The big idea: A planet's gravity makes a well. The deeper into the well you are, the more energy it takes to climb out.
Gravitational potential V is the energy per kilogram at a point. Gravitational potential energy Ep is the energy of a particular object of mass m at that point.
Both are negative, and both are zero infinitely far away (where gravity 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.
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)
Worked example — escape speed from a planet
A planet has mass M = 6.0 × 10²⁴ kg and radius r = 6.4 × 10⁶ m. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². Find the escape speed from its surface.
Solution
- Start with the escape-speed formula:
- Put in the numbers (M = 6.0 × 10²⁴, r = 6.4 × 10⁶):
- Work out inside the root, then take the square root — keep the unit:
Final answer
vesc ≈ 1.1 × 10⁴ m s⁻¹ (about 11 km s⁻¹). Notice the rocket's own mass never entered the calculation — escape speed depends only on the planet.
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How this is tested: Gravitational energy and escape speed are tested on Paper 2 (structured) and as Paper 1A concept MCQs.
- Paper 1A: the sign and shape — Ep is negative and rises to 0 at infinity; what 'escape' means. - Paper 2: plug numbers into v_{esc} = √(2GM/r), or use energy conservation (Ep change vs kinetic energy) to find a launch or impact speed.
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).
IB-style question — (a) potential energy at the surface
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. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². Find the gravitational potential energy of the probe.
Solution
- Start with the potential-energy formula:
- Put in the numbers (M = 7.3 × 10²², m = 1200, r = 1.7 × 10⁶):
- Work it out — it must come out negative:
Final answer
Ep ≈ −3.4 × 10⁹ J. The minus sign is essential — the probe is sitting in the well, below the zero level at infinity.
IB-style question — (b) escape speed from that moon
Using the same moon (M = 7.3 × 10²² kg, r = 1.7 × 10⁶ m), find the escape speed from its surface.
Solution
- Start with the escape-speed formula:
- Put in the numbers (M = 7.3 × 10²², r = 1.7 × 10⁶):
- Work it out — keep the unit:
Final answer
vesc ≈ 2.4 × 10³ m s⁻¹ (about 2.4 km s⁻¹) — much less than Earth's, because the moon is far less massive.