Given the orbital radius of a satellite, how is the orbital period calculated?

Question

I know I can find the orbit radius of a satellite from the equation:

$$r=\sqrt[3]{\frac{T^2GM}{4 \pi^2}}$$

but what determines the orbit period $T$? If I assume a geosynchronous orbit, would that simply mean the orbit period is the same as how long the planet takes to turn?

What is a safe orbit radius / period of a satellite that would, for example, send a lander to the planet?

The reason I ask is that I'm looking for a in this first equation:

$$\Delta V=\sqrt{\frac{\mu_s}{r_1}\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1 \right)^2+\frac{2 \mu_1}{a_1}}-\sqrt{\frac{\mu_1}{a_1}}+\sqrt{\frac{\mu_s}{r_2}\left(\sqrt{\frac{2r_1}{r_1+r_2}}-1 \right)^2+\frac{2 \mu_2}{a_2}}-\sqrt{\frac{\mu_2}{a_2}}$$

$$\Delta v=v \ln \frac{m_0}{m_1}$$

In addition, what role does the mass of the satellite play in this?

If someone could tell me how to calculate it that would be great, thanks

Answer 1

You could just solve your first equation for $T$....

Simplistically:

For a circular orbit orbital velocity is constant at $$\sqrt{(G/r)(M+m)}$$

So for orbital period: you know $r$ (also constant), so calculate the length of the orbit (circumference, assuming it's a circle) and period is just length / velocity

Notice for a given primary the only thing that really matters is $r$.

Mass of the satellite (assuming it's artificial) is kinda negligible ($M+m$) where $M$ is the primary.

You are right about the geosynchronous orbit...but the period is determined by the radius...

props to HDE 22686 for adding the math graphics.

Answer 2

The orbital period of a satellite is solely determined by the semi-major axis of its orbit and the body it’s orbiting, specifically:

$$T = 2\pi \sqrt{a^3/\mu}$$

Where $\mu$ is the gravitational constant of the body being orbited. For Earth, $\mu$ = 5.166 $km^3/hr^2$ (we neglect the mass of the satellite because the Earth weights about 1 hellagram), and $a$ is the semi-major axis of the orbit, which is related to the radius (they are equal for circular orbits).

If you solve this equation with the orbital period $T$ equal to one sidereal day, you can calculate the altitude of a geosynchronous orbit, which is at roughly 42,000 km.

Answer 3

Adam Wuerl has already given a good answer:

$$T = 2\pi \sqrt{a^3/\mu}$$

This equation can be made easier to work with by choice of units.

For example, if we use years and astronomical units, $\mu$ becomes $4\pi^2AU^3/year^2$ which cancels the figure outside the square root sign.

Then we have

$$T = \sqrt{a^3year^2/AU^3}$$

For example, suppose radius is 9 AU. square root of 9 is 3. 3 cubed is 27.

9 AUs, 27 years.

16 AUs, 64 years.

4 AUs, 8 years.

Same trick can be used with other bodies. For example pick your unit of length as radius of a geosynchronous orbit. Use one sidereal day for the time unit. An earth orbit with 4 times geosynch radius would have a period of 8 sidereal days.