<p><a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="noreferrer">Kepler's third law</a> in this situation tells us that
$$ \frac{R^3}{P^2} = \frac{GM}{4\pi^2}, $$
where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period $P \propto R^{3/2}$.</p>

<p>To calculate the orbital speed, we divide the orbital circumference by the orbital period
$$ v = \frac{2\pi R}{P}. $$</p>

<p>Then substituting for $P$ from Kepler's third law, we get
$$ v = \sqrt{\frac{GM}{R}}$$</p>

<p>Thus, the larger the orbital radius, the slower the orbital speed.</p>

<p>Or you can just use Newton's second law and equate the centripetal acceleration to the gravitational force:
$$ m \frac{v^2}{R} = G\frac{Mm}{R^2},$$
where $m$ is the mass of the satellite, which leads directly to the same result.</p>

CC BY-SA 3.0
2017-04-05T08:47:19.763Z
2017-04-05T08:47:19.763Z
ProfRob
<p>For example, a geostationary satellite travels slower than the ISS. What is going on?</p>

CC BY-SA 3.0
2017-04-06T00:43:07.087Z
Have you considered looking at the force balance that keeps satellites in orbit?
AtmosphericPrisonEscape
CC BY-SA 3.0
2017-04-06T07:35:25.513Z
@AtmosphericPrisonEscape That is a bit misleading, there is no force balance as the only force acting on the orbiting object is gravity.
SE - stop firing the good guys
CC BY-SA 3.0
2017-04-06T14:48:25.067Z
@Hohmannfan: That doesn't matter. It doesn't seem to be a university-level question, and the most intuitive 'layman' way of understanding Keplerian orbits is through inertial forces.
AtmosphericPrisonEscape
<satellite>
Why do satellites with greater altitude travel at a slower speed?
Leo Codron
# Why do satellites with greater altitude travel at a slower speed?

(Asked by: username_0 on 2017-04-05)

For example, a geostationary satellite travels slower than the ISS. What is going on?

**Tags:** satellite

> **Comments:**
> **username_1**: Have you considered looking at the force balance that keeps satellites in orbit?  
> **username_2**: @username_1 That is a bit misleading, there is no force balance as the only force acting on the orbiting object is gravity.  
> **username_1**: @Hohmannfan: That doesn't matter. It doesn't seem to be a university-level question, and the most intuitive 'layman' way of understanding Keplerian orbits is through inertial forces.  

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## 1 answers

### Answer by username_3 on 2017-04-05. Score: 3 | Accepted answer

[Kepler's third law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law) in this situation tells us that
$$ \frac{R^3}{P^2} = \frac{GM}{4\pi^2}, $$
where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period $P \propto R^{3/2}$.

To calculate the orbital speed, we divide the orbital circumference by the orbital period
$$ v = \frac{2\pi R}{P}. $$

Then substituting for $P$ from Kepler's third law, we get
$$ v = \sqrt{\frac{GM}{R}}$$

Thus, the larger the orbital radius, the slower the orbital speed.

Or you can just use Newton's second law and equate the centripetal acceleration to the gravitational force:
$$ m \frac{v^2}{R} = G\frac{Mm}{R^2},$$
where $m$ is the mass of the satellite, which leads directly to the same result.


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