Gravitational potential energy and escape speed

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The gravitational potential energy **per kilogram** at a point: $V = -\dfrac{GM}{r}$. Unit: J kg⁻¹. Negative everywhere, zero at infinity.

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$E_{p} = -\dfrac{GMm}{r}$ — for a mass m at distance r from a mass M. Unit: joules (J).

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We set it to **zero at infinity**; anywhere closer in, gravity has already pulled the object 'downhill', so it has less than zero — it sits in a **well**.

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**At infinity** — infinitely far from the mass, where the field has faded to nothing.

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The minimum launch speed needed for an object to escape a planet's gravity — to reach where V = 0 (infinitely far) and just stop there.

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$v_{esc} = \sqrt{\dfrac{2GM}{r}}$ — from energy conservation. r is usually the planet's radius.

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**No** — the mass cancels in v_{esc} = √(2GM/r). It depends only on the planet's mass M and radius r.

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Supplying enough **kinetic energy** to climb out of the gravitational well to where V = 0 (infinitely far away): ½mv² = GMm/r.

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It **doubles** — v_{esc} ∝ √M, so √4 = 2.

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E_{p} becomes **less negative** (rises towards 0), because r increases in E_{p} = -GMm/r.

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V is the energy **per kilogram** (J kg⁻¹); E_{p} = mV is the energy of a specific object of mass m (J).

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**v∞ = √(v² − v_esc²)**: it spends ½mv_esc² climbing out of the well.

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**No.** At any point: v_esc = √(2GM/r) with r the distance from the centre there; it is smaller further out.

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Not even light could leave the body: it would be (close to) a **black hole**.