# How do you calculate the altitude and velocity of a satellite in a geosynchronous orbit of Mars?

Apr 21, 2018

The satellite in Mars geostationary orbit must be $17005 \text{ Kilometers}$ above the surface of the planet and it must be travelling at a speed of $1446 \text{ m/s}$.

#### Explanation:

To calculate the necessary altitude and velocity needed for a geosynchronous orbit of any planet, you must use a few relationships.

You need to know that the centripetal force exerted on an object in circular motion is

${F}_{c} = \frac{{m}_{2} {v}^{2}}{r}$

Additionally, you must know that the force of gravitational attraction between any two objects is

${F}_{g} = \frac{G {M}_{1} {m}_{2}}{r} ^ 2$

Note: ${M}_{1}$ is the mass of the planet while ${m}_{2}$ is the mass of the satellite.

For an object to be in a stable orbit at any altitude, the centripetal force and the gravitational force must be equal. So the first step in solving this problem is to set these expressions equal to each other.

$\frac{{m}_{2} {v}^{2}}{r} = \frac{G {M}_{1} {m}_{2}}{r} ^ 2$

Some simplifying yields

$v = \sqrt{\frac{G {M}_{1}}{r}}$

This is the standard form of the equation. If you looked up orbital velocity equation on google, this is what you'd get.

So this equation is fine for a general orbit, but we are trying to find requirements for a geosynchronous orbit. Another relationship is needed.

Luckily, in this type of orbit, since you stay above the same spot on the ground the entire time, you could relate the angular velocities of the planet and the satellite somehow. What people realized is that in a geosynchronous orbit, the period (seconds/cycle) of rotation for the satellite is the same as the planet.

To relate period, $T$, to velocity, you must remember how far the satellite travels in a single orbit with respect to the distance, $r$, from the planet's center.

$\text{Orbit Length} = 2 \pi r$

Since velocity, $v$ is defined as distance/time, you can now set up your equation with the new variables.

$\frac{2 \pi r}{T} = \sqrt{\frac{G {M}_{1}}{r}}$

Simplifying

$\frac{4 {\pi}^{2} {r}^{2}}{T} ^ 2 = \frac{G {M}_{1}}{r}$

${r}^{3} = \frac{G {M}_{1} {T}^{2}}{4 {\pi}^{2}}$

$r = \sqrt[3]{\frac{G {M}_{1} {T}^{2}}{4 {\pi}^{2}}}$

The mass and rotational period for Mars are known, so you can just plug in the numbers to find the distance from the center of Mars at which you should orbit to become geosynchronous.

$r \approx 20395 \text{ Kilometers}$

or

$r \approx 12673 \text{ Miles}$

NOTE: This distance is not altitude! This is the distance from the center of Mars! To find altitude, subtract the radius of Mars from this number. Remember to always use the distance from the core of the planet in these calculations!

$\text{Altitude"approx17005" Kilometers}$

or

$\text{Altitude"approx10566" Miles}$

So now you have your necessary distance. You still don't know how fast it needs to go, but this is an easy step. Just plug it back into the old velocity equation

$v = \frac{2 \pi r}{T}$

to get

$v \approx 1446 \text{ m/s}$

or

$v \approx 3235 \text{ mph}$

The satellite in Mars geostationary orbit must be stationed $17005 \text{ Kilometers}$ above the surface of the planet and it must be travelling at a speed of $1446 \text{ m/s}$.