Planets are satellites orbiting the Sun

A satellite is defined as any object that orbits around a larger mass, primarily due to the gravitational pull exerted by the larger mass. For example, the Moon is a satellite of the Earth, and planets in our solar system are satellites orbiting the Sun.

Planets often have orbits that are nearly circular, and their motion is governed by principles of uniform circular motion.

Factors determining orbital speed

The gravitational pull between the two objects provides the necessary centripetal force for circular motion. By equating the gravitational force and the centripetal force, we derive the equation for orbital speed:

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

${\text{v}}^2=\frac{\text{G M}}{\text{r}}$

$\text{v = }\sqrt{\frac{\text{G M}}{\text{r}}}$

Where:

• v = orbital speed (m s^-1)
• G = Gravitational constant (6.67 x 10^-11 Nm² kg^-2)
• M = Mass of the central object (kg)
• m = Mass of the orbiting object (kg)
• r = Distance between the two objects, or orbital radius (m)

Factors determining orbital period

The orbital period (T) is the time it takes for an object to complete one full orbit. In the case of circular orbits:

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

By substituting the value of v from the orbital speed equation, we get:

$\text{ T = }2\pi \sqrt{\frac{{\text{r}}^3}{\text{G M}}}$

Where:

T = time period (s)

r = orbital radius (m)

G = gravitational constant (6.67 x 10^-11 Nm² kg^-2)

M = mass of the central object (kg)

Worked example - Calculating orbital radius of the Moon

The time period of the Moon’s orbit around the Earth is 27.3 days. The Earth has a mass of 5.97 x 10²⁴ kg.

Calculate the orbital radius of the moon.

Step 1: Convert days to seconds

1 day equals 24 hours, which is 24 × 3600 seconds.

27.3 days = 2,358,720 s

Step 2: Formula

$\text{T = }2\pi \sqrt{\frac{{\text{r}}^3}{\text{G M}}}$

Step 3: Rearrange the formula to find radius (r)

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

Step 4: Substitution and correct evaluation

${\text{r}}^3=\frac{6.67\times {10}^{-11}\times 5.97\times {10}^{24}\times {(\mathrm{2,358,720})}^2}{4{\pi }^2}=5.61\times {10}^{25}$

$\text{r }=\sqrt[3]{5.61\times {10}^{25}}=382.8\times {10}^6\text{ m}$

Application to geostationary satellites

Geostationary satellites are designed to orbit Earth directly above the equator and rotate synchronously with Earth. Their notable characteristics include:

• Completing one orbit every 24 hours.
• Appearing stationary from any point on Earth's surface.
• Being particularly useful for telecommunications, as the angle of the receiver does not need frequent adjustments.