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Momentum and acceleration due to gravity

  1. Dec 17, 2006 #1
    what object is dangerous for the target, the one that has greatest momentum or the one that has least momentum and why? what is safest?
    I answeres all the parts that contain momentum and velocity and force but i stucked in this one

    what is the value of g (acceleration due to gravity) if the radius doubled?
    is the right formular for this question is Fg= G m1m2/r^2, or do i have to substitute the Fg with m*a, then do the rest
  2. jcsd
  3. Dec 17, 2006 #2


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    Things with momentum are dangerous because it means they're moving. Are you more scared of a knife on a table or a knife thrown at you?
    That equation you posted is correct. Everything should factor out leaving 1/(r^2). How does the gravity change when the r in 1/(r^2) is changed from 1 to 2, or from 2 to 4?
  4. Dec 17, 2006 #3
    so you mean that Fg is the same as g right
    and the object that has greatest momentum is the most dangerous
  5. Dec 17, 2006 #4
    if i doubled the radius then the g= 1/4
  6. Dec 17, 2006 #5


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    No, Fg is not the same as g. From the equation, F=(Gm1m2)/r^2 which gives m2g=(Gm1m2)/r^2 and so g=Gm1/r^2


    Right idea: the new value of g obtained when doubling the radius will be 1/4 times the old value of g. So, write [itex]g_{new}=\frac{1}{4}g [/itex]
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