Anonymous
Anonymous asked in Science & MathematicsPhysics · 5 months ago

As we increase the average altitude of an orbiting satellite, what happens to the satellite’s speed and period?

6 Answers

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  • Tom
    Lv 7
    5 months ago
    Favourite answer

    The speed goes down as objects in orbit slow down the farther away they they are, and the orbit will be LONGER in distance.   A fact that  makes space docking a bit tricky.  As moving up and down to intercept also causes the craft to move (towards or away from the docking target too.

  • Zirp
    Lv 7
    5 months ago

    Well unless the seed goes down and the period gets longer, the satellite will fly away into space

  • 5 months ago

    Johannes Kepler provided the answer.  R^3 is proportional to T^2 for any orbit around a fixed body.  as V is proportional to R/ T this means that V is proportional to R / (sqrt( R^3) = sqrt(R^2/ R^3) = sqrt( 1/R)

  • Jim
    Lv 7
    5 months ago

    Orbital time is a little weird, see pic, proportional to √r³

    Mass of satellite disappears from equation.

    Attachment image
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  • You can think of the orbit as being roughly circular.

    v = sqrt(G * M / r)

    G = gravitational constant

    M = mass of object being orbited

    r = distance between centers of mass

    v = orbital velocity

    The actual formula is a little more complex since it involves the mass of the orbiting object as well, but since the mass of a satellite compared to the mass of the earth is so small, it just makes it easier to not even include it

    v^2 = G * M / r

    G and M stay the same.  We can call this k

    v^2 = k / r

    As r increases, v^2 decreases.  Can you see that?

    r * v^2 = k

    If we increase r by some factor p, then v decreases by a factor of sqrt(p).

    Now, the orbit is 2 * pi * r, and the velocity is v, so the time is 2 * pi * r / v

    v * t = 2 * pi * r

    t = 2pi * r / v

    t^2 = 4 * pi^2 * r^2 / v^2

    t^2 = 4 * pi^2 * r^2 / (k / r)

    t^2 = 4 * pi^2 * r^3 / k

    4 * pi^2 / k = j, which is just some other constant

    t^2 = j * r^3

    Looks a bit like Kepler's 3rd law of planetary motion, doesn't it?  There's a reason for that...

    As the height increases, the period of the orbit increases as well.

    t^(2/3) = j^(1/3) * r

    j^(1/3) is just some constant, we can call it a

    t^(2/3) = a * r

  • Anonymous
    5 months ago

    speed slows down, time to make an orbit increases.

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