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

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

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.

• sorry, what is the value of r here? Apr 2, 2015 at 0:51
• you have to pick either t or r and calculate the other. Because orbital period depends on the radius. Apr 2, 2015 at 1:00
• Thanks for the cool edit @HDE 226868! I'll have to look at what you did! Apr 2, 2015 at 1:10
• Glad to help! Here's the notation used: math.stackexchange.com/help/notation This (en.wikibooks.org/wiki/LaTeX/…) and this (en.wikibooks.org/wiki/LaTeX/Advanced_Mathematics) are also helpful. Apr 2, 2015 at 1:15
• The phrase "sitting just outside the body's atmosphere" has no meaning on Earth as the atmosphere doesn't have a hard boundary. Drag is a major consideration for satellites even as high as the International Space Station, at over 400 km of altitude. But yes, you could theoretically orbit a body with no atmosphere just above the surface. Note that close to the ground gravity can no longer be modeled as simply as features like mountains and valleys and density gradients invalidate the assumptions of the simple equations above. Apr 3, 2015 at 3:37

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.

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

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.