3.7.1.2–3FieldsStretch

Gravitational Potential and Orbits

A satellite doesn’t need an engine to stay in orbit — gravity alone provides exactly the centripetal force needed to keep it circling. From that single idea comes a precise relationship between a satellite’s orbital radius and how long it takes to complete one orbit, one that applies just as well to the Moon around the Earth as to a geostationary TV satellite.

45 min Video by Understanding A Level Physics by Dave 3.7.1 Gravitational fields

AQA A Level Physics: Orbits of Planets and SatellitesWatch the full walkthrough before the notes below.

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What you'll be able to do

Define gravitational potential and use $V = - \frac{GM}{r}$
Calculate the work done moving a mass between two points in a gravitational field
Derive and use the relationship between orbital speed, period and radius for a satellite
Use Kepler’s third law, $T^{2} \propto r^{3}$
Explain what makes an orbit geostationary

Gravitational potential

$Gravitational potential$ , $V$ , at a point is the work done per unit mass to bring a small test mass from infinity (where potential is defined as zero) to that point. Because gravity always does positive work pulling a mass inward, gravitational potential is always $negative$ — it becomes less negative (rises toward zero) as you move further from the source mass, and would only reach zero infinitely far away.

The $potential energy$ of a mass $m$ at a point is simply $m$ multiplied by the potential there. Moving a mass between two points in a gravitational field requires (or releases) energy equal to the mass times the potential difference between those points.

V = - \frac{GM}{r} E_{p} = mV = - \frac{GM m}{r}

Gravitational potential and potential energy (both negative, taking zero at infinity).

1Use

V = - \frac{GM}{r}

V = - \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}}{6.37 \times 1 0 ^{6}}

Answer

V \approx - 6.25 \times 1 0^{7}

J kg⁻¹

Tip — A negative potential doesn’t mean "less than nothing is possible" — it’s just a consequence of choosing zero at infinity. What matters physically is always the DIFFERENCE in potential between two points.

Orbital speed and period

For a satellite in a stable circular orbit, gravity provides exactly the centripetal force needed to keep it moving in a circle — there is no other force involved. Setting the gravitational force equal to the required centripetal force lets you find the orbital speed for any given radius, and combining that with the basic relationship between speed, circumference and period gives the orbital period.

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

Orbital speed and period, derived by equating gravitational and centripetal force.

1Use

v = \frac{GM}{r}

v = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}}{7.0 \times 1 0 ^{6}}

v = 5.69 \times 1 0^{7}

Answer

v \approx 7.54 \times 1 0^{3}

m s⁻¹ (about 7.5 km/s)

Tip — Notice orbital speed decreases as radius increases (v ∝ 1/√r) — a satellite in a higher orbit moves more slowly, not faster, which is why the Moon (very high orbit) creeps around the Earth over weeks while low-orbit satellites circle in under two hours.

Kepler’s third law and geostationary orbits

Squaring the period equation gives $Kepler’s third law$ : for any set of satellites orbiting the same central body, the square of the orbital period is proportional to the cube of the orbital radius, with the constant of proportionality depending only on the mass being orbited.

A $geostationary orbit$ is one where a satellite’s period exactly matches the Earth’s rotation period (24 hours), and the orbit lies directly above the equator — so, from the ground, the satellite appears to stay fixed in the sky at all times. This makes geostationary orbits ideal for communications and weather satellites, which need to always point at the same ground location.

T^{2} = \frac{4 π ^{2}}{GM} r^{3}

Kepler’s third law: T² ∝ r³ for a given central mass M.

1Rearrange

T^{2} = \frac{4 π ^{2}}{GM} r^{3}

r^{3} = \frac{GM T ^{2}}{4 π ^{2}}

r^{3} = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24} \times ( 8.64 \times 1 0 ^{4} ) ^{2}}{4 π ^{2}}

3Numerator:

6.67 \times 1 0^{- 11} \times 5.97 \times 1 0^{24} = 3.98 \times 1 0^{14}

;

\times (8.64 \times 1 0^{4})^{2} = 3.98 \times 1 0^{14} \times 7.47 \times 1 0^{9} = 2.97 \times 1 0^{24}

r^{3} = \frac{2.97 \times 1 0 ^{24}}{39.48} = 7.52 \times 1 0^{22}

, so

r = 3 7.52 \times 1 0^{22}

Answer

r \approx 4.23 \times 1 0^{7}

m (about 42,300 km from the Earth’s centre, or roughly 35,900 km above the surface).

Tip — A geostationary orbit has one specific radius (and must be directly above the equator) — you cannot choose a different altitude and still have the satellite appear stationary in the sky.

Equation recap

V = - \frac{GM}{r}

Gravitational potential.

E_{p} = mV = - \frac{GM m}{r}

Gravitational potential energy.

v = \frac{GM}{r}

Orbital speed.

T^{2} = \frac{4 π ^{2}}{GM} r^{3}

Kepler’s third law.

Common mistakes to avoid

Thinking a higher orbit means a faster orbital speed.

Orbital speed DECREASES with radius (v ∝ 1/√r) — higher orbits are slower, not faster.

Assuming any orbital period can be made geostationary by choosing the right radius.

Geostationary orbit requires the period to exactly match Earth’s rotation (24 h) AND lie in the equatorial plane — this fixes one single, specific radius.

Forgetting the negative sign when calculating gravitational potential or potential energy.

V and Ep are always negative (with the zero reference at infinity) — a positive answer for either usually signals a sign error.

Key takeaways

Gravitational potential V = −GM/r (negative, zero at infinity); potential energy Ep = mV.
Work done moving a mass between two points = m × (change in potential).
Orbital speed v = √(GM/r), decreasing with orbital radius.
Kepler’s third law: T² = (4π²/GM)r³ — same constant of proportionality for all satellites around a given body.
A geostationary orbit has T = 24 h, lies above the equator, and has one fixed orbital radius.

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Gravitational Potential and Orbits

45 min Video by Understanding A Level Physics by Dave 3.7.1 Gravitational fields