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Is a Satellite free falling?

  1. Dec 15, 2014 #1
    it is said that a settelite is a free falling body. For free falling body its velocity should free continuesly increasing. Is it true? I am confused..
     
  2. jcsd
  3. Dec 15, 2014 #2

    Dale

    Staff: Mentor

    It is free falling. The velocity of a free-falling body in a gravitational field should be continuously changing, but not necessarily increasing. The change can be a decrease or a change in direction, not necessarily an increase.
     
  4. Dec 15, 2014 #3
    but gravitational acceleration is positive when the setelite is falling down, and positive acceleration is only produce when magnitude of velocity is increasing.?
     
  5. Dec 15, 2014 #4

    Dale

    Staff: Mentor

    It doesn't really make sense to think of gravitational acceleration as positive or negative. It is a vector which points towards the center of the earth.

    At any point on a satellite's trajectory you can break the acceleration vector into components parallel and perpendicular to the velocity. The parallel component will change the speed and the perpendicular component will change the direction.
     
  6. Dec 15, 2014 #5
    it becomes more confusing for me now
     
  7. Dec 15, 2014 #6

    Dale

    Staff: Mentor

    What is confusing? Do you know what a vector is?
     
  8. Dec 15, 2014 #7

    Bandersnatch

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    How about this then:

    Free-fall is the same as being weightless. You are weightless if the only force acting on you is the force of gravity. That's it. Directions of velocities and accelerations don't matter.

    Whether the satellite is falling exactly towards the planet, flying away from it, going sideways in circles or ellipses - none of it matters. As long as it is under the influence of gravity and only gravity, it's in free-fall.

    If you add engines to the satellite and turn them on, or if you add the effects of atmospheric drag, you no longer have just gravity acting on the satellite, and it is no longer in free-fall.
     
  9. Dec 15, 2014 #8

    yeah i know, a quantity which need both magnitude and direction for complete discription
     
  10. Dec 15, 2014 #9

    Dale

    Staff: Mentor

    OK, so velocity and acceleration each are vectors. Each has a magnitude and each has a direction. Their directions may be different. Does that make sense?
     
  11. Dec 15, 2014 #10

    Dale

    Staff: Mentor

    Also, do you know what it means to break a vector into components?
     
  12. Dec 15, 2014 #11
    yeah resolving in vertical and horizontal components
     
  13. Dec 15, 2014 #12
    yeah when velocity is decreasing the direction of acceleration will be opposite...
     
  14. Dec 15, 2014 #13

    Dale

    Staff: Mentor

    Exactly. The horizontal component is the component perpendicular to gravity and the vertical component is the component parallel to gravity.

    Just like you can break velocity into components parallel and perpendicular to gravity, you could instead break gravity into components parallel and perpendicular to velocity.

    Does that idea make sense?
     
  15. Dec 15, 2014 #14
    yeah now i understand a little bit..
     
  16. Dec 15, 2014 #15

    russ_watters

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    Consider this: earth's gravitational acceleration has a known value. So if you ignore the perpendicular component of the acceleration (the change in direction), the acceleration you get won't match that known value for any trajectory besides straight down.

    [Edit]
    Another: if you are "floating" in a spaceship with no sensors and no windows, is there any way to tell if it is in a circular orbit, elliptical orbit, out in empty space somewhere or plummeting toward pancaking onto a planet?
     
    Last edited: Dec 15, 2014
  17. Dec 15, 2014 #16
    Have you ever herd of the thought experiment with the bullet going faster and faster? Imagine firing a gun. The bullet goes really far but hits the ground eventually. If you could make a super powerful gun that fires the bullet really fast, the bullet would still be falling toward the ground, but the curvature of the earth would come into play (the ground would go "away" from the bullet) and thus the bullet keeps "falling" toward the ground, but we say it's in orbit. The velocity that governs this is given by the eqn:

    vorb= (GM/r)^1/2
    with G=big G
    M= mass of earth
    r= radius of earth

    If you fire the bullet even faster (vorb*2^(1/2))eventually it will escape the gravitational pull of earth and go into intergalactic space, never to be seen again (like the satellite voyager).
    I hope that helps, I never understood what people meant by things free falling until I learned of this thought experiment.
     
    Last edited: Dec 15, 2014
  18. Dec 15, 2014 #17

    A.T.

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  19. Dec 15, 2014 #18

    rcgldr

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    It might also help to consider the situation with a circular orbit. The satellite is always accelerating perpendicular to it's current direction. For an analogy, imagine driving a car in a circular path at constant speed, the car is always accelerating (centripetal acceleration) but it's speed never changes and the radius of the turn never changes (circular path).
     
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