Launch to Orbit, Part 3

In Part 2, I found the velocity required for circular orbit was

$\displaystyle v = \sqrt{\frac{GM}{r}},$

where G is the gravitational constant, M is the mass of the body being orbited, and r is the distance to the center of that mass. I tested this on the ISS’s orbit and found the velocity matched quite well. Now, I’m interested in looking at orbital periods, and calculating aspects of some common orbits.

Problem: Calculate the velocity and periods for some typical circular orbits.

If we know the orbital velocity, v, we can find the orbital period, T, by using the velocity relationship v = s/t. The distance traveled is the circumference of the circle, C = 2πr.

$\displaystyle v = \frac{C}{T}$

$\displaystyle \sqrt{\frac{GM}{r}} = \frac{2\pi r}{T}$

$\displaystyle T\sqrt{\frac{GM}{r}} = 2\pi r$

$\displaystyle T = 2\pi r \sqrt{\frac{r}{GM}}$

$\displaystyle T = 2\pi \sqrt{r^2} \sqrt{\frac{r}{GM}}, \ r\geq 0$

(Of course r is not negative.)

$\displaystyle T = 2\pi \sqrt{\frac{r^2r}{GM}}$

$\displaystyle T = 2\pi \sqrt{\frac{r^3}{GM}} \ \blacksquare$

This looks correct, especially because I recall that one of Kepler’s laws is that the square of of the orbital period is proportional to the cube of the semi-major axis (for a circle, the radius).

Let’s test this on the ISS’s orbit, assuming a circular orbit of 412.5 km as last time.

$\displaystyle T = 2\pi\sqrt{\frac{(6371\times 10^3\ \mathrm{m} + 412.5 \times 10^3 \ \mathrm{m})^3}{(6.674\times 10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}})(5.972\times 10^{24}\ \mathrm{kg})}}$

$\displaystyle T = 2\pi\sqrt{\frac{(6783.5\times 10^3 \ \mathrm{m})^3}{(6.674\times 10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}})(5.972\times 10^{24}\ \mathrm{kg})}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}\cdot\frac{{(10^3)}^3}{10^{-11}\cdot 10^{24}}\cdot\mathrm{\frac{m^3}{N\cdot m^2 \cdot kg^{-2}\cdot kg}}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}\cdot\frac{10^9}{10^{-11}\cdot 10^{24}}\cdot\mathrm{\frac{m^3}{N\cdot m^2 \cdot kg^{-2}\cdot kg}}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}\cdot 10^{-4}\cdot\mathrm{\frac{m^3}{N\cdot m^2 \cdot kg^{-2}\cdot kg}}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}\cdot 10^{-4}\cdot\mathrm{\frac{m^3}{\frac{kg\cdot m}{s^2}\cdot m^2 \cdot kg^{-2}\cdot kg}}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}\cdot 10^{-4}\cdot\mathrm{\frac{m^3\cdot s^2}{kg\cdot m \cdot m^2 \cdot kg^{-2}\cdot kg}}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}\cdot 10^{-4}\cdot\mathrm{\frac{ s^2}{kg\cdot kg^{-2}\cdot kg}}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}\cdot 10^{-4}\cdot\mathrm{s^2}}$

It’s always reassuring to see the units work out.

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}}\cdot\sqrt{ 10^{-4}}\cdot\sqrt{\mathrm{s^2}}$

$\displaystyle T = 2\pi\sqrt{\frac{6783.5^3}{6.674\cdot 5.972}}\cdot 10^{-2}\ \mathrm{s}$

$\displaystyle T = 5.56 \times 10^5 \cdot 10^{-2}\ \mathrm{s}$

$\displaystyle T = 5.56 \times 10^3\ \mathrm{s} \ \blacksquare$

5560 seconds is 92.7 minutes. The Wikipedia article on the ISS lists the orbital period at 92.69 minutes, so our calculation works out.

Let’s try the reverse. What altitude would be needed for a geostationary satellite? A geostationary satellite has an orbital period of exactly one day, so that it orbits at the same rate that Earth rotates. From any point on Earth it always appears to be in the same place in the sky, and so it is especially useful for communication satellites.

First, let’s solve the orbital period equation from above for the radius, r:

$\displaystyle T = 2\pi \sqrt{\frac{r^3}{GM}}$

$\displaystyle \frac{T}{2\pi} = \sqrt{\frac{r^3}{GM}}$

I will square both sides. This will not include any extraneous solutions, since T and r are both positive (as are the constants).

$\displaystyle \frac{T^2}{{(2\pi)}^2} = \frac{r^3}{GM}$

$\displaystyle \frac{T^2}{4\pi^2} = \frac{r^3}{GM}$

$\displaystyle \frac{T^2 GM}{4\pi^2} = r^3$

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

The orbital period should be one sidereal day, or 23 hours, 56 minutes, 4 seconds (86, 164.1 s). This is slightly shorter than a standard solar day since we measure our days based on how long it takes for the Earth to rotate once with respect to the sun. Since the Earth is also moving a little along its orbit each day, after it completes a rotation with respect to the distant starts (the sidereal day), it must rotate a little “extra” to turn to show the same face to the sun.

$\displaystyle r = \sqrt[3]{\frac{{(86141.1\ \mathrm{s})}^2(6.674\times 10^{-11}\ \mathrm{N\cdot m^2 \cdot kg^{-2}})(5.972\times 10^{24}\ \mathrm{kg})}{4\pi^2}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972}{4\pi^2}\cdot 10^{-11}\cdot 10^{24}\cdot\frac{\mathrm{s^2\cdot N \cdot m^2\cdot kg}}{\mathrm{kg^2}}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972}{4\pi^2}\cdot 10^{1}\cdot 10^{12}\cdot\frac{\mathrm{s^2\cdot N \cdot m^2\cdot kg}}{\mathrm{kg^2}}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}\cdot 10^{12}\cdot\frac{\mathrm{s^2\cdot N \cdot m^2\cdot kg}}{\mathrm{kg^2}}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}\cdot 10^{12}\cdot\frac{\mathrm{s^2\cdot \frac{kg \cdot m}{s^2} \cdot m^2\cdot kg}}{\mathrm{kg^2}}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}\cdot 10^{12}\cdot\frac{\mathrm{s^2\cdot kg \cdot m \cdot m^2\cdot kg}}{\mathrm{kg^2\cdot s^2}}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}\cdot 10^{12}\cdot\mathrm{m \cdot m^2}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}\cdot 10^{12}\cdot\mathrm{m^3}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}}\cdot \sqrt[3]{10^{12}}\cdot\sqrt[3]{\mathrm{m^3}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}}\cdot 10^4 \cdot\sqrt[3]{\mathrm{m^3}}$

$\displaystyle r = \sqrt[3]{\frac{86141.1^2\cdot 6.674\cdot 5.972\cdot 10}{4\pi^2}}\cdot 10^4 \ \mathrm{m}$

$\displaystyle r = \sqrt[3]{74914708027}\cdot 10^4 \ \mathrm{m}$

$\displaystyle r = 4216 \cdot 10^4 \ \mathrm{m}$

$\displaystyle r = 42.2 \times 10^6 \ \mathrm{m}$

This is the distance from the center of the Earth. Since geostationary satellites orbit over the equator, let’s subtract the equatorial radius of Earth, 6378.1 km = 6.3781×10⁶ m.

$\displaystyle 42.2 \times 10^6 \ \mathrm{m} - 6.3781\times 10^6\ \mathrm{m}$

$\displaystyle = 35.8 \times 10^6 \ \mathrm{m} \ \blacksquare$

The altitude for geostationary orbits is 35,786 km (Wikipedia), so our answer is correct. Note that this orbit is far wider than the low earth orbit (altitude around 412 km) in which the ISS orbits.

Ancora Imparo

I am still learning

Launch to Orbit, Part 3

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