understand that a satellite in a geostationary orbit remains at the same point above the Earth’s surface, with an orbital period of 24 hours, orbiting from west to east, directly above the Equator

1. Gravitational Force Between Point Masses

The universal law of gravitation states that any two point masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

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F = G\frac{m1 m2}{r^{2}}

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• \$F\$ – magnitude of the gravitational force (N)
• \$G\$ – universal gravitational constant, \$6.674\times10^{-11}\ \text{N m}^{2}\text{kg}^{-2}\$
• \$m1\$, \$m2\$ – masses of the two bodies (kg)
• \$r\$ – centre‑to‑centre distance between the masses (m)

1.1 Key Features

1. Force acts along the line joining the two masses.
2. It is always attractive.
3. The magnitude depends only on \$m1\$, \$m2\$, \$r\$, and \$G\$ – no other properties of the bodies matter.

2. Satellite Motion Under Gravity

A satellite of mass \$m\$ moving in a circular orbit of radius \$r\$ about the Earth experiences a centripetal force provided entirely by gravity:

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\frac{m v^{2}}{r}=G\frac{M_{\oplus} m}{r^{2}}

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where \$M_{\oplus}\$ is the mass of the Earth and \$v\$ is the orbital speed.

2.1 Orbital Speed and Period

Solving for \$v\$ gives the orbital speed:

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v = \sqrt{\frac{G M_{\oplus}}{r}}

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The orbital period \$T\$ is the time to complete one revolution:

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T = \frac{2\pi r}{v}=2\pi\sqrt{\frac{r^{3}}{G M_{\oplus}}}

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3. Geostationary Orbit – Objective

A geostationary satellite remains fixed above a single point on the Earth’s equator. To achieve this, three conditions must be satisfied:

• Orbital period \$T = 24\ \text{h} = 86\,400\ \text{s}\$.
• Orbit lies in the equatorial plane (inclination \$i = 0^{\circ}\$).
• Satellite moves from west to east, in the same direction as Earth’s rotation.

3.1 Determining the Geostationary Radius

Set \$T = 86\,400\ \text{s}\$ in the period formula and solve for \$r\$:

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r = \left(\frac{G M_{\oplus} T^{2}}{4\pi^{2}}\right)^{\!1/3}

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Using \$M_{\oplus}=5.972\times10^{24}\ \text{kg}\$ and \$G=6.674\times10^{-11}\ \text{N m}^{2}\text{kg}^{-2}\$, the calculation yields:

The altitude \$h{\text{geo}}\$ is obtained by subtracting Earth’s mean radius \$R{\oplus}=6.371\times10^{6}\ \text{m}\$ from \$r_{\text{geo}}\$.

3.2 Direction of Motion

Because Earth rotates eastward, a satellite must also travel eastward to remain above the same longitude. The required orbital angular velocity \$\omega\$ equals Earth’s rotational angular velocity:

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\omega = \frac{2\pi}{T}=7.2921\times10^{-5}\ \text{rad s}^{-1}

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4. Summary of Key Points

• The gravitational force between two point masses follows \$F = G m1 m2 / r^{2}\$.
• For a circular orbit, gravity provides the necessary centripetal force.
• A geostationary orbit requires a period of exactly 24 h, an equatorial plane, and eastward motion.
• The resulting orbital radius is about \$42\,164\ \text{km}\$ from Earth’s centre, giving an altitude of \$35\,786\ \text{km}\$ above sea level.