5.4FieldsStretch

Gravitational Potential and Orbits

A satellite needs no engine to stay in orbit — gravity alone provides exactly the centripetal force required. That single fact fixes an exact relationship between a satellite’s orbital radius and how long it takes to go around, whether it’s the Moon or a communications satellite.

45 min Video by Science Ready Module 5.4 Gravitational fields

Orbital Motion & Velocity of Satellites and Planets + ExamplesWatch 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 work done moving a mass between two points in a gravitational field
Derive and use the orbital speed and period relationships
Use Kepler’s third law

Gravitational potential

Gravitational potential, $V$ , is the work done per unit mass bringing a test mass from infinity to a point, always negative (zero at infinity).

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

Gravitational potential and potential energy.

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

Orbital speed and Kepler’s third law

Equating gravitational force to the required centripetal force gives the orbital speed and period for a satellite.

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

Orbital speed, and Kepler’s third law.

1Use

v = \frac{GM}{r}

v = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}}{7.5 \times 1 0 ^{6}}

Answer

v \approx 7280

m s⁻¹

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

Equation recap

V = - \frac{GM}{r}

Gravitational potential.

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 faster orbital speed.

Orbital speed decreases with radius (v ∝ 1/√r).

Forgetting the negative sign for gravitational potential.

V is always negative (zero reference at infinity) — a positive answer signals a sign error.

Assuming any orbital period can be made geostationary at any radius.

Geostationary orbit requires exactly T=24h AND the equatorial plane — this fixes one specific radius.

Key takeaways

Gravitational potential V = −GM/r (negative, zero at infinity).
Orbital speed v = √(GM/r), decreasing with radius.
Kepler’s third law: T² = (4π²/GM)r³.
Geostationary orbit: T=24h, equatorial plane, one fixed radius.

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Gravitational Potential and Orbits

45 min Video by Science Ready Module 5.4 Gravitational fields