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Gravitational potential extends the concept of gravitational field strength by introducing an energy-based description of the field. Combined with orbital mechanics, this topic forms the heart of AQA specification 3.7.2 and is tested extensively in Paper 2.
Key Definition: The gravitational potential (V) at a point in a gravitational field is the work done per unit mass in bringing a small test mass from infinity to that point.
V = −GM/r
where V is the gravitational potential (J kg⁻¹), G = 6.67 × 10⁻¹¹ N m² kg⁻², M is the mass creating the field (kg), and r is the distance from the centre of the mass (m).
Key features:
Common Misconception: Students sometimes think that a more negative potential means "less energy." In fact, a more negative potential means that more work must be done to move the test mass to infinity. The mass is more tightly bound in the gravitational field.
Equipotential surfaces connect all points at the same gravitational potential. For a spherical mass, equipotentials are concentric spheres.
Described diagram — Equipotential surfaces around a planet: Concentric circles (representing spherical surfaces) are drawn around the planet, each labelled with a potential value that becomes less negative (closer to zero) further from the planet. The field lines are radial, pointing inward, and are everywhere perpendicular to the equipotential surfaces.
Key properties:
The gravitational potential energy (E_p) of a mass m at a point where the gravitational potential is V is:
E_p = mV = −GMm/r
This is the work done in bringing the mass m from infinity to the distance r from mass M.
The work done in moving a mass m from one point to another in a gravitational field is:
W = mΔV = m(V₂ − V₁)
If V₂ > V₁ (moving further from the mass, to a less negative potential), W is positive — you must do work against the field. If V₂ < V₁ (moving closer), W is negative — the field does work on the mass.
Question: Calculate the minimum energy required to move a 500 kg satellite from the Earth's surface to an orbit at an altitude of 600 km. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)
Solution:
At the surface: r₁ = 6.37 × 10⁶ m At the orbit: r₂ = 6.37 × 10⁶ + 600 × 10³ = 6.97 × 10⁶ m
V₁ = −GM/r₁ = −(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.37 × 10⁶) = −3.98 × 10¹⁴ / 6.37 × 10⁶ = −6.25 × 10⁷ J kg⁻¹
V₂ = −GM/r₂ = −3.98 × 10¹⁴ / 6.97 × 10⁶ = −5.71 × 10⁷ J kg⁻¹
ΔV = V₂ − V₁ = (−5.71 × 10⁷) − (−6.25 × 10⁷) = 0.54 × 10⁷ = 5.4 × 10⁶ J kg⁻¹
W = mΔV = 500 × 5.4 × 10⁶ = 2.7 × 10⁹ J = 2.7 GJ
Note: This is only the energy needed to raise the satellite against gravity. Additional energy is needed to give it orbital speed.
The gravitational field strength is the negative of the potential gradient:
g = −dV/dr
This means:
Key Definition: The escape velocity is the minimum speed an object must have at the surface of a planet to escape the gravitational field completely (i.e., to reach infinity with zero kinetic energy remaining).
To derive the escape velocity, set the total energy (kinetic + potential) equal to zero (the value at infinity):
½mv² + (−GMm/r) = 0
½mv² = GMm/r
v² = 2GM/r
v_escape = √(2GM/r)
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