Why do the planets closest to the sun move at a greater velocity in their orbits?

  • #1
why does mercury move with such velocity while pluto is much slower?
 

Answers and Replies

  • #2
82
0
Study Kepler's Laws of Planetary Motion, then it will make sense.
 
  • #4
DaveC426913
Gold Member
18,992
2,483
If Pluto had as much velocity as Mercury does, it would be moving too fast for Sun's relatively weak gravity to bend its orbit into an ellipse; the fast-moving Pluto would just sail on by the solar system.

If Mercury moved as slowly as Pluto does, it would be moving too slow to stay in its elliptical orbit under the Sun's very strong pull, and would fall into the sun.

Indeed, (highly simplified) in the early years of the solar system, there would have been many bodies with many different velocities just like that. They flew off or collided with each other fell into the sun until the only ones left were the few we see that found stable orbits and stopped crashing into each other.
 
  • #5
2,967
5
The centripetal acceleration is:
[tex]
a = \frac{v^{2}}{r}
[/tex]

If this acceleration is provided by a gravitational force $F(r)$ from the Sun, that somehow varies with distance, then:
[tex]
F(r) = m a
[/tex]


From here. by knowing how the force varies with distance, we can find how the velocity varies with distance:
[tex]
v(r) = \sqrt{\frac{r F(r)}{m}}
[/tex]
The velocity will not depend on distance iff:
[tex]
F(r) = O(\frac{1}{r})
[/tex]
If the gravitational force falls off with distance faster than this, then the orbital velocity of more distant bodies should decrease. On the other hand, if it falls off more slowly, then it will increase.
 
  • #6
DaveC426913
Gold Member
18,992
2,483
The centripetal acceleration is:
[tex]
a = \frac{v^{2}}{r}
[/tex]

If this acceleration is provided by a gravitational force $F(r)$ from the Sun, that somehow varies with distance, then:
[tex]
F(r) = m a
[/tex]


From here. by knowing how the force varies with distance, we can find how the velocity varies with distance:
[tex]
v(r) = \sqrt{\frac{r F(r)}{m}}
[/tex]
The velocity will not depend on distance iff:
[tex]
F(r) = O(\frac{1}{r})
[/tex]
If the gravitational force falls off with distance faster than this, then the orbital velocity of more distant bodies should decrease. On the other hand, if it falls off more slowly, then it will increase.
Judging by the question being asked, do you think this answer will be useful (comprehensible) to the OP? :rolleyes:
 

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