3.3 - Man-made Satellites

Satellite Orbits and Linear Speed

Satellites need a suitable linear speed for their orbital height. Examples are the International Space Station (ISS) at 408 km and MEASAT at 35 786 km.
For a satellite of mass $m$ in a circular orbit of radius $r$ around Earth of mass $M$ , gravitational force provides centripetal force:

\frac{m v ^{2}}{r} = \frac{GM m}{r ^{2}} \Rightarrow v^{2} = \frac{GM}{r} \Rightarrow v = \frac{GM}{r}

r = R + h, v = \frac{GM}{R + h}

For Earth, $GM$ is constant: orbital speed depends on orbital radius, and the satellite’s mass cancels. A larger circular orbit requires a lower linear speed.
In a stable circular orbit, centripetal acceleration equals gravitational acceleration.
If speed falls below the required value, the satellite drops to a lower orbit and moves towards Earth. Entering the atmosphere at high speed produces heat through air resistance and can burn the satellite.

A Global Positioning System (GPS) satellite has $h = 20200 km = 2.02 \times 1 0^{7} m$ . With $R = 6.37 \times 1 0^{6} m$ :

r = 6.37 \times 1 0^{6} + 2.02 \times 1 0^{7} = 2.657 \times 1 0^{7} m

v = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{2.657 \times 1 0 ^{7}} = 3.87 \times 1 0^{3} m s^{- 1}

Satellite type	Characteristics and uses
Geostationary	Occupies the Geostationary Earth Orbit; moves in the same direction as Earth’s rotation; period 24 hours; remains above the same geographical location. Used for communication; example: MEASAT.
Non-geostationary	Usually has an orbit lower or higher than the Geostationary Earth Orbit; period shorter or longer than 24 hours; passes above different locations. Its direction need not match Earth’s rotation. Uses include Earth imaging, GPS and weather forecasting; examples: TiungSAT, RazakSAT, Pipit and ISS.

v = \frac{GM}{r}, T = \frac{4 π ^{2} r ^{3}}{GM}

Escape velocity is the minimum velocity needed for an object on Earth’s surface to overcome gravitational attraction and escape into outer space.
At distance $r$ from Earth’s centre, its gravitational potential energy is:

U = - \frac{GM m}{r}

At the minimum escape velocity, the object’s kinetic energy balances this potential energy so it can reach an infinite distance:

\frac{1}{2} m v^{2} + (- \frac{GM m}{r}) = 0 \Rightarrow v^{2} = \frac{2 GM}{r}

v = \frac{2 GM}{r}

Escape velocity depends on the central body’s mass $M$ and the object’s distance $r$ from its centre. It does not depend on the escaping object’s mass $m$ .
At Earth’s surface, $r = R$ . Using $M = 5.97 \times 1 0^{24} kg$ and $R = 6.37 \times 1 0^{6} m$ :

v = \frac{2 ( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{6.37 \times 1 0 ^{6}} = 1.12 \times 1 0^{4} m s^{- 1} = 11.2 km s^{- 1}

Earth’s high escape velocity helps it retain its atmosphere. The average speed of atmospheric molecules, $500 m s^{- 1}$ , is much lower than $11200 m s^{- 1}$ .
Commercial aircraft at $250 m s^{- 1}$ and fighter jets reaching $2200 m s^{- 1}$ remain below Earth’s escape velocity.
Rockets need large amounts of fuel to generate the high thrust needed to achieve escape velocity and send spacecraft into space.
Mars’s low escape velocity allows gases to escape more easily; its atmosphere is 100 times less dense than Earth’s. Jupiter’s high escape velocity retains hot gases.
Knowing a body’s escape velocity helps determine how spacecraft can land and take off safely.

g = \frac{GM}{R ^{2}}, v = \frac{2 GM}{R}

Body	Mass and radius
Moon	$M = 7.35 \times 1 0^{22} kg$ ; $R = 1.74 \times 1 0^{6} m$
Sun	$M = 1.99 \times 1 0^{30} kg$ ; $R = 6.96 \times 1 0^{8} m$

Body	$g / m s^{- 2}$	Escape velocity $/ m s^{- 1}$
Moon	$1.62$	$2.37 \times 1 0^{3}$
Sun	$274.0$	$6.18 \times 1 0^{5}$

The Moon’s values are much lower than the Sun’s. The Sun, the largest body in the Solar System, has higher surface gravitational acceleration and escape velocity than the Moon and the planets.