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If I'm in a satellite which is moving in circular orbit of radius $r$ around a planet which has no atmosphere, means no air friction. And if I throw a stone from satellite toward planet then it's obvious that it will land on its surface, same as probe landing. But here is some calculation- $$$$ If mass of planet is $M$ then orbital velocity of satellite will be $$v=\sqrt{\frac{GM}{r}}$$ which will be the velocity of stone also in the orbit. $$$$ Let's say the stone came to the radius $r/2$ Then according to law of conservation of angular momentum the orbital velocity of stone become double i.e. $v'=2v$. $$v'=2 \sqrt{\frac{GM}{r}}$$ To make the stone come further down toward planet the $v'$ should be less than the velocity which overcome the centripetal force i.e. $v'< \sqrt{\frac{GM}{(r/2)}}$ but it's not. This means that the stone will not come down. Then how the probe land on planets with no atmosphere like moon?

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  • $\begingroup$ I should have say that. $\endgroup$ – Brett Leigh Sep 9 '16 at 3:29
  • $\begingroup$ You decrease your orbital velocity with forward a rocket burn - which reduces your orbital height. $\endgroup$ – Martin Beckett Sep 9 '16 at 3:29
  • $\begingroup$ Is this what astronomers do to land a probe? $\endgroup$ – Brett Leigh Sep 9 '16 at 3:30
  • $\begingroup$ NASA are on the case. nasa.gov/ames/… $\endgroup$ – Farcher Sep 9 '16 at 8:39
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If I'm in a satellite which is moving in circular orbit of radius $r$ around a planet which has no atmosphere, means no air friction. And if I throw a stone from satellite toward planet then it's obvious that it will land on its surface, same as probe landing.

This is not the case. What you have done by throwing the stone toward the planet is to put the stone into a slightly elliptical orbit, one with an orbital period slightly greater than that of the satellite. In fact, if you had thrown the stone away from the planet with the same relative speed, the stone would have an identically shaped orbit.

The best way to land a probe released from a vehicle in a circular orbit about a planet is to have the $\Delta v$ supplied to the probe be directed against the vehicle's orbital velocity vector such that the resulting elliptical orbit just barely intersects the planet at periapsis. This was the approach used during the Apollo moon landings and other lunar landers.

The lander separated from the Apollo vehicle and then thrusted against the velocity vector so as to put the lander in a elliptical orbit that just barely intersected the Moon. Following this trajectory all the way to the Moon's surface would have resulted in a 1700 meter/second collision (mostly horizontally). The lander followed this free-fall trajectory until about 500 kilometers from what otherwise would have been a nasty impact, at which point it performed a breaking burn that killed most of the horizontal component of velocity. The lander then had to kill the vertical component of velocity so as to land gently on the Moon.

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To get out of orbit and land on an object you need to reduce your orbital height - ultimately to the planets radius. To do this you need to reduce the orbital speed. Orbital height only depends on speed.

One way to do this is to fire a rocket motor toward the direction you are travelling in - to slow you down. The problem with this is you have to carry a large rocket motor and its fuel. But it is the only option on an airless planet.

An alternative to get out of Earth orbit ( or any other planet with an atmosphere) is to slightly change speed and get into a more eliptical orbit, one part of which goes into the atmosphere. The friction with the atmosphere will reduce your speed, and so your height, which means more atmosphere and more friction, and so more speed loss - until you can land

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    $\begingroup$ Orbital height of a circular orbit depends only on speed. Of a more general (elliptic) orbit, the eccentricity matters. $\endgroup$ – Whit3rd Sep 9 '16 at 6:19

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