The big idea: Throw a ball straight up and it slows, stops, and drops back — near a planet you sit in a gravitational well, and climbing out costs energy.
Gravitational potential V is that energy per kilogram at a point; gravitational potential energy Ep is the energy of a particular object of mass m there.
Both are negative, and both are zero infinitely far away (where gravity has faded to nothing).
A planet of mass M. Its gravitational field points inwards (gravity always attracts), so an object near it sits in a 'potential well'. To escape, it must gain enough energy to climb all the way out to where the field has faded to nothing.
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Why negative?: We set the energy to zero at infinity (right out of the well).
Anywhere closer in, gravity has already pulled the object 'downhill', so it has less than zero energy — hence the minus sign. Closer in = more negative = deeper in the well.
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Gravitational potential is the gravitational potential energy per kilogram at a distance r from a mass M:
- gravitational potential (J kg⁻¹)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻² (given)
- mass of the planet or star (kg)
- distance from the centre of that mass (m)
Multiply by the object's mass m to get the gravitational potential energy of that object:
- gravitational potential energy (J)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻² (given)
- mass of the planet or star (kg)
- mass of the small object placed in the field (kg)
- distance from the centre of M (m)
What 'escape' means: To escape a planet, an object must be launched fast enough to climb out of the well — to reach the place where V = 0 (infinitely far away) and still be moving (or only just stop there).
Using energy conservation, the kinetic energy at launch must equal the depth of the well. This gives the escape speed, and notice the object's own mass cancels out:
- escape speed — the launch speed needed to escape (m s⁻¹)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻² (given)
- mass of the planet or star (kg)
- distance from the centre of M at launch — usually its radius (m)
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A planet has mass M = 6.0 × 10²⁴ kg and radius r = 6.4 × 10⁶ m. Find the escape speed from its surface. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
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How this is tested — gravitational energy and escape speed span both papers:
Paper 1A
- The sign and shape — Ep is negative and rises to 0 at infinity.
- What 'escape' means in energy terms.
Paper 2
- Plug numbers into v = √(2GM/r).
- Or use energy conservation (potential-energy change vs kinetic energy) to find a launch or impact speed.
The classic trap: Thinking escape speed depends on the launched object's mass. It does not — the mass cancels.
The energy method: Many questions are really energy conservation:
kinetic energy gained = depth of the well climbed.
To just escape from the surface: ½mv² = GMm/r. The m cancels, leaving vesc = √(2GM/r).
A 1200 kg probe rests on the surface of a moon of mass M = 7.3 × 10²² kg and radius r = 1.7 × 10⁶ m. (a) Find the gravitational potential energy of the probe. (b) Find the escape speed from the moon's surface. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
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