3 ms·

Can you explain why you have to pay for the Sun's gravitational well on the way to Mercury? From the famous XKCD illustration[1] I get the impression that it mi

Can you explain why you have to pay for the Sun's gravitational well on the way to Mercury? From the famous XKCD illustration[1] I get the impression that it might be hard to "stop at" Mercury since the pull of the Sun is relatively strong. Is there no clever trajectory that comes close enough to Mercury to get caught in its orbit? [1] https://xkcd.com/681/ https://xkcd.com/681/

forand 6y ago
You want to stop when you get to Mercury. In the XKCD illustration you have fallen quite a bit from Earth to Mercury and now have that as kinetic energy. You can dissipate that by hitting Mercury but that would likely have significant negative side effects.
- lmilcin 6y ago
  Loosing the energy on Mercury is only part of the problem. You also have to get there. Contrary to popular belief, going closer to the Sun is as much difficult as going further away from it. Orbits require energy to change them regardless which direction you want to do it. You can go to tables and see going to Mercury costs almost the same as going to Jupiter: https://en.wikipedia.org/wiki/Delta-v_budget https://en.wikipedia.org/wiki/Delta-v_budget There are handy maps of Sol system if you are interested: https://forum.kerbalspaceprogram.com/index.php?/topic/131682-is-there-a-delta-v-chart-for-real-solar-system-mod/ https://forum.kerbalspaceprogram.com/index.php?/topic/131682... You can of course minimize the energy even further using gravity assists, but this would require multi-year journey. There is another trick when you want to go very close to the Sun, which is to go further away first, and then use much less energy to actually stop your orbital speed and fall however deep into Sun well you want. Unfortunately, Mercury is just far enough from the Sun to make it unprofitable and it also takes an awful amount of time to execute.
AllegedAlec 6y ago
A better image (imho): https://i.imgur.com/SqdzxzF.png https://i.imgur.com/SqdzxzF.png
lmilcin 6y ago
I think Randall's image is a bit misleading. I mean it is correct and fun, but for an uninitiated person it might cause impression that going down the well costs zero energy. That of course is true, if you are stationary. Then you just start falling like apple from apple tree. But almost no objects in space are stationary, typically everything orbits something and frequently multiple things at the same time. Think about a rocket that is on an orbit around Earth. It is not free to fall to Earth, it must expend delta V to slow down enough so that its perigee is within Earth atmosphere (and typically well within atmosphere or it will just bounce off it). At the same time, the rocket orbits the Sun, so even if it leaves Earth it still is not free to "fall" into Mercury. It must expend delta V to slow down so that its perihelion lowers at least to Mercury orbit. So the graph on xkcd is correct, but with regards to bowling balls you might want to shoot up or let fall down from a position that is stationary in some celestial body's frame of reference.

forand 6y ago

You want to stop when you get to Mercury. In the XKCD illustration you have fallen quite a bit from Earth to Mercury and now have that as kinetic energy. You can dissipate that by hitting Mercury but that would likely have significant negative side effects.

lmilcin 6y ago
Loosing the energy on Mercury is only part of the problem. You also have to get there. Contrary to popular belief, going closer to the Sun is as much difficult as going further away from it. Orbits require energy to change them regardless which direction you want to do it. You can go to tables and see going to Mercury costs almost the same as going to Jupiter: https://en.wikipedia.org/wiki/Delta-v_budget https://en.wikipedia.org/wiki/Delta-v_budget There are handy maps of Sol system if you are interested: https://forum.kerbalspaceprogram.com/index.php?/topic/131682-is-there-a-delta-v-chart-for-real-solar-system-mod/ https://forum.kerbalspaceprogram.com/index.php?/topic/131682... You can of course minimize the energy even further using gravity assists, but this would require multi-year journey. There is another trick when you want to go very close to the Sun, which is to go further away first, and then use much less energy to actually stop your orbital speed and fall however deep into Sun well you want. Unfortunately, Mercury is just far enough from the Sun to make it unprofitable and it also takes an awful amount of time to execute.

AllegedAlec 6y ago

A better image (imho): https://i.imgur.com/SqdzxzF.png https://i.imgur.com/SqdzxzF.png

lmilcin 6y ago

I think Randall's image is a bit misleading. I mean it is correct and fun, but for an uninitiated person it might cause impression that going down the well costs zero energy. That of course is true, if you are stationary. Then you just start falling like apple from apple tree. But almost no objects in space are stationary, typically everything orbits something and frequently multiple things at the same time. Think about a rocket that is on an orbit around Earth. It is not free to fall to Earth, it must expend delta V to slow down enough so that its perigee is within Earth atmosphere (and typically well within atmosphere or it will just bounce off it). At the same time, the rocket orbits the Sun, so even if it leaves Earth it still is not free to "fall" into Mercury. It must expend delta V to slow down so that its perihelion lowers at least to Mercury orbit. So the graph on xkcd is correct, but with regards to bowling balls you might want to shoot up or let fall down from a position that is stationary in some celestial body's frame of reference.