Physics 9702/42 — May/June 2013

Background Concept

A satellite will appear stationary relative to the Earth only if its angular speed matches the Earth’s angular speed of rotation. This requires the satellite to complete one orbit in the same time that the Earth completes one rotation.

Additionally, for the satellite to stay above the same point on the surface (same latitude and longitude), the orbit must lie in the equatorial plane; otherwise the satellite would move north and south in the sky as seen from the ground.

Understanding the Question

The question asks for what “geostationary orbit” means. For full marks, you must give the key conditions that ensure an observer on Earth sees the satellite in a fixed position in the sky.

Approach

State the defining properties:

1. period equals Earth’s rotational period,
2. orbit is in the equatorial plane,
3. motion is in the same direction as Earth’s rotation,
   (and typically also “circular”). Then link these to “appears fixed above one point”.

Step-by-Step Reasoning

• If the satellite’s period is $24\ \text{h}$, its angular speed equals Earth’s rotation rate, so it returns to the same longitude each day.
• If it orbits in the equatorial plane, it stays above latitude $0^\circ$ and does not drift north/south.
• If it moves west-to-east (same sense as Earth’s rotation), it can match Earth’s rotation as seen from the ground.
• Therefore it appears stationary above a fixed point on the equator.

Key Takeaways

• “Geostationary” means same angular speed as Earth.
• To stay above the same point: equatorial plane + same direction + period 24 h (and usually circular).

Common Mistakes

• Saying only “period is 24 h” but not stating equatorial orbit.
• Omitting “same direction as Earth’s rotation”.
• Confusing “geostationary” (fixed position) with “geosynchronous” (same period but may move in the sky).

Things to Be Careful About

• The orbit radius is measured from the centre of the Earth, not from the surface.
• Many mark schemes expect at least three points: period, equatorial plane, and appears above same point (often also direction/circular).

Background Concept

For a body of mass $m$ moving in a circle of radius $R$ with speed $v$, the required centripetal acceleration is

$$a = \frac{v^2}{R} = \omega^2 R$$

so the required centripetal force is

$$F_c = ma = m\frac{v^2}{R} = m\omega^2 R.$$

For a satellite orbiting a planet, the gravitational attraction provides this centripetal force. Newton’s law of gravitation gives the gravitational force magnitude as

$$F_g = \frac{GMm}{R^2}$$

where $M$ is the planet mass and $R$ is the distance from the planet’s centre (for a spherically symmetric planet).

Also, angular speed and period are linked by

$$\omega = \frac{2\pi}{T}.$$

Understanding the Question

You are told the satellite is in a circular orbit of radius $R$ around a planet of mass $M$ (concentrated at its centre). You must show that $R$ and the orbital period $T$ are related by

$$R^3 = \frac{GMT^2}{4\pi^2}.$$

So the task is a derivation: start from known force laws and kinematics of circular motion and rearrange to reach the given expression.

Approach

1. Write gravitational force $F_g$ between planet and satellite.
2. Write centripetal force needed for circular motion (in terms of $\omega$ or $v$).
3. Set $F_g = F_c$.
4. Use $\omega = 2\pi/T$.
5. Rearrange to make $R^3$ the subject.

Step-by-Step Reasoning

Start with the two forces:

• Gravitational attraction:

$$F_g = \frac{GMm}{R^2}$$

• Centripetal force required:

$$F_c = m\omega^2 R$$

(You could also use $m v^2/R$ and then substitute $v = \omega R$; both are equivalent.)

Because gravity is the only significant force causing the circular motion,

$$\frac{GMm}{R^2} = m\omega^2 R.$$

Cancel $m$ (important: the result does not depend on satellite mass):

$$\frac{GM}{R^2} = \omega^2 R.$$

Now substitute $\omega = 2\pi/T$:

$$\frac{GM}{R^2} = \left(\frac{2\pi}{T}\right)^2 R.$$

Multiply both sides by $R^2$:

$$GM = \left(\frac{2\pi}{T}\right)^2 R^3.$$

Finally rearrange for $R^3$:

$$R^3 = GM\left(\frac{T^2}{4\pi^2}\right) = \frac{GMT^2}{4\pi^2}.$$

This is Kepler’s third law in Newtonian form for circular orbits.

Key Takeaways

• In circular orbits, gravity provides the centripetal force.
• Use $\omega = 2\pi/T$ to connect orbit size to period.
• The satellite mass cancels: $T$ depends on $R$ and the central mass $M$ only.

Common Mistakes

• Using $R$ as the height above the planet surface instead of distance from the centre.
• Forgetting to square $\omega$ or $(2\pi/T)$.
• Not cancelling $m$, leaving an incorrect dependence on satellite mass.
• Mixing $v^2/R$ and $\omega^2 R$ inconsistently.

Things to Be Careful About

• Keep $R$ consistently as the orbital radius from the planet’s centre.
• Ensure algebra gives $R^3$ (not $R^2$) on one side.
• Remember $\left(\frac{2\pi}{T}\right)^2 = \frac{4\pi^2}{T^2}$, so $T^2$ ends up in the numerator in the final expression.

Background Concept

For a circular orbit,

$$R^3 = \frac{GMT^2}{4\pi^2}$$

links the orbital radius $R$ (from the centre of the Earth) to the period $T$. For a geostationary satellite, the key feature is that its period matches the Earth’s rotation period, so we use $T \approx 24\ \text{h}$ (often taken as $86400\ \text{s}$ in exam questions).

Understanding the Question

You are given Earth’s mass $M = 6.0\times 10^{24}\ \text{kg}$ and must use the expression from (b) to calculate the radius of the geostationary orbit.

Known:

• $G = 6.67\times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$
• $M = 6.0\times 10^{24}\ \text{kg}$
• $T = 86400\ \text{s}$

Unknown:

• $R$ in metres.

Approach

1. Convert the period into seconds.
2. Substitute into $R^3 = GMT^2/(4\pi^2)$.
3. Compute $R^3$ in standard form.
4. Take the cube root to find $R$.

Step-by-Step Reasoning

Convert time:

$$T = 24\ \text{h} = 24\times 3600 = 86400\ \text{s}.$$

Substitute values:

$$R^3 = \frac{(6.67\times 10^{-11})(6.0\times 10^{24})(86400)^2}{4\pi^2}.$$

Work through the powers of ten first:

$$(6.67\times 10^{-11})(6.0\times 10^{24}) = 4.00\times 10^{14}.$$

Also,

$$T^2 = (86400)^2 \approx 7.46\times 10^{9}.$$

So the numerator is approximately

$$(4.00\times 10^{14})(7.46\times 10^{9}) \approx 2.99\times 10^{24}.$$

Divide by $4\pi^2 \approx 39.5$:

$$R^3 \approx \frac{2.99\times 10^{24}}{39.5} \approx 7.6\times 10^{22}\ \text{m}^3.$$

Now cube-root:

$$R = (7.6\times 10^{22})^{1/3} \approx 4.2\times 10^{7}\ \text{m}.$$

This is the distance from Earth’s centre. (If you wanted altitude above Earth’s surface, you would subtract Earth’s radius, but the question asks for the orbit radius.)

Key Takeaways

• Always convert $T$ into seconds before substitution.
• $R$ found from this formula is measured from the centre of the Earth.
• Cube roots of numbers in standard form can be estimated by splitting coefficient and power of ten.

Common Mistakes

• Using $T = 24$ directly (forgetting to convert hours to seconds).
• Forgetting the $4\pi^2$ in the denominator.
• Treating $R$ as height above the Earth’s surface rather than distance from the centre.
• Calculator error when taking the cube root.

Things to Be Careful About

• Use consistent significant figures: $M$ is given to 2 s.f., so $R$ should be about 2 s.f. (e.g. $4.2\times 10^{7}\ \text{m}$).
• Ensure $R^3$ has units of $\text{m}^3$ so that $R$ comes out in $\text{m}$.
• Some contexts use the sidereal day ($\approx 86164\ \text{s}$); unless stated, A Level questions typically accept $86400\ \text{s}$.