Is there any way to land a rover on the Moon without using any thrusters?The upcoming landing of Chang'e 3 on the Moon - Is there detailed information about it anywhere?Can you buy land on the moon?Is there enough man-made junk on the moon to help a stranded person get home?Are there “rover-friendly region maps” for the Moon?Could SpaceX land on the moon today?Is there any way to tell which NASA rover this “rover” might be modeled after?What are the drawbacks using an intelligent bulldozer on Mars for the assistance of a rover?Is there any Scientific Benefit to the Lower Albedo levels on the Darkside of the Moon?Differences in the design of a commercial Moon rover and a commercial Mars rover?Is there any way to determine the fate of Chandrayaan-2?
Maintenance tips to prolong engine lifespan for short trips
Resume: How to quantify my contributions as a software engineer?
A medieval fantasy adventurer lights a torch in a 100% pure oxygen room. What happens?
I asked for a graduate student position from a professor. He replied "welcome". What does that mean?
Why did it become so much more expensive to start a university?
Gravity on an Orbital Ring
Are there any non-WEB re-implementation of TeX Core in recent years?
Napkin Folding Problem / Rumpled Ruble Problem
Is there an inconsistency about Natasha Romanoff's middle name in the MCU?
Are Democrats more likely to believe Astrology is a science?
Why is the T-1000 humanoid?
Is English tonal for some words, like "permit"?
How seriously should I take a CBP interview where I was told I have a red flag and could only stay for 30 days?
How can I locate a missing person abroad?
How is Team Scooby Doo (Mystery Inc.) funded?
Does a gnoll speak both Gnoll and Abyssal, or is Gnoll a dialect of Abyssal?
What are uses of the byte after BRK instruction on 6502?
How to say "quirky" in German without sounding derogatory?
Writing a love interest for my hero
I was promised a work PC but still awaiting approval 3 months later so using my own laptop - Is it fair to ask employer for laptop insurance?
Glue or not to glue boots
Do ibuprofen or paracetamol cause hearing loss?
How can I fix a framing mistake so I can drywall?
Might have gotten a coworker sick, should I address this?
Is there any way to land a rover on the Moon without using any thrusters?
The upcoming landing of Chang'e 3 on the Moon - Is there detailed information about it anywhere?Can you buy land on the moon?Is there enough man-made junk on the moon to help a stranded person get home?Are there “rover-friendly region maps” for the Moon?Could SpaceX land on the moon today?Is there any way to tell which NASA rover this “rover” might be modeled after?What are the drawbacks using an intelligent bulldozer on Mars for the assistance of a rover?Is there any Scientific Benefit to the Lower Albedo levels on the Darkside of the Moon?Differences in the design of a commercial Moon rover and a commercial Mars rover?Is there any way to determine the fate of Chandrayaan-2?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Is there any way to land a rover on the moon without using any thruster, with the help of Thermocol, Cotton, Bubble wrap or any other packaging material, like we can receive from online shop?
the-moon landing rovers
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
asked 9 hours ago
editiniteditinit
1312 bronze badges
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment
|
$begingroup$
Is there any way to land a rover on the moon without using any thruster, with the help of Thermocol, Cotton, Bubble wrap or any other packaging material, like we can receive from online shop?
the-moon landing rovers
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
asked 9 hours ago
editiniteditinit
1312 bronze badges
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
In theory, sure. You just need a big enough crumple zone. But since weight is the limiting factor, I doubt that it would be feasible.
$endgroup$
– Polygnome
9 hours ago
1
$begingroup$
NO thrusters for landing or deorbiting? You need to slow down from an orbital velocity to reduce altitude and land, so would you edit your question to make it more clear? If it can use deorbiting retrothrusters, but not to land it can follow some of the ways described here pages.citebite.com/s3e3l8g4j0poy
$endgroup$
– Rajath Pai
8 hours ago
add a comment
|
$begingroup$
Is there any way to land a rover on the moon without using any thruster, with the help of Thermocol, Cotton, Bubble wrap or any other packaging material, like we can receive from online shop?
the-moon landing rovers
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
asked 9 hours ago
editiniteditinit
1312 bronze badges
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is there any way to land a rover on the moon without using any thruster, with the help of Thermocol, Cotton, Bubble wrap or any other packaging material, like we can receive from online shop?
the-moon landing rovers
the-moon landing rovers
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
asked 9 hours ago
editiniteditinit
1312 bronze badges
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
asked 9 hours ago
editiniteditinit
1312 bronze badges
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
edited 8 hours ago
peterh
2,1313 gold badges17 silver badges35 bronze badges
2,1313 gold badges17 silver badges35 bronze badges
asked 9 hours ago
editiniteditinit
1312 bronze badges
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
editiniteditinit
1312 bronze badges
asked 9 hours ago
editiniteditinit
1312 bronze badges
1312 bronze badges
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
editinit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
In theory, sure. You just need a big enough crumple zone. But since weight is the limiting factor, I doubt that it would be feasible.
$endgroup$
– Polygnome
9 hours ago
1
$begingroup$
NO thrusters for landing or deorbiting? You need to slow down from an orbital velocity to reduce altitude and land, so would you edit your question to make it more clear? If it can use deorbiting retrothrusters, but not to land it can follow some of the ways described here pages.citebite.com/s3e3l8g4j0poy
$endgroup$
– Rajath Pai
8 hours ago
add a comment
|
3
$begingroup$
In theory, sure. You just need a big enough crumple zone. But since weight is the limiting factor, I doubt that it would be feasible.
$endgroup$
– Polygnome
9 hours ago
1
$begingroup$
NO thrusters for landing or deorbiting? You need to slow down from an orbital velocity to reduce altitude and land, so would you edit your question to make it more clear? If it can use deorbiting retrothrusters, but not to land it can follow some of the ways described here pages.citebite.com/s3e3l8g4j0poy
$endgroup$
– Rajath Pai
8 hours ago
3
3
$begingroup$
In theory, sure. You just need a big enough crumple zone. But since weight is the limiting factor, I doubt that it would be feasible.
$endgroup$
– Polygnome
9 hours ago
$begingroup$
In theory, sure. You just need a big enough crumple zone. But since weight is the limiting factor, I doubt that it would be feasible.
$endgroup$
– Polygnome
9 hours ago
1
1
$begingroup$
NO thrusters for landing or deorbiting? You need to slow down from an orbital velocity to reduce altitude and land, so would you edit your question to make it more clear? If it can use deorbiting retrothrusters, but not to land it can follow some of the ways described here pages.citebite.com/s3e3l8g4j0poy
$endgroup$
– Rajath Pai
8 hours ago
$begingroup$
NO thrusters for landing or deorbiting? You need to slow down from an orbital velocity to reduce altitude and land, so would you edit your question to make it more clear? If it can use deorbiting retrothrusters, but not to land it can follow some of the ways described here pages.citebite.com/s3e3l8g4j0poy
$endgroup$
– Rajath Pai
8 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
It is not practical to use this approach from orbital (1.6km/s) or escape velocity (2.4km/s), for two major reasons. The first is the acceleration reason.
The kinematic for bringing objects to a stop under constant acceleration is
$$d = fracv^22a$$
from which we can quickly solve for the acceleration to be
$$a = fracv^22d$$
Even with 10 meters of padding, as @zeta-band used, the acceleration would be $283000fracms^2 = 29000g$, which is roughly the acceleration experienced by electronics in artillery shells. So in theory we could build the electronics to survive the impact.
However, this is where the second issue arises. Even if a crumple zone of arbitrary size were to be used, there would be mechanical and material issues associated with the speed of sound in the material. The problem, in a nutshell, is that in ultra-high velocity impacts, even strong materials like steel splash and crack rather than deform neatly. In order to be effective, a crumple zone must effectively reduce the speed of sound in the material, spreading the shock of impact. However the velocities involved are too high. Our crumple zone would not act like the crumple zone in a car. In fact, the pressure wave indicating the start of the impact with the ground may only reach the payload ~50% faster than the ground itself, and that is if we use a sturdy crumple zone out of a material like aluminum or steel (former better for weight).
Items like bubble wrap, cardboard, foam cushions, things that we consider soft and suitable for packing do not have a high speed of sound. They would not even transmit the ground impact force until the spacecraft itself hit the ground.
Finally, I want to debunk the idea that one could "land" (come in on a strong tangent, rather than straight down) on a long runway on the moon. In theory, this is possible. Using ultra-strong alloys, one could (barely) make wheels that could spin up to the required velocity of 2.4 km/s. However the gyroscopic issues will be severe, not to mention the wheel balancing (and what happens when a wheel breaks?). One could even go simple and just slide it out on the longest slip n' slide ever built on or off this world. But hypervelocity issues strike again. Any rubbing surface at these speeds won't just heat up, they will plasmify. Atoms in the materials will impact so hard they simply get dislodged off the material completely. Even diamonds will degrade. Its not that any of these methods are impossible in principle, they are just impractical and the engineering challenges to make them work are monumentous.
So the upshot is that there must be some form of propulsion to slow the spacecraft down to land on the moon.
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
$endgroup$
$begingroup$
What is the speed of sound on the moon? It is a variable value on Earth not a static one.
$endgroup$
– James Jenkins
7 hours ago
2
$begingroup$
@JamesJenkins I reference the speed of sound in the material.
$endgroup$
– Quietghost
7 hours ago
2
$begingroup$
Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure.
$endgroup$
– James Jenkins
7 hours ago
add a comment
|
$begingroup$
Well, lunar escape velocity is 2.38 km/s. So this is about the speed that a rover dropped in from orbit (with no sideways velocity) will hit at. So let's take a guess at how many g's deceleration will be. Assume it has 10 meters of crumple to stop in. It will take it about 10/2380 seconds to stop. Which is .0042 seconds. Deceleration will be 2380 / .00042 = 566,666 m/s squared. Which is about 57,823 g's.
That's pretty tough to survive and still function.
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
$endgroup$
2
$begingroup$
This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=fracv^22d = 283000 m/s^2 approx 29000g$.
$endgroup$
– Quietghost
7 hours ago
1
$begingroup$
I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit.
$endgroup$
– Polygnome
7 hours ago
add a comment
|
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "508"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
editinit is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f38709%2fis-there-any-way-to-land-a-rover-on-the-moon-without-using-any-thrusters%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not practical to use this approach from orbital (1.6km/s) or escape velocity (2.4km/s), for two major reasons. The first is the acceleration reason.
The kinematic for bringing objects to a stop under constant acceleration is
$$d = fracv^22a$$
from which we can quickly solve for the acceleration to be
$$a = fracv^22d$$
Even with 10 meters of padding, as @zeta-band used, the acceleration would be $283000fracms^2 = 29000g$, which is roughly the acceleration experienced by electronics in artillery shells. So in theory we could build the electronics to survive the impact.
However, this is where the second issue arises. Even if a crumple zone of arbitrary size were to be used, there would be mechanical and material issues associated with the speed of sound in the material. The problem, in a nutshell, is that in ultra-high velocity impacts, even strong materials like steel splash and crack rather than deform neatly. In order to be effective, a crumple zone must effectively reduce the speed of sound in the material, spreading the shock of impact. However the velocities involved are too high. Our crumple zone would not act like the crumple zone in a car. In fact, the pressure wave indicating the start of the impact with the ground may only reach the payload ~50% faster than the ground itself, and that is if we use a sturdy crumple zone out of a material like aluminum or steel (former better for weight).
Items like bubble wrap, cardboard, foam cushions, things that we consider soft and suitable for packing do not have a high speed of sound. They would not even transmit the ground impact force until the spacecraft itself hit the ground.
Finally, I want to debunk the idea that one could "land" (come in on a strong tangent, rather than straight down) on a long runway on the moon. In theory, this is possible. Using ultra-strong alloys, one could (barely) make wheels that could spin up to the required velocity of 2.4 km/s. However the gyroscopic issues will be severe, not to mention the wheel balancing (and what happens when a wheel breaks?). One could even go simple and just slide it out on the longest slip n' slide ever built on or off this world. But hypervelocity issues strike again. Any rubbing surface at these speeds won't just heat up, they will plasmify. Atoms in the materials will impact so hard they simply get dislodged off the material completely. Even diamonds will degrade. Its not that any of these methods are impossible in principle, they are just impractical and the engineering challenges to make them work are monumentous.
So the upshot is that there must be some form of propulsion to slow the spacecraft down to land on the moon.
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
$endgroup$
$begingroup$
What is the speed of sound on the moon? It is a variable value on Earth not a static one.
$endgroup$
– James Jenkins
7 hours ago
2
$begingroup$
@JamesJenkins I reference the speed of sound in the material.
$endgroup$
– Quietghost
7 hours ago
2
$begingroup$
Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure.
$endgroup$
– James Jenkins
7 hours ago
add a comment
|
$begingroup$
It is not practical to use this approach from orbital (1.6km/s) or escape velocity (2.4km/s), for two major reasons. The first is the acceleration reason.
The kinematic for bringing objects to a stop under constant acceleration is
$$d = fracv^22a$$
from which we can quickly solve for the acceleration to be
$$a = fracv^22d$$
Even with 10 meters of padding, as @zeta-band used, the acceleration would be $283000fracms^2 = 29000g$, which is roughly the acceleration experienced by electronics in artillery shells. So in theory we could build the electronics to survive the impact.
However, this is where the second issue arises. Even if a crumple zone of arbitrary size were to be used, there would be mechanical and material issues associated with the speed of sound in the material. The problem, in a nutshell, is that in ultra-high velocity impacts, even strong materials like steel splash and crack rather than deform neatly. In order to be effective, a crumple zone must effectively reduce the speed of sound in the material, spreading the shock of impact. However the velocities involved are too high. Our crumple zone would not act like the crumple zone in a car. In fact, the pressure wave indicating the start of the impact with the ground may only reach the payload ~50% faster than the ground itself, and that is if we use a sturdy crumple zone out of a material like aluminum or steel (former better for weight).
Items like bubble wrap, cardboard, foam cushions, things that we consider soft and suitable for packing do not have a high speed of sound. They would not even transmit the ground impact force until the spacecraft itself hit the ground.
Finally, I want to debunk the idea that one could "land" (come in on a strong tangent, rather than straight down) on a long runway on the moon. In theory, this is possible. Using ultra-strong alloys, one could (barely) make wheels that could spin up to the required velocity of 2.4 km/s. However the gyroscopic issues will be severe, not to mention the wheel balancing (and what happens when a wheel breaks?). One could even go simple and just slide it out on the longest slip n' slide ever built on or off this world. But hypervelocity issues strike again. Any rubbing surface at these speeds won't just heat up, they will plasmify. Atoms in the materials will impact so hard they simply get dislodged off the material completely. Even diamonds will degrade. Its not that any of these methods are impossible in principle, they are just impractical and the engineering challenges to make them work are monumentous.
So the upshot is that there must be some form of propulsion to slow the spacecraft down to land on the moon.
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
$endgroup$
$begingroup$
What is the speed of sound on the moon? It is a variable value on Earth not a static one.
$endgroup$
– James Jenkins
7 hours ago
2
$begingroup$
@JamesJenkins I reference the speed of sound in the material.
$endgroup$
– Quietghost
7 hours ago
2
$begingroup$
Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure.
$endgroup$
– James Jenkins
7 hours ago
add a comment
|
$begingroup$
It is not practical to use this approach from orbital (1.6km/s) or escape velocity (2.4km/s), for two major reasons. The first is the acceleration reason.
The kinematic for bringing objects to a stop under constant acceleration is
$$d = fracv^22a$$
from which we can quickly solve for the acceleration to be
$$a = fracv^22d$$
Even with 10 meters of padding, as @zeta-band used, the acceleration would be $283000fracms^2 = 29000g$, which is roughly the acceleration experienced by electronics in artillery shells. So in theory we could build the electronics to survive the impact.
However, this is where the second issue arises. Even if a crumple zone of arbitrary size were to be used, there would be mechanical and material issues associated with the speed of sound in the material. The problem, in a nutshell, is that in ultra-high velocity impacts, even strong materials like steel splash and crack rather than deform neatly. In order to be effective, a crumple zone must effectively reduce the speed of sound in the material, spreading the shock of impact. However the velocities involved are too high. Our crumple zone would not act like the crumple zone in a car. In fact, the pressure wave indicating the start of the impact with the ground may only reach the payload ~50% faster than the ground itself, and that is if we use a sturdy crumple zone out of a material like aluminum or steel (former better for weight).
Items like bubble wrap, cardboard, foam cushions, things that we consider soft and suitable for packing do not have a high speed of sound. They would not even transmit the ground impact force until the spacecraft itself hit the ground.
Finally, I want to debunk the idea that one could "land" (come in on a strong tangent, rather than straight down) on a long runway on the moon. In theory, this is possible. Using ultra-strong alloys, one could (barely) make wheels that could spin up to the required velocity of 2.4 km/s. However the gyroscopic issues will be severe, not to mention the wheel balancing (and what happens when a wheel breaks?). One could even go simple and just slide it out on the longest slip n' slide ever built on or off this world. But hypervelocity issues strike again. Any rubbing surface at these speeds won't just heat up, they will plasmify. Atoms in the materials will impact so hard they simply get dislodged off the material completely. Even diamonds will degrade. Its not that any of these methods are impossible in principle, they are just impractical and the engineering challenges to make them work are monumentous.
So the upshot is that there must be some form of propulsion to slow the spacecraft down to land on the moon.
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
$endgroup$
It is not practical to use this approach from orbital (1.6km/s) or escape velocity (2.4km/s), for two major reasons. The first is the acceleration reason.
The kinematic for bringing objects to a stop under constant acceleration is
$$d = fracv^22a$$
from which we can quickly solve for the acceleration to be
$$a = fracv^22d$$
Even with 10 meters of padding, as @zeta-band used, the acceleration would be $283000fracms^2 = 29000g$, which is roughly the acceleration experienced by electronics in artillery shells. So in theory we could build the electronics to survive the impact.
However, this is where the second issue arises. Even if a crumple zone of arbitrary size were to be used, there would be mechanical and material issues associated with the speed of sound in the material. The problem, in a nutshell, is that in ultra-high velocity impacts, even strong materials like steel splash and crack rather than deform neatly. In order to be effective, a crumple zone must effectively reduce the speed of sound in the material, spreading the shock of impact. However the velocities involved are too high. Our crumple zone would not act like the crumple zone in a car. In fact, the pressure wave indicating the start of the impact with the ground may only reach the payload ~50% faster than the ground itself, and that is if we use a sturdy crumple zone out of a material like aluminum or steel (former better for weight).
Items like bubble wrap, cardboard, foam cushions, things that we consider soft and suitable for packing do not have a high speed of sound. They would not even transmit the ground impact force until the spacecraft itself hit the ground.
Finally, I want to debunk the idea that one could "land" (come in on a strong tangent, rather than straight down) on a long runway on the moon. In theory, this is possible. Using ultra-strong alloys, one could (barely) make wheels that could spin up to the required velocity of 2.4 km/s. However the gyroscopic issues will be severe, not to mention the wheel balancing (and what happens when a wheel breaks?). One could even go simple and just slide it out on the longest slip n' slide ever built on or off this world. But hypervelocity issues strike again. Any rubbing surface at these speeds won't just heat up, they will plasmify. Atoms in the materials will impact so hard they simply get dislodged off the material completely. Even diamonds will degrade. Its not that any of these methods are impossible in principle, they are just impractical and the engineering challenges to make them work are monumentous.
So the upshot is that there must be some form of propulsion to slow the spacecraft down to land on the moon.
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
edited 7 hours ago
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
answered 7 hours ago
QuietghostQuietghost
1,1354 silver badges16 bronze badges
1,1354 silver badges16 bronze badges
$begingroup$
What is the speed of sound on the moon? It is a variable value on Earth not a static one.
$endgroup$
– James Jenkins
7 hours ago
2
$begingroup$
@JamesJenkins I reference the speed of sound in the material.
$endgroup$
– Quietghost
7 hours ago
2
$begingroup$
Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure.
$endgroup$
– James Jenkins
7 hours ago
add a comment
|
$begingroup$
What is the speed of sound on the moon? It is a variable value on Earth not a static one.
$endgroup$
– James Jenkins
7 hours ago
2
$begingroup$
@JamesJenkins I reference the speed of sound in the material.
$endgroup$
– Quietghost
7 hours ago
2
$begingroup$
Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
What is the speed of sound on the moon? It is a variable value on Earth not a static one.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
What is the speed of sound on the moon? It is a variable value on Earth not a static one.
$endgroup$
– James Jenkins
7 hours ago
2
2
$begingroup$
@JamesJenkins I reference the speed of sound in the material.
$endgroup$
– Quietghost
7 hours ago
$begingroup$
@JamesJenkins I reference the speed of sound in the material.
$endgroup$
– Quietghost
7 hours ago
2
2
$begingroup$
Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure.
$endgroup$
– James Jenkins
7 hours ago
add a comment
|
$begingroup$
Well, lunar escape velocity is 2.38 km/s. So this is about the speed that a rover dropped in from orbit (with no sideways velocity) will hit at. So let's take a guess at how many g's deceleration will be. Assume it has 10 meters of crumple to stop in. It will take it about 10/2380 seconds to stop. Which is .0042 seconds. Deceleration will be 2380 / .00042 = 566,666 m/s squared. Which is about 57,823 g's.
That's pretty tough to survive and still function.
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
$endgroup$
2
$begingroup$
This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=fracv^22d = 283000 m/s^2 approx 29000g$.
$endgroup$
– Quietghost
7 hours ago
1
$begingroup$
I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit.
$endgroup$
– Polygnome
7 hours ago
add a comment
|
$begingroup$
Well, lunar escape velocity is 2.38 km/s. So this is about the speed that a rover dropped in from orbit (with no sideways velocity) will hit at. So let's take a guess at how many g's deceleration will be. Assume it has 10 meters of crumple to stop in. It will take it about 10/2380 seconds to stop. Which is .0042 seconds. Deceleration will be 2380 / .00042 = 566,666 m/s squared. Which is about 57,823 g's.
That's pretty tough to survive and still function.
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
$endgroup$
2
$begingroup$
This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=fracv^22d = 283000 m/s^2 approx 29000g$.
$endgroup$
– Quietghost
7 hours ago
1
$begingroup$
I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit.
$endgroup$
– Polygnome
7 hours ago
add a comment
|
$begingroup$
Well, lunar escape velocity is 2.38 km/s. So this is about the speed that a rover dropped in from orbit (with no sideways velocity) will hit at. So let's take a guess at how many g's deceleration will be. Assume it has 10 meters of crumple to stop in. It will take it about 10/2380 seconds to stop. Which is .0042 seconds. Deceleration will be 2380 / .00042 = 566,666 m/s squared. Which is about 57,823 g's.
That's pretty tough to survive and still function.
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
$endgroup$
Well, lunar escape velocity is 2.38 km/s. So this is about the speed that a rover dropped in from orbit (with no sideways velocity) will hit at. So let's take a guess at how many g's deceleration will be. Assume it has 10 meters of crumple to stop in. It will take it about 10/2380 seconds to stop. Which is .0042 seconds. Deceleration will be 2380 / .00042 = 566,666 m/s squared. Which is about 57,823 g's.
That's pretty tough to survive and still function.
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
answered 8 hours ago
zeta-bandzeta-band
3222 silver badges7 bronze badges
3222 silver badges7 bronze badges
2
$begingroup$
This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=fracv^22d = 283000 m/s^2 approx 29000g$.
$endgroup$
– Quietghost
7 hours ago
1
$begingroup$
I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit.
$endgroup$
– Polygnome
7 hours ago
add a comment
|
2
$begingroup$
This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=fracv^22d = 283000 m/s^2 approx 29000g$.
$endgroup$
– Quietghost
7 hours ago
1
$begingroup$
I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit.
$endgroup$
– Polygnome
7 hours ago
2
2
$begingroup$
This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=fracv^22d = 283000 m/s^2 approx 29000g$.
$endgroup$
– Quietghost
7 hours ago
$begingroup$
This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=fracv^22d = 283000 m/s^2 approx 29000g$.
$endgroup$
– Quietghost
7 hours ago
1
1
$begingroup$
I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well.
$endgroup$
– James Jenkins
7 hours ago
$begingroup$
You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit.
$endgroup$
– Polygnome
7 hours ago
$begingroup$
You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit.
$endgroup$
– Polygnome
7 hours ago
add a comment
|
editinit is a new contributor. Be nice, and check out our Code of Conduct.
editinit is a new contributor. Be nice, and check out our Code of Conduct.
editinit is a new contributor. Be nice, and check out our Code of Conduct.
editinit is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Space Exploration Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f38709%2fis-there-any-way-to-land-a-rover-on-the-moon-without-using-any-thrusters%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
In theory, sure. You just need a big enough crumple zone. But since weight is the limiting factor, I doubt that it would be feasible.
$endgroup$
– Polygnome
9 hours ago
1
$begingroup$
NO thrusters for landing or deorbiting? You need to slow down from an orbital velocity to reduce altitude and land, so would you edit your question to make it more clear? If it can use deorbiting retrothrusters, but not to land it can follow some of the ways described here pages.citebite.com/s3e3l8g4j0poy
$endgroup$
– Rajath Pai
8 hours ago