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Sun's gravitational force

  1. Apr 8, 2010 #1
    This book that I'm reading mentions something where the gravitational pull of the Sun is proportional to that planet's mass/its orbital radius. What is this proportionality constant?
  2. jcsd
  3. Apr 8, 2010 #2


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    Not quite , for an object in orbit the centrifugal force throwing it outwards is equal to the gravitational force pulling it into the object in the centre.
  4. Apr 8, 2010 #3
    Oh, okay. So really, F=(4pi^2mr)/T2 should be used.
  5. Apr 8, 2010 #4


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    The gravitational pull on any planet by the Sun is proportional to that planet's mass divided by the square of its orbital radius. The actual formula is

    [tex] F_g = G\frac{M_{sun}M_{planet}}{r_{orbit}^2}[/tex]

    where G is the gravitational constant and is equal to 6.673e-11 when M is measured in Kg and r in meters.
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