1. Limited time only! Sign up for a free 30min personal 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!

Freefall of an object in Space (towards earth)

  1. Dec 4, 2014 #1
    Hi guys,

    My professor showed us in class how to find the final velocity of an object as it hits the the atmosphere of earth(Rf=6.37e6) and I think he might have made a mistake. I have been trying to solve the same problem but I get different answer than the one my professor showed in class.
    The Ri=6.37e6 + 10^5. Vi=0.

    The way I solved it - 1/2mvf^2 −GmM/Rf=−GmM/Ri (I canceled out the m)
    My answer is 695 m/s.

    It seems like my professor didn't cancel out the first m and he got 435 or something around that number.
    I wonder if there is a reason why he didn't cancel the first m, or rather he just forgot to do so.

    Thank you so much!
     
  2. jcsd
  3. Dec 5, 2014 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You might want to check your values for "G" and for "M." If your prof came up with 435, you've both missed something.
     
  4. Dec 5, 2014 #3
    My answer is 695, not 435. I think my professor canceled out the m from the right side but didn't from the left side of the equation.
    the G = 6.67e-11
    M = 5.98e24
    m= 10 kg - but we canceled it out anyway so it doesn't matter.
     
  5. Dec 5, 2014 #4

    Nugatory

    User Avatar

    Staff: Mentor

    Hits the atmosphere of the earth starting from where? "Free-fall" makes it sound like the object is falling in from infinity, but your calculation doesn't make sense for that problem. The formula you're using looks like you're trying to calculate the speed of an object dropped from the top of the atmosphere to the surface of the earth (if there were no air resistance), and in that case Bystander is right - neither your 695 m/sec nor the professor's 435 m/sec is close.

    As an aside: The difference between ##(6.37\times{10}^6)^2## and ##(6.37\times{10}^6+1\times{10}^5)^2## is so small as to be irrelevant (especially with only one significant digit in the atmospheric thickness) that you can save yourself some work and take the gravitational force to be the same at both altitudes.
     
  6. Dec 5, 2014 #5
    I see what you did there, I have no idea why the professor solved it this way then. that's how he showed in class. the the object is moving towards earth from that distance and we need to figure out what would be the speed when Rf= 6.37e6 meters. (starting from the Ri=6.37e6+10^5).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Freefall of an object in Space (towards earth)
  1. Objects in space (Replies: 3)

Loading...