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

### 6 Answers

- 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.