Sci-fi ships falling on planets











up vote
1
down vote

favorite












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?










share|cite|improve this question







New contributor




LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    up vote
    1
    down vote

    favorite












    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?










    share|cite|improve this question







    New contributor




    LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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?










      share|cite|improve this question







      New contributor




      LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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?







      gravity






      share|cite|improve this question







      New contributor




      LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 3 hours ago









      LudovicoN

      1084




      1084




      New contributor




      LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      LudovicoN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          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.






          share|cite|improve this answer




























            up vote
            2
            down vote













            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).






            share|cite|improve this answer





















            • Isn't "parking" a ship the same as having a fixed orbit?
              – LudovicoN
              2 hours ago






            • 2




              @LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
              – Mark H
              2 hours ago










            • 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"
              – R. Rankin
              1 hour ago






            • 1




              @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.
              – Mark H
              37 mins ago


















            up vote
            1
            down vote













            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.






            share|cite|improve this answer





















              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',
              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
              });


              }
              });






              LudovicoN is a new contributor. Be nice, and check out our Code of Conduct.










              draft saved

              draft discarded


















              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








              up vote
              2
              down vote



              accepted










              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.






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted










                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.






                share|cite|improve this answer























                  up vote
                  2
                  down vote



                  accepted







                  up vote
                  2
                  down vote



                  accepted






                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  S. McGrew

                  5,7932924




                  5,7932924






















                      up vote
                      2
                      down vote













                      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).






                      share|cite|improve this answer





















                      • Isn't "parking" a ship the same as having a fixed orbit?
                        – LudovicoN
                        2 hours ago






                      • 2




                        @LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
                        – Mark H
                        2 hours ago










                      • 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"
                        – R. Rankin
                        1 hour ago






                      • 1




                        @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.
                        – Mark H
                        37 mins ago















                      up vote
                      2
                      down vote













                      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).






                      share|cite|improve this answer





















                      • Isn't "parking" a ship the same as having a fixed orbit?
                        – LudovicoN
                        2 hours ago






                      • 2




                        @LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
                        – Mark H
                        2 hours ago










                      • 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"
                        – R. Rankin
                        1 hour ago






                      • 1




                        @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.
                        – Mark H
                        37 mins ago













                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      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).






                      share|cite|improve this answer












                      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).







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 hours ago









                      Mark H

                      11.7k22339




                      11.7k22339












                      • Isn't "parking" a ship the same as having a fixed orbit?
                        – LudovicoN
                        2 hours ago






                      • 2




                        @LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
                        – Mark H
                        2 hours ago










                      • 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"
                        – R. Rankin
                        1 hour ago






                      • 1




                        @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.
                        – Mark H
                        37 mins ago


















                      • Isn't "parking" a ship the same as having a fixed orbit?
                        – LudovicoN
                        2 hours ago






                      • 2




                        @LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
                        – Mark H
                        2 hours ago










                      • 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"
                        – R. Rankin
                        1 hour ago






                      • 1




                        @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.
                        – Mark H
                        37 mins ago
















                      Isn't "parking" a ship the same as having a fixed orbit?
                      – LudovicoN
                      2 hours ago




                      Isn't "parking" a ship the same as having a fixed orbit?
                      – LudovicoN
                      2 hours ago




                      2




                      2




                      @LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
                      – Mark H
                      2 hours ago




                      @LudovicoN With real spaceships, that is correct. In Star Trek, it seems like parking requires constant power so they can stay in one place.
                      – Mark H
                      2 hours ago












                      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"
                      – R. Rankin
                      1 hour ago




                      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"
                      – R. Rankin
                      1 hour ago




                      1




                      1




                      @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.
                      – Mark H
                      37 mins ago




                      @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.
                      – Mark H
                      37 mins ago










                      up vote
                      1
                      down vote













                      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.






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        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.






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          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.






                          share|cite|improve this answer












                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          ErickShock

                          565




                          565






















                              LudovicoN is a new contributor. Be nice, and check out our Code of Conduct.










                              draft saved

                              draft discarded


















                              LudovicoN is a new contributor. Be nice, and check out our Code of Conduct.













                              LudovicoN is a new contributor. Be nice, and check out our Code of Conduct.












                              LudovicoN is a new contributor. Be nice, and check out our Code of Conduct.
















                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              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





















































                              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







                              Popular posts from this blog

                              Create new schema in PostgreSQL using DBeaver

                              Deepest pit of an array with Javascript: test on Codility

                              Costa Masnaga