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!

Finding height of revolving body at speed ##\sqrt{2gR}##

  1. May 2, 2015 #1
    1. The problem statement, all variables and given/known data
    There was a question in which an object was released upwards from earth with a velocity of ## 2\sqrt {gR}##.
    Using conservation of energy I found that the speed of the body in the orbit was ## \sqrt {2gR} ##
    Everything's fine till now. But then I wanted to find the height at which the body was travelling from the centre of the earth.

    2. Relevant equations
    Equating centripetal force and gravitational force
    v is the speed of body
    m is mass of body.
    R is radius of earth.
    The body is at a height of h from the center of the earth.
    To find h.

    3. The attempt at a solution
    h is coming out to be less that R. How is that possible?
     

    Attached Files:

  2. jcsd
  3. May 2, 2015 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't know how you worked out the orbital height or velocity.

    Try to work out the orbital height for the given initial velocity. I think you'll see that there's something wrong with the question.
     
  4. May 2, 2015 #3

    cnh1995

    User Avatar
    Homework Helper

    I haven't done any calculations but h can be less than R..R is the radius of the earth (distance from center to surface) and h is the height at which the object is above the surface...
     
  5. May 4, 2015 #4
    But the object isn't revolving at the given initial velocity.
     

    Attached Files:

  6. May 4, 2015 #5
    I have taken h to be the distance from the center. Then how can it be less than R?
     
  7. May 4, 2015 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Dear Ap,

    Where does your first equation come from ?
    To me it looks as if it comes from assuming a circular orbit around the earh.
    But from the problem statement I get the impression the object is sent off in a vertical direction. With an enormous speed. It can get very far away from earth ! How far ? Your turn !
     
  8. May 4, 2015 #7
    Why can't we consider it to be launched in circular orbit like a satellite? The orbit assumed to be circular?
     
  9. May 4, 2015 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Actually, that was part II of my approach.

    But let's not go too fast and follow your assumption: the object is launched with a given speed.

    In order to bring the object from where it starts off to the height of its 'orbit', it must loose some of its kinetic energy to overcome the gravitational attraction by the earth. In your first equation you use the same speed as the object had at launch, so this trade-in of kinetic energy for potential energy from gravity isn't taken into account. Can you think of a relationship that does handle that a little better ?
     
  10. May 4, 2015 #9
    The given initial speed was ## 2\sqrt{gR}##.In the second attachment I obtained the orbit speed.
     
  11. May 4, 2015 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, sorry. I read over that ##2\sqrt{gR}## altogether.

    But don't you get a little suspicious if the launching speed is ##\sqrt 2## times the escape velocity ? What induces you to think there will be an orbit at all ?
     
  12. May 4, 2015 #11
    If the launching speed is more than escape velocity, doesn't the extra energy add to the KE of body in orbit, i.e velocity?

    A body can orbit if its kinetic energy is equal to its gravitational potential energy. What happens if it has even more energy?
     
  13. May 4, 2015 #12

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    As you calculated, it loses half its kinetic energy to escape from the influence of the earth gravitational field. If there are no other celestial bodies (sun !?) around (*) to play a role, the remaining kinetic energy is constant and the thing flies off into infinity at the escape velocity, 11.2 m/s

    The "part II" I alluded to is that it doesn't matter in which direction the launch takes place (as long as it's upward :smile:), because it's a calculation with energies.

    (*) there is also the rotation of the earth around its axis to take into account, as well as the rotation of the earth around the sun, and there is some friction with the atmosphere, but we seem to be allowed to ignore those here.
     
  14. May 4, 2015 #13
    Then how do we increase the speed of any orbiter?
     
  15. May 4, 2015 #14

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First of all, correct my 11.2 m/s to 11.2 km/s. A mere factor 1000 :wink:.

    Increasing the speed of an orbiter ? You don't mean something like "by accelerating the thing, e.g. with a rocket engine" , I suppose.

    There's the orbit itself to consider: for circular orbits there are simple expressions. Once in a circular orbit, accelerating)/decelerating) causes the orbit to become elliptical (Kepler orbit), i.e. initially the thing goes higher or lower. Proper acceleration/deceleration at a later point can make the orbit circular again.
     
  16. May 4, 2015 #15
    If it has escape velocity, why would it escape ? Writing error?
     
  17. May 4, 2015 #16

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't understand this question. If it has escape velocity, it has enough kinetic energy to overcome the graviational field. So it can go infinitely far.
     
  18. May 4, 2015 #17
    why wouldn't it orbit?
     
  19. May 4, 2015 #18

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The attracting force from the earth isn't enough to curve the trajectory sufficiently.
    Or: by the time the radial deceleration has dropped to zero, there is still outward radial speed left over.
     
  20. May 4, 2015 #19
    Then how do we make things orbit?
     
  21. May 4, 2015 #20

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Launching them with a lower speed !
    (Actually, speed isn't the issue: energy is what counts -- as we have seen )
    11.2 km/s is awfully fast (33 x speed of sound) !
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted