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!

Question about satellites falling 'around' Earth

  1. Aug 18, 2010 #1
    I've studied the basics of this in a physics textbook, and have searched on this forum, but one thing about this escapes me.

    My understanding is that a satellite is elevated somewhere above the Earth's atmosphere where air resistance is negligible. The satellite is then accelerated to a very high speed where it is moving laterally as fast as it is falling toward the center of the Earth so it continues to orbit around the Earth.

    What I don't understand is if the acceleration of gravity, being 9.8m/s^2 (or slightly less since it's farther from the center of the Earth) is causing it to fall toward Earth faster and faster, how does a constant lateral speed keep it in orbit? In other words, if it falls 5 meters in the first second, it'll fall exponentially faster each additional second as it picks up acceleration with no terminal velocity (due to the lack of air resistance) until it is falling faster toward Earth than it is moving laterally in order to 'escape' Earth, so how do satellites stay in orbit without falling back down toward us?

    Of course my reasoning is incorrect somewhere, so please help me understand!
     
    Last edited: Aug 18, 2010
  2. jcsd
  3. Aug 18, 2010 #2
    What you're missing is that the acceleration of gravity does have a constant magnitude, but not a constant direction. After the satellite has fallen 5 meters in the first second, it will have moved more than 7 km to the side, it will be at the same distance from the earth, and the force of gravity will still be perpendicular to the velocity.

    Your exponential (should be quadratic) acceleration is only valid if gravity has a constant magnitude AND direction.
     
  4. Aug 18, 2010 #3
    Galileo worked this one out. (around 1590)

    He imagined standing on top of a mountain and throwing a stone, which falls to earth some distance away.
    If he throws it harder, the stone still falls, but because of the curvature of the earth, it falls much further away than if the earth were flat.
    If he throws the stone really hard, it keeps falling but because the earth also keeps curving away, it never strikes the ground.
     
  5. Aug 18, 2010 #4

    russ_watters

    User Avatar

    Staff: Mentor

    Speed is a acceleration times time and distance is acceleration times time squared, right? (meaning distance changes at an ever increasing rate). The horizon is curved: it also falls away from the satellite as a function of time squared.
     
  6. Aug 18, 2010 #5
    I'm not sure I understand. Are you saying that it will fall the same distance toward the Earth in each passing second, instead of traveling more units in each passing second?

    So because the direction of acceleration is constantly changing, it resets to a constant 9.8 (edit: mistake, I should have said 'resets to a constant [for example] 5 meter drop per second)?
     
    Last edited: Aug 19, 2010
  7. Aug 18, 2010 #6
    One vector is straight ahead, in the direction that the orbiting object is going at any single moment. The other vector is straight toward the center of the earth. The resultant vector is a seemingly infinite circle. There's nothing stopping the object's motion except it keeps getting led around in circles by its leash like a retarded dog tied to a yard stake.
     
  8. Aug 18, 2010 #7
    Hi Yzp123,

    To start with.. 9.8 is not a constant value, but kinda avg value. g is not 9.8 everywhere on the surface of the earth and it also varies as you acquire altitude. g = MG/R^2, so it depends on the distance R.

    To answer to your specific question, the satellite (or for that matter any object rotating earth, or any other body) will not only have a velocity towards earth, but will have a tangential velocity.

    So, imagine a stone tied on a rubber band and you are rotating it circularly. The faster you rotate it, the rubber band will expand more. At one point, when you have stabilized the rotation speed, the rubber will stop expanding, but will not collapse.

    Here, the pull of the rubber band is the gravitational pull (the only difference is, gravity reduces as distance increases, but for the explanation purpose, just think of the pull). The stone is currently experiencing a centrifugal force i.e. it would like to escape into space breaking the rubber band, but the force makes it stay there, but.. in order to escape, it don't act exactly opposite to the gravity, but tangentially, i.e. in an angle of 90 degree.

    So, each time, the stone (satellite) tries to move tangentially, the rubber (gravity) will pull it in. These fractions of motion i.e down wards + tangentially, results in a circular motion around your hand (earth or the source of gravity).

    Hope this is of some help.
     
  9. Aug 18, 2010 #8
    The satellite doesn't "pick up acceleration". Acceleration stays constant at g.

    Second, acceleration isn't a change in speed, it's a change in velocity. Velocity consists of speed and direction. Therefore, acceleration is either a change in speed, direction or a combination of both. In a satellite's case, it's purely direction, while the speed stays the same. That's why it doesn't fall faster and faster towards the ground.

    An easy example would be to run around in a circle. You're constantly accelerating towards the center, but your speed is not necessarily changing.
     
  10. Aug 18, 2010 #9

    russ_watters

    User Avatar

    Staff: Mentor

    Consider other situations where the force is always perpendicular to the direction of motion, such as an object on a string that you whirl around in a circle. The force and acceleration are always constant and always directed inwards.
     
  11. Aug 19, 2010 #10
    I think it's starting to make sense to me. So the 10th second after the satellite's release it doesn't fall a larger number of units (toward the Earth vector) than it does in the 1st second after release?

    However, can the fall toward Earth still be described in 9.x m/s^2, or is this only for one direction? Maybe willem was saying that this is no longer valid without direction.
     
  12. Aug 19, 2010 #11
    What do you mean with "the fall can be described in 9.x m/s^2". If you mean "works the same as constant acceleration used for projectile motion for small distances on earth such as thrown baseballs and the like", the answer is no.
     
  13. Aug 19, 2010 #12
    Hmm. But the acceleration of a satellite is still a constant acceleration, right?

    Instead of the speed changing, the direction is changing, so can we still say it's 9.8 m/s^2, but whereas a baseball falling at that acceleration would have its speed changing without a direction change, a satellite would have a direction change without its speed increasing, still making the same 9.8 m/s^2?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about satellites falling 'around' Earth
Loading...