As we increase the average altitude of an orbiting satellite, what happens to the satellite’s speed and period?
- TomLv 75 months agoFavourite 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.
- ZirpLv 75 months ago
Well unless the seed goes down and the period gets longer, the satellite will fly away into space
- Andrew SmithLv 75 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)
- JimLv 75 months ago
Orbital time is a little weird, see pic, proportional to √r³
Mass of satellite disappears from equation.
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- 5 months ago
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
- Anonymous5 months ago
speed slows down, time to make an orbit increases.