1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Falling into the Sun

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    (b) If an object of 0.1 Solar masses fell into the Sun (starting at rest at infinity) calculate what its final speed and energy would be.

    Mass of the Sun = 1.99 x [tex]10^{30} kg [/tex]
    Radius of the Sun = 6.96 x [tex]10^8 m [/tex]

    3. The attempt at a solution

    Having done the rest of the question (parts a, c and d), I returned to this part but am having trouble. A couple of years ago we studied free-fall time, but in this question we really aren't given much data. If I had the time, I could backwards-engineer the free fall time to get the velocity, but we don't. Having looked through my notes, I found the free fall velocity:

    [tex]v_{ff} = (\frac{2GM}{R})^\frac{1}{2}[/tex]

    We thought that integration may be the way to go, but we wouldn't know what with respect to - if we did it to r, then we could put in infinity and zero, but that goes a bit weird. I believe that the R in that equation should be the radius of the object, that we don't know! Also, we want to find the velocity, so changing it to [tex]\frac{dr}{dt}[/tex] won't actually help us. Anyone know where we'd begin?
     
  2. jcsd
  3. Dec 9, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Treat this as a conservation of energy problem. (That's where your equation for free fall speed comes from.) What's the change in PE when that mass goes from ∞ to the Sun's radius?
     
  4. Dec 9, 2008 #3
    Ah, I think I see what you're getting at. KE + PE = 0 therefore

    [tex] \frac{1}{2}v^2 = \frac{GM_{0}}{R} - \frac{GM_{0}}{R_{0}} [/tex]

    meaning that our velocity would be like the equation from my original post but where R is the radius of the Sun, [tex]R_{0}[/tex] is infinity, making that term disappear, and [tex]M_{0}[/tex] is the mass of the falling object, giving a [tex]v_{ff}[/tex] of ~195.3 km/s. Did I do that right?
     
  5. Dec 9, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    M0 is not the mass of the falling object. (Note that the mass of the object canceled out from your equation.)

    This also assumes that you can ignore the motion of the Sun itself. Probably not the best idea, since the falling object has significant mass. To be a bit more accurate, figure out the speeds of both falling mass and sun. (Assuming both start out at rest.)
     
  6. Dec 9, 2008 #5
    Thankfully, this question is only a couple of marks, we've been advised to ignore the Sun's motion for this. Going from the equation and making [tex]M_{0}[/tex] the mass of the Sun, gives the freefall velocity the same value as the escape velocity for the Sun. Would that be correct?
     
  7. Dec 9, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. Solving for the escape velocity is just the inverse of this problem. Same equation!
     
  8. Dec 9, 2008 #7
    Thanks ever so much! We were trying to go straight from the equation without taking into account where it comes from. We also have a slight tendency to overthink things! Thanks for your help! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Falling into the Sun
  1. The sun (Replies: 15)

  2. The mass of the Sun (Replies: 7)

Loading...