Sci-fi ships falling on planets
$begingroup$
I hope the question is suitable for this forum....
Watching Star Trek: The Next Generation, I have found at least a couple cases where a navigation malfunction on a shuttle makes it fall towards the nearby planet in cuestion of minutes (note that the ship is just passing by, not getting out of the planet).
I'm not an expert physicist, but judging by the way we move our probes through the solar system and that even asteroids just pass near earth without blinking I understand that you have to spiral around the planet for some time before entering it and crashing on its surface... even the Tiangong-1 chinese station took a long time to fall.
I know the situation is created in the sake of drama but I wonder about the possibilities for a shuttle to get captured by a planet's gravity and be forced to crash almost instantly, making useless every rescue effort.
Any of you could enlighten me?
newtonian-gravity angular-momentum orbital-motion satellites
$endgroup$
add a comment |
$begingroup$
I hope the question is suitable for this forum....
Watching Star Trek: The Next Generation, I have found at least a couple cases where a navigation malfunction on a shuttle makes it fall towards the nearby planet in cuestion of minutes (note that the ship is just passing by, not getting out of the planet).
I'm not an expert physicist, but judging by the way we move our probes through the solar system and that even asteroids just pass near earth without blinking I understand that you have to spiral around the planet for some time before entering it and crashing on its surface... even the Tiangong-1 chinese station took a long time to fall.
I know the situation is created in the sake of drama but I wonder about the possibilities for a shuttle to get captured by a planet's gravity and be forced to crash almost instantly, making useless every rescue effort.
Any of you could enlighten me?
newtonian-gravity angular-momentum orbital-motion satellites
$endgroup$
1
$begingroup$
there is a great & fun discussion on that topic here : tvtropes.org/pmwiki/pmwiki.php/Main/GravitySucks and here tvtropes.org/pmwiki/pmwiki.php/Analysis/GravitySucks
$endgroup$
– Manu de Hanoi
Dec 13 '18 at 5:54
add a comment |
$begingroup$
I hope the question is suitable for this forum....
Watching Star Trek: The Next Generation, I have found at least a couple cases where a navigation malfunction on a shuttle makes it fall towards the nearby planet in cuestion of minutes (note that the ship is just passing by, not getting out of the planet).
I'm not an expert physicist, but judging by the way we move our probes through the solar system and that even asteroids just pass near earth without blinking I understand that you have to spiral around the planet for some time before entering it and crashing on its surface... even the Tiangong-1 chinese station took a long time to fall.
I know the situation is created in the sake of drama but I wonder about the possibilities for a shuttle to get captured by a planet's gravity and be forced to crash almost instantly, making useless every rescue effort.
Any of you could enlighten me?
newtonian-gravity angular-momentum orbital-motion satellites
$endgroup$
I hope the question is suitable for this forum....
Watching Star Trek: The Next Generation, I have found at least a couple cases where a navigation malfunction on a shuttle makes it fall towards the nearby planet in cuestion of minutes (note that the ship is just passing by, not getting out of the planet).
I'm not an expert physicist, but judging by the way we move our probes through the solar system and that even asteroids just pass near earth without blinking I understand that you have to spiral around the planet for some time before entering it and crashing on its surface... even the Tiangong-1 chinese station took a long time to fall.
I know the situation is created in the sake of drama but I wonder about the possibilities for a shuttle to get captured by a planet's gravity and be forced to crash almost instantly, making useless every rescue effort.
Any of you could enlighten me?
newtonian-gravity angular-momentum orbital-motion satellites
newtonian-gravity angular-momentum orbital-motion satellites
edited Dec 13 '18 at 5:08
Qmechanic♦
104k121881195
104k121881195
asked Dec 13 '18 at 0:09
LudovicoNLudovicoN
1134
1134
1
$begingroup$
there is a great & fun discussion on that topic here : tvtropes.org/pmwiki/pmwiki.php/Main/GravitySucks and here tvtropes.org/pmwiki/pmwiki.php/Analysis/GravitySucks
$endgroup$
– Manu de Hanoi
Dec 13 '18 at 5:54
add a comment |
1
$begingroup$
there is a great & fun discussion on that topic here : tvtropes.org/pmwiki/pmwiki.php/Main/GravitySucks and here tvtropes.org/pmwiki/pmwiki.php/Analysis/GravitySucks
$endgroup$
– Manu de Hanoi
Dec 13 '18 at 5:54
1
1
$begingroup$
there is a great & fun discussion on that topic here : tvtropes.org/pmwiki/pmwiki.php/Main/GravitySucks and here tvtropes.org/pmwiki/pmwiki.php/Analysis/GravitySucks
$endgroup$
– Manu de Hanoi
Dec 13 '18 at 5:54
$begingroup$
there is a great & fun discussion on that topic here : tvtropes.org/pmwiki/pmwiki.php/Main/GravitySucks and here tvtropes.org/pmwiki/pmwiki.php/Analysis/GravitySucks
$endgroup$
– Manu de Hanoi
Dec 13 '18 at 5:54
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your intuition is right. If a spacecraft is moving past a planet, its has angular momentum that must be dissipated before it can fall to the planet's surface. The spacecraft will, unless affected by an atmosphere or driven by its engines, follow an elliptical orbit that can be calculated from its velocity, position, and the planet's mass. If that orbit does not intersect the planet's surface, the spacecraft will not hit the planet -- it will just loop around. If the velocity is great enough (given a certain position) and not directed toward the planet, the trajectory will be a parabola or hyperbola and the spacecraft will just keep going on a curved path that carries it to infinity.
$endgroup$
add a comment |
$begingroup$
Judging by the episodes of Star Trek I've seen, the crews of the Enterprise and other ships prefer to hover above a planet instead of orbiting. That is, they park the ship so that it remains unmoving above the planet, requiring engine power to keep the ship from falling to the surface due to gravity. In this situation, a navigation malfunction that shuts off the engines would result in the ship falling down to the planet.
The ships could save a lot of fuel if they parked with enough sideways velocity to orbit the planet like a space station or moon. Then they would only need to expend a little fuel to counteract atmospheric drag if they were sufficiently close to the planet (see the orbit corrections on the International Space Station).
$endgroup$
$begingroup$
Isn't "parking" a ship the same as having a fixed orbit?
$endgroup$
– LudovicoN
Dec 13 '18 at 1:25
4
$begingroup$
@LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
$endgroup$
– Mark H
Dec 13 '18 at 1:29
$begingroup$
If they're orbiting around the planets own axis of revolution (at the equator), they may just be in a geosynced orbit (revolving with the same angular velocity as the planet). In this case they would need no power to stay "in one place"
$endgroup$
– R. Rankin
Dec 13 '18 at 2:26
5
$begingroup$
@R.Rankin If the Enterprise were in geosynchronous orbit, then losing engine power wouldn't result in falling towards the planet. You forget that the laws of drama supersede the laws of physics in fiction.
$endgroup$
– Mark H
Dec 13 '18 at 2:59
add a comment |
$begingroup$
I'm not an expert in orbital mechanics neither in aerospace flights, but I'll try my best. In the ideal scenario, where drag forces are absent, the ship's trajectory around a planet may be an elipse, a parabola or a hyperbola. Take for instance the following image that depicts the kinds of trajectories a ship under gravitational influence of a body located at point $F$ may take (taken from Wikipedia)
A "pass-by" trajectory can either be like the green or the blue one, and a closed orbit can either be the red or the gray one. If the smallest distance between the point $F$ and one of the trajectories is smaller than the planet's radius, the ship will obviously hit the surface. So if the ship's crew wants to avoid this tragic fate they must adjust their trajectory to one that passes further away from the planet.
Now if we take drag forces into consideration, the crew should also avoid entering the planet's atmosphere (with this I mean the "thick" layers of the atmosphere). Depending on how deep they penetrate and how fast they're travelling the atmospheric drag forces can rip apart the entire ship (even if their altitude is tens of kilometers away from the surface!). But even if they avoid being burned alive the atmosphere may still have slowed the ship down to a point it'll inevitably hit the ground. So the "safest approach distance" must be such that the atmosphere is "thin enough" to not break the ship apart and neither change the trajectory to a collision one.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
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/3.0/"u003ecc by-sa 3.0 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
});
}
});
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%2fphysics.stackexchange.com%2fquestions%2f446951%2fsci-fi-ships-falling-on-planets%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your intuition is right. If a spacecraft is moving past a planet, its has angular momentum that must be dissipated before it can fall to the planet's surface. The spacecraft will, unless affected by an atmosphere or driven by its engines, follow an elliptical orbit that can be calculated from its velocity, position, and the planet's mass. If that orbit does not intersect the planet's surface, the spacecraft will not hit the planet -- it will just loop around. If the velocity is great enough (given a certain position) and not directed toward the planet, the trajectory will be a parabola or hyperbola and the spacecraft will just keep going on a curved path that carries it to infinity.
$endgroup$
add a comment |
$begingroup$
Your intuition is right. If a spacecraft is moving past a planet, its has angular momentum that must be dissipated before it can fall to the planet's surface. The spacecraft will, unless affected by an atmosphere or driven by its engines, follow an elliptical orbit that can be calculated from its velocity, position, and the planet's mass. If that orbit does not intersect the planet's surface, the spacecraft will not hit the planet -- it will just loop around. If the velocity is great enough (given a certain position) and not directed toward the planet, the trajectory will be a parabola or hyperbola and the spacecraft will just keep going on a curved path that carries it to infinity.
$endgroup$
add a comment |
$begingroup$
Your intuition is right. If a spacecraft is moving past a planet, its has angular momentum that must be dissipated before it can fall to the planet's surface. The spacecraft will, unless affected by an atmosphere or driven by its engines, follow an elliptical orbit that can be calculated from its velocity, position, and the planet's mass. If that orbit does not intersect the planet's surface, the spacecraft will not hit the planet -- it will just loop around. If the velocity is great enough (given a certain position) and not directed toward the planet, the trajectory will be a parabola or hyperbola and the spacecraft will just keep going on a curved path that carries it to infinity.
$endgroup$
Your intuition is right. If a spacecraft is moving past a planet, its has angular momentum that must be dissipated before it can fall to the planet's surface. The spacecraft will, unless affected by an atmosphere or driven by its engines, follow an elliptical orbit that can be calculated from its velocity, position, and the planet's mass. If that orbit does not intersect the planet's surface, the spacecraft will not hit the planet -- it will just loop around. If the velocity is great enough (given a certain position) and not directed toward the planet, the trajectory will be a parabola or hyperbola and the spacecraft will just keep going on a curved path that carries it to infinity.
answered Dec 13 '18 at 1:10
S. McGrewS. McGrew
7,74721234
7,74721234
add a comment |
add a comment |
$begingroup$
Judging by the episodes of Star Trek I've seen, the crews of the Enterprise and other ships prefer to hover above a planet instead of orbiting. That is, they park the ship so that it remains unmoving above the planet, requiring engine power to keep the ship from falling to the surface due to gravity. In this situation, a navigation malfunction that shuts off the engines would result in the ship falling down to the planet.
The ships could save a lot of fuel if they parked with enough sideways velocity to orbit the planet like a space station or moon. Then they would only need to expend a little fuel to counteract atmospheric drag if they were sufficiently close to the planet (see the orbit corrections on the International Space Station).
$endgroup$
$begingroup$
Isn't "parking" a ship the same as having a fixed orbit?
$endgroup$
– LudovicoN
Dec 13 '18 at 1:25
4
$begingroup$
@LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
$endgroup$
– Mark H
Dec 13 '18 at 1:29
$begingroup$
If they're orbiting around the planets own axis of revolution (at the equator), they may just be in a geosynced orbit (revolving with the same angular velocity as the planet). In this case they would need no power to stay "in one place"
$endgroup$
– R. Rankin
Dec 13 '18 at 2:26
5
$begingroup$
@R.Rankin If the Enterprise were in geosynchronous orbit, then losing engine power wouldn't result in falling towards the planet. You forget that the laws of drama supersede the laws of physics in fiction.
$endgroup$
– Mark H
Dec 13 '18 at 2:59
add a comment |
$begingroup$
Judging by the episodes of Star Trek I've seen, the crews of the Enterprise and other ships prefer to hover above a planet instead of orbiting. That is, they park the ship so that it remains unmoving above the planet, requiring engine power to keep the ship from falling to the surface due to gravity. In this situation, a navigation malfunction that shuts off the engines would result in the ship falling down to the planet.
The ships could save a lot of fuel if they parked with enough sideways velocity to orbit the planet like a space station or moon. Then they would only need to expend a little fuel to counteract atmospheric drag if they were sufficiently close to the planet (see the orbit corrections on the International Space Station).
$endgroup$
$begingroup$
Isn't "parking" a ship the same as having a fixed orbit?
$endgroup$
– LudovicoN
Dec 13 '18 at 1:25
4
$begingroup$
@LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
$endgroup$
– Mark H
Dec 13 '18 at 1:29
$begingroup$
If they're orbiting around the planets own axis of revolution (at the equator), they may just be in a geosynced orbit (revolving with the same angular velocity as the planet). In this case they would need no power to stay "in one place"
$endgroup$
– R. Rankin
Dec 13 '18 at 2:26
5
$begingroup$
@R.Rankin If the Enterprise were in geosynchronous orbit, then losing engine power wouldn't result in falling towards the planet. You forget that the laws of drama supersede the laws of physics in fiction.
$endgroup$
– Mark H
Dec 13 '18 at 2:59
add a comment |
$begingroup$
Judging by the episodes of Star Trek I've seen, the crews of the Enterprise and other ships prefer to hover above a planet instead of orbiting. That is, they park the ship so that it remains unmoving above the planet, requiring engine power to keep the ship from falling to the surface due to gravity. In this situation, a navigation malfunction that shuts off the engines would result in the ship falling down to the planet.
The ships could save a lot of fuel if they parked with enough sideways velocity to orbit the planet like a space station or moon. Then they would only need to expend a little fuel to counteract atmospheric drag if they were sufficiently close to the planet (see the orbit corrections on the International Space Station).
$endgroup$
Judging by the episodes of Star Trek I've seen, the crews of the Enterprise and other ships prefer to hover above a planet instead of orbiting. That is, they park the ship so that it remains unmoving above the planet, requiring engine power to keep the ship from falling to the surface due to gravity. In this situation, a navigation malfunction that shuts off the engines would result in the ship falling down to the planet.
The ships could save a lot of fuel if they parked with enough sideways velocity to orbit the planet like a space station or moon. Then they would only need to expend a little fuel to counteract atmospheric drag if they were sufficiently close to the planet (see the orbit corrections on the International Space Station).
answered Dec 13 '18 at 1:22
Mark HMark H
12.3k22441
12.3k22441
$begingroup$
Isn't "parking" a ship the same as having a fixed orbit?
$endgroup$
– LudovicoN
Dec 13 '18 at 1:25
4
$begingroup$
@LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
$endgroup$
– Mark H
Dec 13 '18 at 1:29
$begingroup$
If they're orbiting around the planets own axis of revolution (at the equator), they may just be in a geosynced orbit (revolving with the same angular velocity as the planet). In this case they would need no power to stay "in one place"
$endgroup$
– R. Rankin
Dec 13 '18 at 2:26
5
$begingroup$
@R.Rankin If the Enterprise were in geosynchronous orbit, then losing engine power wouldn't result in falling towards the planet. You forget that the laws of drama supersede the laws of physics in fiction.
$endgroup$
– Mark H
Dec 13 '18 at 2:59
add a comment |
$begingroup$
Isn't "parking" a ship the same as having a fixed orbit?
$endgroup$
– LudovicoN
Dec 13 '18 at 1:25
4
$begingroup$
@LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
$endgroup$
– Mark H
Dec 13 '18 at 1:29
$begingroup$
If they're orbiting around the planets own axis of revolution (at the equator), they may just be in a geosynced orbit (revolving with the same angular velocity as the planet). In this case they would need no power to stay "in one place"
$endgroup$
– R. Rankin
Dec 13 '18 at 2:26
5
$begingroup$
@R.Rankin If the Enterprise were in geosynchronous orbit, then losing engine power wouldn't result in falling towards the planet. You forget that the laws of drama supersede the laws of physics in fiction.
$endgroup$
– Mark H
Dec 13 '18 at 2:59
$begingroup$
Isn't "parking" a ship the same as having a fixed orbit?
$endgroup$
– LudovicoN
Dec 13 '18 at 1:25
$begingroup$
Isn't "parking" a ship the same as having a fixed orbit?
$endgroup$
– LudovicoN
Dec 13 '18 at 1:25
4
4
$begingroup$
@LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
$endgroup$
– Mark H
Dec 13 '18 at 1:29
$begingroup$
@LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
$endgroup$
– Mark H
Dec 13 '18 at 1:29
$begingroup$
If they're orbiting around the planets own axis of revolution (at the equator), they may just be in a geosynced orbit (revolving with the same angular velocity as the planet). In this case they would need no power to stay "in one place"
$endgroup$
– R. Rankin
Dec 13 '18 at 2:26
$begingroup$
If they're orbiting around the planets own axis of revolution (at the equator), they may just be in a geosynced orbit (revolving with the same angular velocity as the planet). In this case they would need no power to stay "in one place"
$endgroup$
– R. Rankin
Dec 13 '18 at 2:26
5
5
$begingroup$
@R.Rankin If the Enterprise were in geosynchronous orbit, then losing engine power wouldn't result in falling towards the planet. You forget that the laws of drama supersede the laws of physics in fiction.
$endgroup$
– Mark H
Dec 13 '18 at 2:59
$begingroup$
@R.Rankin If the Enterprise were in geosynchronous orbit, then losing engine power wouldn't result in falling towards the planet. You forget that the laws of drama supersede the laws of physics in fiction.
$endgroup$
– Mark H
Dec 13 '18 at 2:59
add a comment |
$begingroup$
I'm not an expert in orbital mechanics neither in aerospace flights, but I'll try my best. In the ideal scenario, where drag forces are absent, the ship's trajectory around a planet may be an elipse, a parabola or a hyperbola. Take for instance the following image that depicts the kinds of trajectories a ship under gravitational influence of a body located at point $F$ may take (taken from Wikipedia)
A "pass-by" trajectory can either be like the green or the blue one, and a closed orbit can either be the red or the gray one. If the smallest distance between the point $F$ and one of the trajectories is smaller than the planet's radius, the ship will obviously hit the surface. So if the ship's crew wants to avoid this tragic fate they must adjust their trajectory to one that passes further away from the planet.
Now if we take drag forces into consideration, the crew should also avoid entering the planet's atmosphere (with this I mean the "thick" layers of the atmosphere). Depending on how deep they penetrate and how fast they're travelling the atmospheric drag forces can rip apart the entire ship (even if their altitude is tens of kilometers away from the surface!). But even if they avoid being burned alive the atmosphere may still have slowed the ship down to a point it'll inevitably hit the ground. So the "safest approach distance" must be such that the atmosphere is "thin enough" to not break the ship apart and neither change the trajectory to a collision one.
$endgroup$
add a comment |
$begingroup$
I'm not an expert in orbital mechanics neither in aerospace flights, but I'll try my best. In the ideal scenario, where drag forces are absent, the ship's trajectory around a planet may be an elipse, a parabola or a hyperbola. Take for instance the following image that depicts the kinds of trajectories a ship under gravitational influence of a body located at point $F$ may take (taken from Wikipedia)
A "pass-by" trajectory can either be like the green or the blue one, and a closed orbit can either be the red or the gray one. If the smallest distance between the point $F$ and one of the trajectories is smaller than the planet's radius, the ship will obviously hit the surface. So if the ship's crew wants to avoid this tragic fate they must adjust their trajectory to one that passes further away from the planet.
Now if we take drag forces into consideration, the crew should also avoid entering the planet's atmosphere (with this I mean the "thick" layers of the atmosphere). Depending on how deep they penetrate and how fast they're travelling the atmospheric drag forces can rip apart the entire ship (even if their altitude is tens of kilometers away from the surface!). But even if they avoid being burned alive the atmosphere may still have slowed the ship down to a point it'll inevitably hit the ground. So the "safest approach distance" must be such that the atmosphere is "thin enough" to not break the ship apart and neither change the trajectory to a collision one.
$endgroup$
add a comment |
$begingroup$
I'm not an expert in orbital mechanics neither in aerospace flights, but I'll try my best. In the ideal scenario, where drag forces are absent, the ship's trajectory around a planet may be an elipse, a parabola or a hyperbola. Take for instance the following image that depicts the kinds of trajectories a ship under gravitational influence of a body located at point $F$ may take (taken from Wikipedia)
A "pass-by" trajectory can either be like the green or the blue one, and a closed orbit can either be the red or the gray one. If the smallest distance between the point $F$ and one of the trajectories is smaller than the planet's radius, the ship will obviously hit the surface. So if the ship's crew wants to avoid this tragic fate they must adjust their trajectory to one that passes further away from the planet.
Now if we take drag forces into consideration, the crew should also avoid entering the planet's atmosphere (with this I mean the "thick" layers of the atmosphere). Depending on how deep they penetrate and how fast they're travelling the atmospheric drag forces can rip apart the entire ship (even if their altitude is tens of kilometers away from the surface!). But even if they avoid being burned alive the atmosphere may still have slowed the ship down to a point it'll inevitably hit the ground. So the "safest approach distance" must be such that the atmosphere is "thin enough" to not break the ship apart and neither change the trajectory to a collision one.
$endgroup$
I'm not an expert in orbital mechanics neither in aerospace flights, but I'll try my best. In the ideal scenario, where drag forces are absent, the ship's trajectory around a planet may be an elipse, a parabola or a hyperbola. Take for instance the following image that depicts the kinds of trajectories a ship under gravitational influence of a body located at point $F$ may take (taken from Wikipedia)
A "pass-by" trajectory can either be like the green or the blue one, and a closed orbit can either be the red or the gray one. If the smallest distance between the point $F$ and one of the trajectories is smaller than the planet's radius, the ship will obviously hit the surface. So if the ship's crew wants to avoid this tragic fate they must adjust their trajectory to one that passes further away from the planet.
Now if we take drag forces into consideration, the crew should also avoid entering the planet's atmosphere (with this I mean the "thick" layers of the atmosphere). Depending on how deep they penetrate and how fast they're travelling the atmospheric drag forces can rip apart the entire ship (even if their altitude is tens of kilometers away from the surface!). But even if they avoid being burned alive the atmosphere may still have slowed the ship down to a point it'll inevitably hit the ground. So the "safest approach distance" must be such that the atmosphere is "thin enough" to not break the ship apart and neither change the trajectory to a collision one.
answered Dec 13 '18 at 1:24
ErickShockErickShock
28218
28218
add a comment |
add a comment |
Thanks for contributing an answer to Physics 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%2fphysics.stackexchange.com%2fquestions%2f446951%2fsci-fi-ships-falling-on-planets%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
1
$begingroup$
there is a great & fun discussion on that topic here : tvtropes.org/pmwiki/pmwiki.php/Main/GravitySucks and here tvtropes.org/pmwiki/pmwiki.php/Analysis/GravitySucks
$endgroup$
– Manu de Hanoi
Dec 13 '18 at 5:54