Apollo Missions - Travel time to moon












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I must be missing something - escape velocity is about 25,000mph. The distance to the moon about 240,000 miles. That computes to about a 9 hour flight. Why did it take 3 days for Apollo missions to reach the moon?










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    $begingroup$
    Possible duplicate of Did Apollo's velocity slow down after TLI due to Earth's gravity?
    $endgroup$
    – Russell Borogove
    3 hours ago
















2












$begingroup$


I must be missing something - escape velocity is about 25,000mph. The distance to the moon about 240,000 miles. That computes to about a 9 hour flight. Why did it take 3 days for Apollo missions to reach the moon?










share|improve this question









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Marvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    Possible duplicate of Did Apollo's velocity slow down after TLI due to Earth's gravity?
    $endgroup$
    – Russell Borogove
    3 hours ago














2












2








2





$begingroup$


I must be missing something - escape velocity is about 25,000mph. The distance to the moon about 240,000 miles. That computes to about a 9 hour flight. Why did it take 3 days for Apollo missions to reach the moon?










share|improve this question









New contributor




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







$endgroup$




I must be missing something - escape velocity is about 25,000mph. The distance to the moon about 240,000 miles. That computes to about a 9 hour flight. Why did it take 3 days for Apollo missions to reach the moon?







orbital-mechanics the-moon apollo-program time escape-velocity






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edited 5 hours ago









Nathan Tuggy

3,81842637




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asked 6 hours ago









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  • 1




    $begingroup$
    Possible duplicate of Did Apollo's velocity slow down after TLI due to Earth's gravity?
    $endgroup$
    – Russell Borogove
    3 hours ago














  • 1




    $begingroup$
    Possible duplicate of Did Apollo's velocity slow down after TLI due to Earth's gravity?
    $endgroup$
    – Russell Borogove
    3 hours ago








1




1




$begingroup$
Possible duplicate of Did Apollo's velocity slow down after TLI due to Earth's gravity?
$endgroup$
– Russell Borogove
3 hours ago




$begingroup$
Possible duplicate of Did Apollo's velocity slow down after TLI due to Earth's gravity?
$endgroup$
– Russell Borogove
3 hours ago










2 Answers
2






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8












$begingroup$

Escape velocity is the velocity at a given altitude (usually the surface) that is enough to leave the body's sphere of influence with a positive net velocity. But if you leave a body at exactly escape velocity, your velocity bleeds off as you climb in exchange for gaining gravitational potential energy, and your velocity tends to the limit of zero at sufficiently large distances. (That is, your $V_{inf}$ is 0.)



Apollo did not leave at escape velocity. Rather, the trans-lunar injection gave considerable extra speed in order to make the trip much faster, so the lowest velocity was reached a few tens of thousands of miles short of the moon (where the gravity from the moon and the earth are equal), in the low thousands of miles per hour.






share|improve this answer









$endgroup$





















    1












    $begingroup$

    Developing on Nathan's answer, let's do some math. For simplicity we suppose here that we are not really going to the Moon, that only Earth's gravity us relevant.



    We leave at escape velocity, 25,000 miles per hour from an altitude of 4000 miles above the Earth's center, and climb straight up. As we do so our velocity decreases against Earth's gravity but remains matched to escape velocity at that altitude above Earth's center. Thus at an altitude of 6250 miles (2250 miles above Earth's surface) Earth's gravity has slowed the rocket down to 20,000 miles per hour which is the escape velocity at that altitude.



    Escape velocity is proportional to the $-1/2$ power of altitude above the center, and for 25,000 mph at 4000 miles the proportionality constant is (to three significant figures) $1.58×10^6$ (miles)$^{3/2}$/hr. So to get from an altitude of 4000 miles to 250,000 miles, at escape velocity, we need this much time:



    $int_{4000}^{240000} dfrac{dz}{1.58×10^6z^{-1/2}}=49.5text{ hr}$



    This is roughly correct but misses the fact that at the Moon's altitude we would still be going several thousand miles per hour upwards and the Moon's gravity would not have been strong enough to catch us from such speed. Nathan correctly points out that we went off slower than escape velocity so that the Moon, which is still bound to Earth, could reel us in. Hence the extra day. Similarly, when taking off from the Moon we had to go slowly enough for Earth to pull us in rather than sending us off like a slingshot; technically blasting from the Moon was not up to its escape velocity either.



    So ultimately the reason Apollo missions had such long transit times between the Earth and Moon was not any limit on rocket power but by the limits of gravity within which a lunar mission has to work.






    share|improve this answer











    $endgroup$









    • 2




      $begingroup$
      But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location.
      $endgroup$
      – Uwe
      1 hour ago






    • 2




      $begingroup$
      The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down.
      $endgroup$
      – Mark
      37 mins ago











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    2 Answers
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    2 Answers
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    active

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    8












    $begingroup$

    Escape velocity is the velocity at a given altitude (usually the surface) that is enough to leave the body's sphere of influence with a positive net velocity. But if you leave a body at exactly escape velocity, your velocity bleeds off as you climb in exchange for gaining gravitational potential energy, and your velocity tends to the limit of zero at sufficiently large distances. (That is, your $V_{inf}$ is 0.)



    Apollo did not leave at escape velocity. Rather, the trans-lunar injection gave considerable extra speed in order to make the trip much faster, so the lowest velocity was reached a few tens of thousands of miles short of the moon (where the gravity from the moon and the earth are equal), in the low thousands of miles per hour.






    share|improve this answer









    $endgroup$


















      8












      $begingroup$

      Escape velocity is the velocity at a given altitude (usually the surface) that is enough to leave the body's sphere of influence with a positive net velocity. But if you leave a body at exactly escape velocity, your velocity bleeds off as you climb in exchange for gaining gravitational potential energy, and your velocity tends to the limit of zero at sufficiently large distances. (That is, your $V_{inf}$ is 0.)



      Apollo did not leave at escape velocity. Rather, the trans-lunar injection gave considerable extra speed in order to make the trip much faster, so the lowest velocity was reached a few tens of thousands of miles short of the moon (where the gravity from the moon and the earth are equal), in the low thousands of miles per hour.






      share|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        Escape velocity is the velocity at a given altitude (usually the surface) that is enough to leave the body's sphere of influence with a positive net velocity. But if you leave a body at exactly escape velocity, your velocity bleeds off as you climb in exchange for gaining gravitational potential energy, and your velocity tends to the limit of zero at sufficiently large distances. (That is, your $V_{inf}$ is 0.)



        Apollo did not leave at escape velocity. Rather, the trans-lunar injection gave considerable extra speed in order to make the trip much faster, so the lowest velocity was reached a few tens of thousands of miles short of the moon (where the gravity from the moon and the earth are equal), in the low thousands of miles per hour.






        share|improve this answer









        $endgroup$



        Escape velocity is the velocity at a given altitude (usually the surface) that is enough to leave the body's sphere of influence with a positive net velocity. But if you leave a body at exactly escape velocity, your velocity bleeds off as you climb in exchange for gaining gravitational potential energy, and your velocity tends to the limit of zero at sufficiently large distances. (That is, your $V_{inf}$ is 0.)



        Apollo did not leave at escape velocity. Rather, the trans-lunar injection gave considerable extra speed in order to make the trip much faster, so the lowest velocity was reached a few tens of thousands of miles short of the moon (where the gravity from the moon and the earth are equal), in the low thousands of miles per hour.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 5 hours ago









        Nathan TuggyNathan Tuggy

        3,81842637




        3,81842637























            1












            $begingroup$

            Developing on Nathan's answer, let's do some math. For simplicity we suppose here that we are not really going to the Moon, that only Earth's gravity us relevant.



            We leave at escape velocity, 25,000 miles per hour from an altitude of 4000 miles above the Earth's center, and climb straight up. As we do so our velocity decreases against Earth's gravity but remains matched to escape velocity at that altitude above Earth's center. Thus at an altitude of 6250 miles (2250 miles above Earth's surface) Earth's gravity has slowed the rocket down to 20,000 miles per hour which is the escape velocity at that altitude.



            Escape velocity is proportional to the $-1/2$ power of altitude above the center, and for 25,000 mph at 4000 miles the proportionality constant is (to three significant figures) $1.58×10^6$ (miles)$^{3/2}$/hr. So to get from an altitude of 4000 miles to 250,000 miles, at escape velocity, we need this much time:



            $int_{4000}^{240000} dfrac{dz}{1.58×10^6z^{-1/2}}=49.5text{ hr}$



            This is roughly correct but misses the fact that at the Moon's altitude we would still be going several thousand miles per hour upwards and the Moon's gravity would not have been strong enough to catch us from such speed. Nathan correctly points out that we went off slower than escape velocity so that the Moon, which is still bound to Earth, could reel us in. Hence the extra day. Similarly, when taking off from the Moon we had to go slowly enough for Earth to pull us in rather than sending us off like a slingshot; technically blasting from the Moon was not up to its escape velocity either.



            So ultimately the reason Apollo missions had such long transit times between the Earth and Moon was not any limit on rocket power but by the limits of gravity within which a lunar mission has to work.






            share|improve this answer











            $endgroup$









            • 2




              $begingroup$
              But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location.
              $endgroup$
              – Uwe
              1 hour ago






            • 2




              $begingroup$
              The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down.
              $endgroup$
              – Mark
              37 mins ago
















            1












            $begingroup$

            Developing on Nathan's answer, let's do some math. For simplicity we suppose here that we are not really going to the Moon, that only Earth's gravity us relevant.



            We leave at escape velocity, 25,000 miles per hour from an altitude of 4000 miles above the Earth's center, and climb straight up. As we do so our velocity decreases against Earth's gravity but remains matched to escape velocity at that altitude above Earth's center. Thus at an altitude of 6250 miles (2250 miles above Earth's surface) Earth's gravity has slowed the rocket down to 20,000 miles per hour which is the escape velocity at that altitude.



            Escape velocity is proportional to the $-1/2$ power of altitude above the center, and for 25,000 mph at 4000 miles the proportionality constant is (to three significant figures) $1.58×10^6$ (miles)$^{3/2}$/hr. So to get from an altitude of 4000 miles to 250,000 miles, at escape velocity, we need this much time:



            $int_{4000}^{240000} dfrac{dz}{1.58×10^6z^{-1/2}}=49.5text{ hr}$



            This is roughly correct but misses the fact that at the Moon's altitude we would still be going several thousand miles per hour upwards and the Moon's gravity would not have been strong enough to catch us from such speed. Nathan correctly points out that we went off slower than escape velocity so that the Moon, which is still bound to Earth, could reel us in. Hence the extra day. Similarly, when taking off from the Moon we had to go slowly enough for Earth to pull us in rather than sending us off like a slingshot; technically blasting from the Moon was not up to its escape velocity either.



            So ultimately the reason Apollo missions had such long transit times between the Earth and Moon was not any limit on rocket power but by the limits of gravity within which a lunar mission has to work.






            share|improve this answer











            $endgroup$









            • 2




              $begingroup$
              But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location.
              $endgroup$
              – Uwe
              1 hour ago






            • 2




              $begingroup$
              The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down.
              $endgroup$
              – Mark
              37 mins ago














            1












            1








            1





            $begingroup$

            Developing on Nathan's answer, let's do some math. For simplicity we suppose here that we are not really going to the Moon, that only Earth's gravity us relevant.



            We leave at escape velocity, 25,000 miles per hour from an altitude of 4000 miles above the Earth's center, and climb straight up. As we do so our velocity decreases against Earth's gravity but remains matched to escape velocity at that altitude above Earth's center. Thus at an altitude of 6250 miles (2250 miles above Earth's surface) Earth's gravity has slowed the rocket down to 20,000 miles per hour which is the escape velocity at that altitude.



            Escape velocity is proportional to the $-1/2$ power of altitude above the center, and for 25,000 mph at 4000 miles the proportionality constant is (to three significant figures) $1.58×10^6$ (miles)$^{3/2}$/hr. So to get from an altitude of 4000 miles to 250,000 miles, at escape velocity, we need this much time:



            $int_{4000}^{240000} dfrac{dz}{1.58×10^6z^{-1/2}}=49.5text{ hr}$



            This is roughly correct but misses the fact that at the Moon's altitude we would still be going several thousand miles per hour upwards and the Moon's gravity would not have been strong enough to catch us from such speed. Nathan correctly points out that we went off slower than escape velocity so that the Moon, which is still bound to Earth, could reel us in. Hence the extra day. Similarly, when taking off from the Moon we had to go slowly enough for Earth to pull us in rather than sending us off like a slingshot; technically blasting from the Moon was not up to its escape velocity either.



            So ultimately the reason Apollo missions had such long transit times between the Earth and Moon was not any limit on rocket power but by the limits of gravity within which a lunar mission has to work.






            share|improve this answer











            $endgroup$



            Developing on Nathan's answer, let's do some math. For simplicity we suppose here that we are not really going to the Moon, that only Earth's gravity us relevant.



            We leave at escape velocity, 25,000 miles per hour from an altitude of 4000 miles above the Earth's center, and climb straight up. As we do so our velocity decreases against Earth's gravity but remains matched to escape velocity at that altitude above Earth's center. Thus at an altitude of 6250 miles (2250 miles above Earth's surface) Earth's gravity has slowed the rocket down to 20,000 miles per hour which is the escape velocity at that altitude.



            Escape velocity is proportional to the $-1/2$ power of altitude above the center, and for 25,000 mph at 4000 miles the proportionality constant is (to three significant figures) $1.58×10^6$ (miles)$^{3/2}$/hr. So to get from an altitude of 4000 miles to 250,000 miles, at escape velocity, we need this much time:



            $int_{4000}^{240000} dfrac{dz}{1.58×10^6z^{-1/2}}=49.5text{ hr}$



            This is roughly correct but misses the fact that at the Moon's altitude we would still be going several thousand miles per hour upwards and the Moon's gravity would not have been strong enough to catch us from such speed. Nathan correctly points out that we went off slower than escape velocity so that the Moon, which is still bound to Earth, could reel us in. Hence the extra day. Similarly, when taking off from the Moon we had to go slowly enough for Earth to pull us in rather than sending us off like a slingshot; technically blasting from the Moon was not up to its escape velocity either.



            So ultimately the reason Apollo missions had such long transit times between the Earth and Moon was not any limit on rocket power but by the limits of gravity within which a lunar mission has to work.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 51 mins ago

























            answered 1 hour ago









            Oscar LanziOscar Lanzi

            2635




            2635








            • 2




              $begingroup$
              But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location.
              $endgroup$
              – Uwe
              1 hour ago






            • 2




              $begingroup$
              The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down.
              $endgroup$
              – Mark
              37 mins ago














            • 2




              $begingroup$
              But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location.
              $endgroup$
              – Uwe
              1 hour ago






            • 2




              $begingroup$
              The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down.
              $endgroup$
              – Mark
              37 mins ago








            2




            2




            $begingroup$
            But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location.
            $endgroup$
            – Uwe
            1 hour ago




            $begingroup$
            But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location.
            $endgroup$
            – Uwe
            1 hour ago




            2




            2




            $begingroup$
            The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down.
            $endgroup$
            – Mark
            37 mins ago




            $begingroup$
            The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down.
            $endgroup$
            – Mark
            37 mins ago










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