft may be sent to an initial orbit, then wait for the right time to make an orbital change. Ideal Hohmann transfers aren't used in these cases, but rather a spiral transfer is used which can take 2 or 3 days for a cargo capsule to "catch up" to the space station... And you can't just "speed up" to catch up the space station, because as your velocity changes so does your orbital altitude. The Hohmann transfer is most useful (and obvious) when getting from Earth to Mars (or another planet). The spacecraft starts at earth's orbit (the lower orbit) and then makes the transfer to Mar's orbit (the higher orbit). The same is true for getting to the moon. The moon-bound spacecraft first gets into a parking orbit and then makes the transfer to the moon's orbit when the timing is right. You may see movies that show a "figure eight" flight path to the moon and back. It might look like that in a static picture, but in reality a Hohmann transfer orbit is used.
@@amitwaghamare2624 its really hard to be jealous when you landed on the moon. Nevertheless, he was pointing out that India's accomplishment was built on the works of others, so the first guy shouldn't act as though it was a work done on its own. In fact, many feats of the space race were "first attempts"
At the higher orbit , further from the gravity well, the orbital velocity is lower , so when is the ΔV changed to circularize the second orbit? Equations : V^2/r = G*Me/r^2 mass of spacecraft cancels out V=( (G*Me)/r)^0.5 So at low Earth orbit G=6.67E-11 Me=5.98E24 kg r1= 6.6E6 meters r2=1.32E7 meters Low Earth orbit 200 km above the Earth’s surface V= 7,773.9 m/s High Earth orbit 6600 km V= 5,497m/s
Good question Thales, the difference is you need to compare the velocity at the apogee of the elliptical transfer orbit to the final orbit, not the lower orbit. It's counter-intuitive because your calculations above are spot on - so why do I need to speed up to go slower? :-) So, using your equations above, if I am doing 7.8km/s and I add (for example) a delta V of 2km/s my orbit becomes elliptical (doing 9.8km/s at perigees and 2km/s at apogee). The apogee of my new elliptical orbit is at the altitude of my desired higher orbit but, by the time I get there I have traded a bunch of kinetic energy for potential energy and my velocity has dropped to 2km/s as I prepare to fall back to perigee. But, if I add 3.5km/s of delta v I achieve the 5.5km/s you calculate above and my orbit remains circular instead of returning to my previous altitude (the perigee of my elliptical orbit). I hope that helps, it's challenging to describe without a whiteboard.
In the elliptical orbit at apogee is the spacecraft capable of docking with a station in the higher orbit, assuming the timing is correct for the two to be very close? If so wouldn't this result in the new docked pair to enter an elliptical orbit because they're net velocity is lower than that required for the original higher circular orbit? I'm dying to know the answer, thank you.
It's actually possible to do that but in real life engineers don't need to do that way as it is more complicated and risky(because once they're docked, the ISS let's supose, needs to accelerate as the mass is bigger now as there are two objects, spacecraft and ISS. Adding this the precise timing, you see it's not easy this way). Instead the spacecraft burns once for reaching the higher orbit (letting ISS pass to have it in front of you in your way, to prepare for docking) and once to STAY at this higher orbit. Then, with the ISS in front of the spacecraft, slightly increase the velocity to reach the docking point.
You need to change the velocity of the rocket, or give it a Delta-V. F=ma comes into play, your thrust force creates an acceleration a = Thrust Force / Mass (units = m/sec2 or ft/sec2). If you burn your motor for X seconds you get a velocity change. Velocity Change = accel x time = m/sec2 x sec = m/sec...
The first burst should be fired at the perigee of the transfer ellipse as oriented at time mark 0:55 . You shouldn't be doing any velocity change in respect to apogee or perigee of the original orbit because this Hohmann transfer method only applies to 2 circular orbits, or 2 orbits with eccentricity of 0. The calculations would change if the orbits where not in fact circular.
A prograde burn at perigee will raise your orbit. A retrograde burn at apogee will lower your orbit. That's because you're slower the further you are away from the Earth or planet
The rocket needs to be in a retrograde position, (nose facing away from the direction of the orbit). Then your burns should look the same, closing the orbit
Hi Adeel. It isn't the same, but you might use Hohmann transfers to reach a cycler orbit. Importantly, the Mars-Earth (Aldrin) cycler is heliocentric (i.e. it orbits the sun) so you need to consider leaving Earth orbit and establishing a separate Sun orbit in order to 'cycle' between Earth and Mars. Once you're in the cycler orbit you only need to correct for perturbations ('wobbles' in your orbit) to stay there.
@@stevenhenry3714 Thank you for taking the time to share this information with me. So basically you are suggesting. 1) Leave earth 2) Establish an orbit around the sun 3) Wait for the right alignment and then head off towards one's destination?
I'm not sure if I'm 100% correct but this is what I got: the Hohmann transfer orbit is basically using an elliptical orbit to transfer a satellite from one existing smaller and lower orbit to a bigger and higher one. To first leave the small orbit it's sped up until it leaves the org track, but by leaving this it creates a new elliptical orbit. After leaving the org track it slows down, but when it reaches a certain point, aka the apogee (since when it left the org track it created an elliptical orbit so we have apogee and perigee now yay), it speeds up again and turns slightly away from the orbit it was directed to earlier. now it's fully transferred. 1. speeds up to leave the org circular orbit and forms an elliptical orbit by doing so 2. speeds up again when it reaches apogee 3. change direction by a bit after speeding up 3. at new orbit I hope it helped and gl with your studies! :DD
I have accidentally done this in KSP when i figured out that i didnt have enough delta v to return to kerbin normally like they do in the videos but then i remembered where the most efficient place to burn was
When I have this kind of problem I pull out my fire extinguisher and start flying from space station to space station. #GravityWasStupid
This is taught in every kerbal second grade
What's kerbal
@@mita_pita He's referring to the game Kerbal Space Program (KSP), in which a real-life space program is simulated. Unlike the reality, rockets are programmed and controlled by humanoids called Kerbals, not humans. So that is
Hahahahaha
Nice, thank you - I read a web page that didn't explain it well; this was lovely and exactly what I needed. Thank you again.
why we need a hoffman transfer, I mean can't the spacecraft get directly from the earth to the desired orbit, without orbiting a smaller one ?
Yes, a spacecraft can lift off from the earth and ascend to a specific orbital altitude directly. It gets a little more complicated if you are trying to meet up with something already in orbit (i.e. space station) when precise timing is required and you might not be able to launch at precisely the right time. The spacecraft may be sent to an initial orbit, then wait for the right time to make an orbital change. Ideal Hohmann transfers aren't used in these cases, but rather a spiral transfer is used which can take 2 or 3 days for a cargo capsule to "catch up" to the space station... And you can't just "speed up" to catch up the space station, because as your velocity changes so does your orbital altitude. The Hohmann transfer is most useful (and obvious) when getting from Earth to Mars (or another planet). The spacecraft starts at earth's orbit (the lower orbit) and then makes the transfer to Mar's orbit (the higher orbit). The same is true for getting to the moon. The moon-bound spacecraft first gets into a parking orbit and then makes the transfer to the moon's orbit when the timing is right. You may see movies that show a "figure eight" flight path to the moon and back. It might look like that in a static picture, but in reality a Hohmann transfer orbit is used.
@@labratscientific1127 thank you so much I really appreciate that
nice and informative sir
Actually the 2nd thing is viable too if you burn radially inward...
This is how India accomplished to become the first country to do a successful interplanetary mission in its first attempt.😁
Well, they had plenty of historical examples to draw off of
@@chain3519 So you mean everyone should be able to do it now right...
@@vhwnos he jealous
@@amitwaghamare2624 its really hard to be jealous when you landed on the moon. Nevertheless, he was pointing out that India's accomplishment was built on the works of others, so the first guy shouldn't act as though it was a work done on its own. In fact, many feats of the space race were "first attempts"
At the higher orbit , further from the gravity well, the orbital velocity is lower , so when is the ΔV changed to circularize the second orbit?
Equations :
V^2/r = G*Me/r^2
mass of spacecraft cancels out
V=( (G*Me)/r)^0.5
So at low Earth orbit
G=6.67E-11
Me=5.98E24 kg
r1= 6.6E6 meters
r2=1.32E7 meters
Low Earth orbit 200 km above the Earth’s surface
V= 7,773.9 m/s
High Earth orbit 6600 km
V= 5,497m/s
Good question Thales, the difference is you need to compare the velocity at the apogee of the elliptical transfer orbit to the final orbit, not the lower orbit. It's counter-intuitive because your calculations above are spot on - so why do I need to speed up to go slower? :-)
So, using your equations above, if I am doing 7.8km/s and I add (for example) a delta V of 2km/s my orbit becomes elliptical (doing 9.8km/s at perigees and 2km/s at apogee). The apogee of my new elliptical orbit is at the altitude of my desired higher orbit but, by the time I get there I have traded a bunch of kinetic energy for potential energy and my velocity has dropped to 2km/s as I prepare to fall back to perigee. But, if I add 3.5km/s of delta v I achieve the 5.5km/s you calculate above and my orbit remains circular instead of returning to my previous altitude (the perigee of my elliptical orbit). I hope that helps, it's challenging to describe without a whiteboard.
In the elliptical orbit at apogee is the spacecraft capable of docking with a station in the higher orbit, assuming the timing is correct for the two to be very close? If so wouldn't this result in the new docked pair to enter an elliptical orbit because they're net velocity is lower than that required for the original higher circular orbit? I'm dying to know the answer, thank you.
It's actually possible to do that but in real life engineers don't need to do that way as it is more complicated and risky(because once they're docked, the ISS let's supose, needs to accelerate as the mass is bigger now as there are two objects, spacecraft and ISS. Adding this the precise timing, you see it's not easy this way). Instead the spacecraft burns once for reaching the higher orbit (letting ISS pass to have it in front of you in your way, to prepare for docking) and once to STAY at this higher orbit. Then, with the ISS in front of the spacecraft, slightly increase the velocity to reach the docking point.
this is just me changing orbit in ksp no ifs ands or buts this is all i use when prograde and retrograde
The principal is relatively easy to understand. The bit that does my head in is how you work out how long you burn the engines for.
You need to change the velocity of the rocket, or give it a Delta-V. F=ma comes into play, your thrust force creates an acceleration a = Thrust Force / Mass (units = m/sec2 or ft/sec2). If you burn your motor for X seconds you get a velocity change. Velocity Change = accel x time = m/sec2 x sec = m/sec...
Do you fire the rocket motors at Apogee of the original orbit in a prograde orientation against the Earth?
The first burst should be fired at the perigee of the transfer ellipse as oriented at time mark 0:55 . You shouldn't be doing any velocity change in respect to apogee or perigee of the original orbit because this Hohmann transfer method only applies to 2 circular orbits, or 2 orbits with eccentricity of 0. The calculations would change if the orbits where not in fact circular.
@@tempestandacomputer6951 thanks!
A prograde burn at perigee will raise your orbit. A retrograde burn at apogee will lower your orbit. That's because you're slower the further you are away from the Earth or planet
thanks
How do I transfer from high orbit to lower orbit?
The rocket needs to be in a retrograde position, (nose facing away from the direction of the orbit). Then your burns should look the same, closing the orbit
basically burn the other way
burn backwards
I know exactly what's going on because I'm a ksp major
Is this the same as a cycler? en.wikipedia.org/wiki/Mars_cycler
Hi Adeel. It isn't the same, but you might use Hohmann transfers to reach a cycler orbit. Importantly, the Mars-Earth (Aldrin) cycler is heliocentric (i.e. it orbits the sun) so you need to consider leaving Earth orbit and establishing a separate Sun orbit in order to 'cycle' between Earth and Mars. Once you're in the cycler orbit you only need to correct for perturbations ('wobbles' in your orbit) to stay there.
@@stevenhenry3714 Thank you for taking the time to share this information with me. So basically you are suggesting. 1) Leave earth 2) Establish an orbit around the sun 3) Wait for the right alignment and then head off towards one's destination?
@@AdeelKhan1 Yep, that's about it :-)
Nice name
Ali fazal was copying Raju Srivastava
Nice video. Very interesting. However please don't eat while recording. For a few of us is uterly impossible to stand.
still makes no goddamn sense to me
I'm not sure if I'm 100% correct but this is what I got: the Hohmann transfer orbit is basically using an elliptical orbit to transfer a satellite from one existing smaller and lower orbit to a bigger and higher one.
To first leave the small orbit it's sped up until it leaves the org track, but by leaving this it creates a new elliptical orbit. After leaving the org track it slows down, but when it reaches a certain point, aka the apogee (since when it left the org track it created an elliptical orbit so we have apogee and perigee now yay), it speeds up again and turns slightly away from the orbit it was directed to earlier. now it's fully transferred.
1. speeds up to leave the org circular orbit and forms an elliptical orbit by doing so
2. speeds up again when it reaches apogee
3. change direction by a bit after speeding up
3. at new orbit
I hope it helped and gl with your studies! :DD