How fast are trips to the Moon for unmanned spacecraft typically?
$begingroup$
The period of an orbit around the Earth is given by:
$$T = 2 pi sqrt{a^3/GM}$$
For Earth GM is about 3.986E+14 m^3/s^2. Put in $a$ = ((6378+400)*1000) for the ISS in LEO (as a test) and you get 5553 seconds, or 92.6 minutes.
Put in $a$ = (384400*1000) for the Moon's orbit and you get 27.5 days, close enough. (We didn't correct for reduced mass).
A ellipse with a perigee in LEO and an apogee at the Moon ($a$ = (195589*1000)) has a period of 10.0 days and a half period of 5.0 days.
But the Apollo times from LEO to the Moon were only about 3.0 to 3.3 days, and Chang'e-4 will only be about 3.9 days Liftoff at 18:23 on 7 December UTC and lunar orbit on ~1500 hour UTC on December 11th. See also this answer.
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
Or is there something about the mass of the Moon that makes the trip naturally quicker?
orbital-mechanics planning cislunar
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
$endgroup$
|
show 3 more comments
$begingroup$
The period of an orbit around the Earth is given by:
$$T = 2 pi sqrt{a^3/GM}$$
For Earth GM is about 3.986E+14 m^3/s^2. Put in $a$ = ((6378+400)*1000) for the ISS in LEO (as a test) and you get 5553 seconds, or 92.6 minutes.
Put in $a$ = (384400*1000) for the Moon's orbit and you get 27.5 days, close enough. (We didn't correct for reduced mass).
A ellipse with a perigee in LEO and an apogee at the Moon ($a$ = (195589*1000)) has a period of 10.0 days and a half period of 5.0 days.
But the Apollo times from LEO to the Moon were only about 3.0 to 3.3 days, and Chang'e-4 will only be about 3.9 days Liftoff at 18:23 on 7 December UTC and lunar orbit on ~1500 hour UTC on December 11th. See also this answer.
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
Or is there something about the mass of the Moon that makes the trip naturally quicker?
orbital-mechanics planning cislunar
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
$endgroup$
$begingroup$
Is the formula for the orbit around the Earth applicable for a Earth-Moon transfer trajectory? The trajectory closer to the Moon is influenced by both the gravity of Earth and Moon.
$endgroup$
– Uwe
Dec 8 '18 at 15:33
$begingroup$
@Uwe I had just added one more sentence at the bottom, seems like that occurred to both of us at the same time. Maybe you can just write a short answer?
$endgroup$
– uhoh
Dec 8 '18 at 15:35
2
$begingroup$
Not a full answer but a few things to consider - the 10-day ellipse doesn't count for the Moon's gravity which is significant on approach. Also, some Apollo and other missions aimed for an apogee not at the Moon's altitude for free return
$endgroup$
– Jack
Dec 8 '18 at 15:36
2
$begingroup$
@uhoh: I should be able to do an alternative calculation of the trajectory as a three body problem to write an answer. But I don't know an easy and short way to do that. May be patched conics could be used.
$endgroup$
– Uwe
Dec 8 '18 at 15:43
2
$begingroup$
Lunar Orbiters 1-5 were all about 3.8-3.9 days launch-to-lunar-orbit. A quick skim of the Soviet Luna series has most of them at around 3.5 days.
$endgroup$
– Russell Borogove
Dec 8 '18 at 16:25
|
show 3 more comments
$begingroup$
The period of an orbit around the Earth is given by:
$$T = 2 pi sqrt{a^3/GM}$$
For Earth GM is about 3.986E+14 m^3/s^2. Put in $a$ = ((6378+400)*1000) for the ISS in LEO (as a test) and you get 5553 seconds, or 92.6 minutes.
Put in $a$ = (384400*1000) for the Moon's orbit and you get 27.5 days, close enough. (We didn't correct for reduced mass).
A ellipse with a perigee in LEO and an apogee at the Moon ($a$ = (195589*1000)) has a period of 10.0 days and a half period of 5.0 days.
But the Apollo times from LEO to the Moon were only about 3.0 to 3.3 days, and Chang'e-4 will only be about 3.9 days Liftoff at 18:23 on 7 December UTC and lunar orbit on ~1500 hour UTC on December 11th. See also this answer.
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
Or is there something about the mass of the Moon that makes the trip naturally quicker?
orbital-mechanics planning cislunar
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
$endgroup$
The period of an orbit around the Earth is given by:
$$T = 2 pi sqrt{a^3/GM}$$
For Earth GM is about 3.986E+14 m^3/s^2. Put in $a$ = ((6378+400)*1000) for the ISS in LEO (as a test) and you get 5553 seconds, or 92.6 minutes.
Put in $a$ = (384400*1000) for the Moon's orbit and you get 27.5 days, close enough. (We didn't correct for reduced mass).
A ellipse with a perigee in LEO and an apogee at the Moon ($a$ = (195589*1000)) has a period of 10.0 days and a half period of 5.0 days.
But the Apollo times from LEO to the Moon were only about 3.0 to 3.3 days, and Chang'e-4 will only be about 3.9 days Liftoff at 18:23 on 7 December UTC and lunar orbit on ~1500 hour UTC on December 11th. See also this answer.
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
Or is there something about the mass of the Moon that makes the trip naturally quicker?
orbital-mechanics planning cislunar
orbital-mechanics planning cislunar
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
edited Dec 8 '18 at 17:23
uhoh
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
asked Dec 8 '18 at 15:27
uhohuhoh
36k18127451
36k18127451
$begingroup$
Is the formula for the orbit around the Earth applicable for a Earth-Moon transfer trajectory? The trajectory closer to the Moon is influenced by both the gravity of Earth and Moon.
$endgroup$
– Uwe
Dec 8 '18 at 15:33
$begingroup$
@Uwe I had just added one more sentence at the bottom, seems like that occurred to both of us at the same time. Maybe you can just write a short answer?
$endgroup$
– uhoh
Dec 8 '18 at 15:35
2
$begingroup$
Not a full answer but a few things to consider - the 10-day ellipse doesn't count for the Moon's gravity which is significant on approach. Also, some Apollo and other missions aimed for an apogee not at the Moon's altitude for free return
$endgroup$
– Jack
Dec 8 '18 at 15:36
2
$begingroup$
@uhoh: I should be able to do an alternative calculation of the trajectory as a three body problem to write an answer. But I don't know an easy and short way to do that. May be patched conics could be used.
$endgroup$
– Uwe
Dec 8 '18 at 15:43
2
$begingroup$
Lunar Orbiters 1-5 were all about 3.8-3.9 days launch-to-lunar-orbit. A quick skim of the Soviet Luna series has most of them at around 3.5 days.
$endgroup$
– Russell Borogove
Dec 8 '18 at 16:25
|
show 3 more comments
$begingroup$
Is the formula for the orbit around the Earth applicable for a Earth-Moon transfer trajectory? The trajectory closer to the Moon is influenced by both the gravity of Earth and Moon.
$endgroup$
– Uwe
Dec 8 '18 at 15:33
$begingroup$
@Uwe I had just added one more sentence at the bottom, seems like that occurred to both of us at the same time. Maybe you can just write a short answer?
$endgroup$
– uhoh
Dec 8 '18 at 15:35
2
$begingroup$
Not a full answer but a few things to consider - the 10-day ellipse doesn't count for the Moon's gravity which is significant on approach. Also, some Apollo and other missions aimed for an apogee not at the Moon's altitude for free return
$endgroup$
– Jack
Dec 8 '18 at 15:36
2
$begingroup$
@uhoh: I should be able to do an alternative calculation of the trajectory as a three body problem to write an answer. But I don't know an easy and short way to do that. May be patched conics could be used.
$endgroup$
– Uwe
Dec 8 '18 at 15:43
2
$begingroup$
Lunar Orbiters 1-5 were all about 3.8-3.9 days launch-to-lunar-orbit. A quick skim of the Soviet Luna series has most of them at around 3.5 days.
$endgroup$
– Russell Borogove
Dec 8 '18 at 16:25
$begingroup$
Is the formula for the orbit around the Earth applicable for a Earth-Moon transfer trajectory? The trajectory closer to the Moon is influenced by both the gravity of Earth and Moon.
$endgroup$
– Uwe
Dec 8 '18 at 15:33
$begingroup$
Is the formula for the orbit around the Earth applicable for a Earth-Moon transfer trajectory? The trajectory closer to the Moon is influenced by both the gravity of Earth and Moon.
$endgroup$
– Uwe
Dec 8 '18 at 15:33
$begingroup$
@Uwe I had just added one more sentence at the bottom, seems like that occurred to both of us at the same time. Maybe you can just write a short answer?
$endgroup$
– uhoh
Dec 8 '18 at 15:35
$begingroup$
@Uwe I had just added one more sentence at the bottom, seems like that occurred to both of us at the same time. Maybe you can just write a short answer?
$endgroup$
– uhoh
Dec 8 '18 at 15:35
2
2
$begingroup$
Not a full answer but a few things to consider - the 10-day ellipse doesn't count for the Moon's gravity which is significant on approach. Also, some Apollo and other missions aimed for an apogee not at the Moon's altitude for free return
$endgroup$
– Jack
Dec 8 '18 at 15:36
$begingroup$
Not a full answer but a few things to consider - the 10-day ellipse doesn't count for the Moon's gravity which is significant on approach. Also, some Apollo and other missions aimed for an apogee not at the Moon's altitude for free return
$endgroup$
– Jack
Dec 8 '18 at 15:36
2
2
$begingroup$
@uhoh: I should be able to do an alternative calculation of the trajectory as a three body problem to write an answer. But I don't know an easy and short way to do that. May be patched conics could be used.
$endgroup$
– Uwe
Dec 8 '18 at 15:43
$begingroup$
@uhoh: I should be able to do an alternative calculation of the trajectory as a three body problem to write an answer. But I don't know an easy and short way to do that. May be patched conics could be used.
$endgroup$
– Uwe
Dec 8 '18 at 15:43
2
2
$begingroup$
Lunar Orbiters 1-5 were all about 3.8-3.9 days launch-to-lunar-orbit. A quick skim of the Soviet Luna series has most of them at around 3.5 days.
$endgroup$
– Russell Borogove
Dec 8 '18 at 16:25
$begingroup$
Lunar Orbiters 1-5 were all about 3.8-3.9 days launch-to-lunar-orbit. A quick skim of the Soviet Luna series has most of them at around 3.5 days.
$endgroup$
– Russell Borogove
Dec 8 '18 at 16:25
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
The 5 spacecraft of the US Lunar Orbiter program took 3.8-3.9 days to reach lunar orbit.
The Ranger and Surveyor missions were on faster trajectories -- around 2.7 days from launch to lunar impact or landing.
The early Soviet Luna series missions that I've looked at took 3.3-3.6 days to reach the moon -- some were orbiters, some were (intentional or unintentional) direct impactors, some were direct soft-landers. The later sample-return missions/attempts (e.g. Luna 20) took 4.5 day flights.
I'm not sure what motivated these different trajectories. Ranger and Surveyor both had solar panels, so shouldn't have been "racing the clock" to complete their missions before batteries ran down. There may have been some drive to make the flights short simply because these early complex spacecraft were not very reliable; you'd want to get to the moon before things broke down.
I don't know if the early lunar projects started from a desired flight time, factored with the available launcher, to determine the maximum spacecraft mass; or if the final design of the spacecraft factored with the launcher to determine the minimum flight time, or some iterated back-and-forth combination thereof.
The Apollo landing flights generally took about 3.3 days from launch to lunar orbit. Apollo 8 was faster, not being burdened with a 15-ton LM, and reached lunar orbit in about 2.9 days.
Some more recent missions have taken much more leisurely approaches, possibly because progressive orbit-raising can be done with a smaller, lighter motor and the hardware is reliable enough to allow the longer flight time.
ISRO's Chandrayaan-1 took two weeks plus another week to reach its final orbit -- obviously not a Hohmann-esque single translunar insertion burn!
Likewise, JAXA's SELENE/Kaguya took more than two weeks to reach lunar orbit.
Or is there something about the mass of the Moon that makes the trip naturally quicker?
At some point in a translunar trajectory, the Moon's gravity does start accelerating the spacecraft toward the destination; for the Apollo missions it would be about 2.5 days into the flight when the moon's gravity began to dominate over the Earth's. So it's not surprising that the flight times would be significantly less than 5 days, but the exact math for the "Hohmann to massive destination" calculation is beyond me.
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
$endgroup$
$begingroup$
wow, somebody's been busy. Thanks!
$endgroup$
– uhoh
Dec 8 '18 at 17:19
add a comment |
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$begingroup$
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
The 5 spacecraft of the US Lunar Orbiter program took 3.8-3.9 days to reach lunar orbit.
The Ranger and Surveyor missions were on faster trajectories -- around 2.7 days from launch to lunar impact or landing.
The early Soviet Luna series missions that I've looked at took 3.3-3.6 days to reach the moon -- some were orbiters, some were (intentional or unintentional) direct impactors, some were direct soft-landers. The later sample-return missions/attempts (e.g. Luna 20) took 4.5 day flights.
I'm not sure what motivated these different trajectories. Ranger and Surveyor both had solar panels, so shouldn't have been "racing the clock" to complete their missions before batteries ran down. There may have been some drive to make the flights short simply because these early complex spacecraft were not very reliable; you'd want to get to the moon before things broke down.
I don't know if the early lunar projects started from a desired flight time, factored with the available launcher, to determine the maximum spacecraft mass; or if the final design of the spacecraft factored with the launcher to determine the minimum flight time, or some iterated back-and-forth combination thereof.
The Apollo landing flights generally took about 3.3 days from launch to lunar orbit. Apollo 8 was faster, not being burdened with a 15-ton LM, and reached lunar orbit in about 2.9 days.
Some more recent missions have taken much more leisurely approaches, possibly because progressive orbit-raising can be done with a smaller, lighter motor and the hardware is reliable enough to allow the longer flight time.
ISRO's Chandrayaan-1 took two weeks plus another week to reach its final orbit -- obviously not a Hohmann-esque single translunar insertion burn!
Likewise, JAXA's SELENE/Kaguya took more than two weeks to reach lunar orbit.
Or is there something about the mass of the Moon that makes the trip naturally quicker?
At some point in a translunar trajectory, the Moon's gravity does start accelerating the spacecraft toward the destination; for the Apollo missions it would be about 2.5 days into the flight when the moon's gravity began to dominate over the Earth's. So it's not surprising that the flight times would be significantly less than 5 days, but the exact math for the "Hohmann to massive destination" calculation is beyond me.
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
$endgroup$
$begingroup$
wow, somebody's been busy. Thanks!
$endgroup$
– uhoh
Dec 8 '18 at 17:19
add a comment |
$begingroup$
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
The 5 spacecraft of the US Lunar Orbiter program took 3.8-3.9 days to reach lunar orbit.
The Ranger and Surveyor missions were on faster trajectories -- around 2.7 days from launch to lunar impact or landing.
The early Soviet Luna series missions that I've looked at took 3.3-3.6 days to reach the moon -- some were orbiters, some were (intentional or unintentional) direct impactors, some were direct soft-landers. The later sample-return missions/attempts (e.g. Luna 20) took 4.5 day flights.
I'm not sure what motivated these different trajectories. Ranger and Surveyor both had solar panels, so shouldn't have been "racing the clock" to complete their missions before batteries ran down. There may have been some drive to make the flights short simply because these early complex spacecraft were not very reliable; you'd want to get to the moon before things broke down.
I don't know if the early lunar projects started from a desired flight time, factored with the available launcher, to determine the maximum spacecraft mass; or if the final design of the spacecraft factored with the launcher to determine the minimum flight time, or some iterated back-and-forth combination thereof.
The Apollo landing flights generally took about 3.3 days from launch to lunar orbit. Apollo 8 was faster, not being burdened with a 15-ton LM, and reached lunar orbit in about 2.9 days.
Some more recent missions have taken much more leisurely approaches, possibly because progressive orbit-raising can be done with a smaller, lighter motor and the hardware is reliable enough to allow the longer flight time.
ISRO's Chandrayaan-1 took two weeks plus another week to reach its final orbit -- obviously not a Hohmann-esque single translunar insertion burn!
Likewise, JAXA's SELENE/Kaguya took more than two weeks to reach lunar orbit.
Or is there something about the mass of the Moon that makes the trip naturally quicker?
At some point in a translunar trajectory, the Moon's gravity does start accelerating the spacecraft toward the destination; for the Apollo missions it would be about 2.5 days into the flight when the moon's gravity began to dominate over the Earth's. So it's not surprising that the flight times would be significantly less than 5 days, but the exact math for the "Hohmann to massive destination" calculation is beyond me.
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
$endgroup$
$begingroup$
wow, somebody's been busy. Thanks!
$endgroup$
– uhoh
Dec 8 '18 at 17:19
add a comment |
$begingroup$
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
The 5 spacecraft of the US Lunar Orbiter program took 3.8-3.9 days to reach lunar orbit.
The Ranger and Surveyor missions were on faster trajectories -- around 2.7 days from launch to lunar impact or landing.
The early Soviet Luna series missions that I've looked at took 3.3-3.6 days to reach the moon -- some were orbiters, some were (intentional or unintentional) direct impactors, some were direct soft-landers. The later sample-return missions/attempts (e.g. Luna 20) took 4.5 day flights.
I'm not sure what motivated these different trajectories. Ranger and Surveyor both had solar panels, so shouldn't have been "racing the clock" to complete their missions before batteries ran down. There may have been some drive to make the flights short simply because these early complex spacecraft were not very reliable; you'd want to get to the moon before things broke down.
I don't know if the early lunar projects started from a desired flight time, factored with the available launcher, to determine the maximum spacecraft mass; or if the final design of the spacecraft factored with the launcher to determine the minimum flight time, or some iterated back-and-forth combination thereof.
The Apollo landing flights generally took about 3.3 days from launch to lunar orbit. Apollo 8 was faster, not being burdened with a 15-ton LM, and reached lunar orbit in about 2.9 days.
Some more recent missions have taken much more leisurely approaches, possibly because progressive orbit-raising can be done with a smaller, lighter motor and the hardware is reliable enough to allow the longer flight time.
ISRO's Chandrayaan-1 took two weeks plus another week to reach its final orbit -- obviously not a Hohmann-esque single translunar insertion burn!
Likewise, JAXA's SELENE/Kaguya took more than two weeks to reach lunar orbit.
Or is there something about the mass of the Moon that makes the trip naturally quicker?
At some point in a translunar trajectory, the Moon's gravity does start accelerating the spacecraft toward the destination; for the Apollo missions it would be about 2.5 days into the flight when the moon's gravity began to dominate over the Earth's. So it's not surprising that the flight times would be significantly less than 5 days, but the exact math for the "Hohmann to massive destination" calculation is beyond me.
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
$endgroup$
Are most trips to the Moon actually around 3 or 4 days, and not the 5.0 days of a half-ellipse?
The 5 spacecraft of the US Lunar Orbiter program took 3.8-3.9 days to reach lunar orbit.
The Ranger and Surveyor missions were on faster trajectories -- around 2.7 days from launch to lunar impact or landing.
The early Soviet Luna series missions that I've looked at took 3.3-3.6 days to reach the moon -- some were orbiters, some were (intentional or unintentional) direct impactors, some were direct soft-landers. The later sample-return missions/attempts (e.g. Luna 20) took 4.5 day flights.
I'm not sure what motivated these different trajectories. Ranger and Surveyor both had solar panels, so shouldn't have been "racing the clock" to complete their missions before batteries ran down. There may have been some drive to make the flights short simply because these early complex spacecraft were not very reliable; you'd want to get to the moon before things broke down.
I don't know if the early lunar projects started from a desired flight time, factored with the available launcher, to determine the maximum spacecraft mass; or if the final design of the spacecraft factored with the launcher to determine the minimum flight time, or some iterated back-and-forth combination thereof.
The Apollo landing flights generally took about 3.3 days from launch to lunar orbit. Apollo 8 was faster, not being burdened with a 15-ton LM, and reached lunar orbit in about 2.9 days.
Some more recent missions have taken much more leisurely approaches, possibly because progressive orbit-raising can be done with a smaller, lighter motor and the hardware is reliable enough to allow the longer flight time.
ISRO's Chandrayaan-1 took two weeks plus another week to reach its final orbit -- obviously not a Hohmann-esque single translunar insertion burn!
Likewise, JAXA's SELENE/Kaguya took more than two weeks to reach lunar orbit.
Or is there something about the mass of the Moon that makes the trip naturally quicker?
At some point in a translunar trajectory, the Moon's gravity does start accelerating the spacecraft toward the destination; for the Apollo missions it would be about 2.5 days into the flight when the moon's gravity began to dominate over the Earth's. So it's not surprising that the flight times would be significantly less than 5 days, but the exact math for the "Hohmann to massive destination" calculation is beyond me.
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
edited Dec 8 '18 at 17:20
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
answered Dec 8 '18 at 16:50
Russell BorogoveRussell Borogove
84.6k3284367
84.6k3284367
$begingroup$
wow, somebody's been busy. Thanks!
$endgroup$
– uhoh
Dec 8 '18 at 17:19
add a comment |
$begingroup$
wow, somebody's been busy. Thanks!
$endgroup$
– uhoh
Dec 8 '18 at 17:19
$begingroup$
wow, somebody's been busy. Thanks!
$endgroup$
– uhoh
Dec 8 '18 at 17:19
$begingroup$
wow, somebody's been busy. Thanks!
$endgroup$
– uhoh
Dec 8 '18 at 17:19
add a comment |
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Is the formula for the orbit around the Earth applicable for a Earth-Moon transfer trajectory? The trajectory closer to the Moon is influenced by both the gravity of Earth and Moon.
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– Uwe
Dec 8 '18 at 15:33
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@Uwe I had just added one more sentence at the bottom, seems like that occurred to both of us at the same time. Maybe you can just write a short answer?
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– uhoh
Dec 8 '18 at 15:35
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Not a full answer but a few things to consider - the 10-day ellipse doesn't count for the Moon's gravity which is significant on approach. Also, some Apollo and other missions aimed for an apogee not at the Moon's altitude for free return
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– Jack
Dec 8 '18 at 15:36
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@uhoh: I should be able to do an alternative calculation of the trajectory as a three body problem to write an answer. But I don't know an easy and short way to do that. May be patched conics could be used.
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– Uwe
Dec 8 '18 at 15:43
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Lunar Orbiters 1-5 were all about 3.8-3.9 days launch-to-lunar-orbit. A quick skim of the Soviet Luna series has most of them at around 3.5 days.
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– Russell Borogove
Dec 8 '18 at 16:25