Gravitation and satellite motion

Covers AS91524

Key ideas

Newton's law of universal gravitation — every mass attracts every other mass:

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

Each variable, with sub-bullets:
- $G$ — the gravitational constant, $6.67 \times 10^{-11}$ N m² kg⁻² (given in the exam)
- $M, m$ — the two masses (kg)
- $r$ — the distance between their centres (m), not their surfaces
The force follows an inverse-square law: triple the separation and the force drops to a ninth.

Satellites in circular orbit

For a satellite, gravity provides the centripetal force. Setting them equal:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

The satellite mass $m$ cancels, giving the orbital speed:

$v = \sqrt{\frac{GM}{r}}$

A larger orbit ( $r$ ) means a slower speed and a longer period. The period follows from $v = \dfrac{2\pi r}{T}$ .
A geostationary satellite has a period of exactly one day, so it stays above the same point on the equator.

A satellite in a circular orbit around Earth, with the gravitational force arrow pointing from the satellite toward Earth's centre, labelled as the centripetal force, and the orbital velocity arrow drawn tangent to the circle.

Worked ExampleSpeed and period of a low orbit

A satellite orbits Earth at an altitude of $400$ km. Earth's mass is $M = 5.97 \times 10^{24}$ kg and its radius is $6.37 \times 10^{6}$ m. Find the orbital speed and period.

Step 1 — Orbital radius is measured from Earth's centre

$r = 6.37\times10^{6} + 4.00\times10^{5} = 6.77 \times 10^{6} \text{ m}$

Step 2 — Orbital speed (gravity = centripetal force)

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{6.77\times10^{6}}} = 7.67 \times 10^{3} \text{ m s}^{-1}$

Step 3 — Period from the circumference

$T = \frac{2\pi r}{v} = \frac{2\pi (6.77\times10^{6})}{7.67\times10^{3}} = 5.55 \times 10^{3} \text{ s} \approx 92 \text{ min}$

Tips

Always work with the distance from the centre. Add the planet's radius to the altitude; using the altitude alone is the most common gravitation error.
Keep every value unrounded on your calculator until the final line — the inverse-square and square-root steps magnify early rounding badly.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Two $1.0 \times 10^{3}$ kg masses are $2.0$ m apart (centre to centre).

Show that the gravitational force between them is about $1.7 \times 10^{-5}$ N. ( $G = 6.67 \times 10^{-11}$ N m² kg⁻².)

Merit

A satellite orbits a planet of mass $6.4 \times 10^{23}$ kg at a radius of $3.8 \times 10^{6}$ m.

Calculate its orbital speed. ( $G = 6.67 \times 10^{-11}$ N m² kg⁻².)

Excellence

Explain why a geostationary satellite must orbit at one particular radius, and why it can only sit above the equator.

Gravitation and satellite motion

Covers AS91524

Key ideas

Newton's law of universal gravitation — every mass attracts every other mass:

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

Each variable, with sub-bullets:
- $G$ — the gravitational constant, $6.67 \times 10^{-11}$ N m² kg⁻² (given in the exam)
- $M, m$ — the two masses (kg)
- $r$ — the distance between their centres (m), not their surfaces
The force follows an inverse-square law: triple the separation and the force drops to a ninth.

Satellites in circular orbit

For a satellite, gravity provides the centripetal force. Setting them equal:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

The satellite mass $m$ cancels, giving the orbital speed:

$v = \sqrt{\frac{GM}{r}}$

A larger orbit ( $r$ ) means a slower speed and a longer period. The period follows from $v = \dfrac{2\pi r}{T}$ .
A geostationary satellite has a period of exactly one day, so it stays above the same point on the equator.

Worked ExampleSpeed and period of a low orbit

A satellite orbits Earth at an altitude of $400$ km. Earth's mass is $M = 5.97 \times 10^{24}$ kg and its radius is $6.37 \times 10^{6}$ m. Find the orbital speed and period.

Step 1 — Orbital radius is measured from Earth's centre

$r = 6.37\times10^{6} + 4.00\times10^{5} = 6.77 \times 10^{6} \text{ m}$

Step 2 — Orbital speed (gravity = centripetal force)

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{6.77\times10^{6}}} = 7.67 \times 10^{3} \text{ m s}^{-1}$

Step 3 — Period from the circumference

$T = \frac{2\pi r}{v} = \frac{2\pi (6.77\times10^{6})}{7.67\times10^{3}} = 5.55 \times 10^{3} \text{ s} \approx 92 \text{ min}$

Tips

Always work with the distance from the centre. Add the planet's radius to the altitude; using the altitude alone is the most common gravitation error.
Keep every value unrounded on your calculator until the final line — the inverse-square and square-root steps magnify early rounding badly.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Two $1.0 \times 10^{3}$ kg masses are $2.0$ m apart (centre to centre).

Show that the gravitational force between them is about $1.7 \times 10^{-5}$ N. ( $G = 6.67 \times 10^{-11}$ N m² kg⁻².)

Merit

A satellite orbits a planet of mass $6.4 \times 10^{23}$ kg at a radius of $3.8 \times 10^{6}$ m.

Calculate its orbital speed. ( $G = 6.67 \times 10^{-11}$ N m² kg⁻².)

Excellence

Explain why a geostationary satellite must orbit at one particular radius, and why it can only sit above the equator.