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What happens when the drag force exceeds the weight of an object falling into earth?


Calculating wind force and drag force on a falling objectWhy aren't Roche limit and the difference in gravitational acceleration the same?How to calculate the distance an accelerating object will be pulled by another object in a given amount of time?How does a small object move with constant velocity when drag force is equal to its weight?Acceleration of a ball thrown into the airPower in 1st GearDrag force equals weightWhy wouldn't the weight of an object affect the magnitude of the drag force on that object?Finding the drag force from the final velocity of a falling objectThrust needed from an engine to reach orbit













4












$begingroup$


Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?



At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?



Thanks in advance!










share|cite|improve this question







New contributor



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






$endgroup$
















    4












    $begingroup$


    Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?



    At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?



    Thanks in advance!










    share|cite|improve this question







    New contributor



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






    $endgroup$














      4












      4








      4





      $begingroup$


      Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?



      At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?



      Thanks in advance!










      share|cite|improve this question







      New contributor



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






      $endgroup$




      Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?



      At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?



      Thanks in advance!







      gravity acceleration drag






      share|cite|improve this question







      New contributor



      Laura Iglesias 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



      Laura Iglesias 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



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








      asked 2 hours ago









      Laura IglesiasLaura Iglesias

      232




      232




      New contributor



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




      New contributor




      Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$



          Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.



          The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.



          Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.



          The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.






          share|cite|improve this answer









          $endgroup$




















            4












            $begingroup$

            Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.



            In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.






            share|cite|improve this answer











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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$



              Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.



              The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.



              Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.



              The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$



                Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.



                The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.



                Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.



                The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$



                  Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.



                  The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.



                  Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.



                  The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.






                  share|cite|improve this answer









                  $endgroup$



                  The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$



                  Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.



                  The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.



                  Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.



                  The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  S. McGrewS. McGrew

                  9,82921239




                  9,82921239





















                      4












                      $begingroup$

                      Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.



                      In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.






                      share|cite|improve this answer











                      $endgroup$

















                        4












                        $begingroup$

                        Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.



                        In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.






                        share|cite|improve this answer











                        $endgroup$















                          4












                          4








                          4





                          $begingroup$

                          Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.



                          In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.






                          share|cite|improve this answer











                          $endgroup$



                          Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.



                          In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 1 hour ago

























                          answered 2 hours ago









                          AllureAllure

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