Geostationary Orbits | CAIE 9702 Paper 4 Notes

Chapters

Paper 4

Physics9702

Geostationary Orbits

TV and phone dishes need a satellite to stay in the same direction. This type is called a .

Definition

1 mark

What is meant by geostationary satellite?

Model answer: A satellite that stays above the same point on the Earth's surface.

Top view: the satellite and Earth turn together. The satellite stays on the same radial line as one point on the Equator, so it appears fixed above that point. The orbit is drawn to scale: r ≈ 6.6R.

The three conditions

It must have a period of exactly 24 hours (one Earth day).
It must orbit above the Equator.
It must move west to east, in the same direction as the Earth's rotation.
All three are needed. An orbit that is polar, or goes east to west, is not geostationary (9702/42/M/J/23 Q1(c)–Q1(d)(ii)).

Finding the radius

Only one radius gives a 24-hour period. Change 24 h to 86 400 s before using the orbital-period equation.
$T^2 = \dfrac{4\pi^2 r^3}{GM}$

Worked example

The geostationary radius above the Earth

Find the radius of a geostationary orbit around the Earth, and the height above the surface. Earth mass 6.0 × 10²⁴ kg, radius 6.4 × 10⁶ m.

Change the period to seconds.
$T = 24 \times 3600 = 86400 \text{ s}$
Calculate the numerator.
$GMT^2=3.0\times10^{24}\ \text{m}^3$
Divide by four pi squared, then take the cube root.
$r^3=\dfrac{3.0\times10^{24}}{4\pi^2}$
$r^3=7.6\times10^{22}\ \text{m}^3$
$r=4.2\times10^7\ \text{m}$
Subtract the Earth's radius to find the height above the surface.
$h=r-R$
$h=4.2\times10^7-6.4\times10^6$
$h=3.6\times10^7\ \text{m}$
This is about 36 000 km.

Answer

r=4.2\times10^7\ \text{m}

Answer

h=3.6\times10^7\ \text{m}

Your turn9702-style

A satellite must stay above the same point on Mars. Find its orbit radius. Mars mass 6.4 × 10²³ kg, rotation period 24.6 h.

Show worked answer

Change the rotation period to seconds.
$T = 24.6 \times 3600 = 88560 \text{ s}$
Use the period to find the orbit radius.
$r^3 = \dfrac{GMT^2}{4\pi^2}$
$r = 2.0 \times 10^{7} \text{ m}$

Answer: r = 2.0 × 10⁷ m from the centre of Mars.

Common mistake

Put the period in seconds (24 h = 86 400 s). The radius r is measured from the centre. Do not use 24 without changing hours to seconds. Do not use the height above the surface as r.

Geostationary = 24 h + above the Equator + west to east. For Earth, the required radius is about 4.2 × 10⁷ m from the centre (about 36 000 km above the surface).

Geostationary Orbits

TV and phone dishes need a satellite to stay in the same direction. This type is called a .

Definition

1 mark

What is meant by geostationary satellite?

Model answer: A satellite that stays above the same point on the Earth's surface.

The three conditions

It must have a period of exactly 24 hours (one Earth day).
It must orbit above the Equator.
It must move west to east, in the same direction as the Earth's rotation.
All three are needed. An orbit that is polar, or goes east to west, is not geostationary (9702/42/M/J/23 Q1(c)–Q1(d)(ii)).

Finding the radius

Only one radius gives a 24-hour period. Change 24 h to 86 400 s before using the orbital-period equation.
$T^2 = \dfrac{4\pi^2 r^3}{GM}$

Worked example

The geostationary radius above the Earth

Find the radius of a geostationary orbit around the Earth, and the height above the surface. Earth mass 6.0 × 10²⁴ kg, radius 6.4 × 10⁶ m.

Change the period to seconds.
$T = 24 \times 3600 = 86400 \text{ s}$
Calculate the numerator.
$GMT^2=3.0\times10^{24}\ \text{m}^3$
Divide by four pi squared, then take the cube root.
$r^3=\dfrac{3.0\times10^{24}}{4\pi^2}$
$r^3=7.6\times10^{22}\ \text{m}^3$
$r=4.2\times10^7\ \text{m}$
Subtract the Earth's radius to find the height above the surface.
$h=r-R$
$h=4.2\times10^7-6.4\times10^6$
$h=3.6\times10^7\ \text{m}$
This is about 36 000 km.

Answer

r=4.2\times10^7\ \text{m}

Answer

h=3.6\times10^7\ \text{m}

Your turn9702-style

A satellite must stay above the same point on Mars. Find its orbit radius. Mars mass 6.4 × 10²³ kg, rotation period 24.6 h.

Show worked answer

Change the rotation period to seconds.
$T = 24.6 \times 3600 = 88560 \text{ s}$
Use the period to find the orbit radius.
$r^3 = \dfrac{GMT^2}{4\pi^2}$
$r = 2.0 \times 10^{7} \text{ m}$

Answer: r = 2.0 × 10⁷ m from the centre of Mars.

Common mistake

Put the period in seconds (24 h = 86 400 s). The radius r is measured from the centre. Do not use 24 without changing hours to seconds. Do not use the height above the surface as r.

Geostationary = 24 h + above the Equator + west to east. For Earth, the required radius is about 4.2 × 10⁷ m from the centre (about 36 000 km above the surface).