Derive and apply the concept of gravitational potential energy in a radial gravitational field, U = -G M m / r, including the concept of escape velocity

Escape velocity is the minimum speed an object needs at a distance $r$ from the centre of a mass $M$ to just barely escape to infinity with zero kinetic energy remaining.

Conservation of energy from the surface to infinity (where both $U$ and $K$ are zero):

$\frac{1}{2} m v_{\text{esc}}^2 - \frac{G M m}{r} = 0$.

Solving for $v_{\text{esc}}$:

$v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}$.

Substituting:

$v_{\text{esc}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}}$
$v_{\text{esc}} = \sqrt{\frac{7.96 \times 10^{14}}{6.37 \times 10^6}}$
$v_{\text{esc}} = \sqrt{1.25 \times 10^8}$
$v_{\text{esc}} = 1.12 \times 10^4$ m/s, or about $11.2$ km/s.

Markers reward the derivation from conservation of energy, the correct identification of the boundary conditions (zero energy at infinity), and the final answer with units. Note that escape velocity does not depend on the mass of the escaping object.

Gravitational potential energy in a radial field is defined as:

$U = -\frac{G M m}{r}$.

The negative sign arises from the choice of zero potential energy at infinity (the natural reference point for radial fields). Because gravity is always attractive, the gravitational force does positive work on a mass as it moves from infinity inward, reducing its potential energy below zero.

Physically, this means a mass that is bound to a gravitational source (such as a planet in orbit) has less energy than a free mass at infinity. To escape the field, work must be done against gravity equal to $\frac{G M m}{r}$, which raises the potential energy to zero at infinity.

The familiar formula $U = m g h$ is the linear approximation valid near a planet's surface where $g$ is approximately constant. At astronomical scales, $g$ varies with $r$ and the full radial formula must be used.

Markers reward the explicit reference point at infinity, the link between attractive force and negative potential energy, and the comparison with $mgh$.

Use $U = -\dfrac{GMm}{r}$ directly, keeping the negative sign.

$U = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(800)}{8.0 \times 10^6} = -3.98 \times 10^{10}\ \text{J}$.

Marks: one for the correct formula with values substituted, one for the answer stated to three significant figures with the correct sign and unit (a positive answer with no explanation earns no mark for sign).

From conservation of energy (total energy zero at infinity), $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$.

$v_{\text{esc}} = \sqrt{\dfrac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} = 1.12 \times 10^4\ \text{m s}^{-1}$ (about $11.2\ \text{km s}^{-1}$).

The escaping object's mass cancels out of $\tfrac{1}{2}mv_{\text{esc}}^2 = \dfrac{GMm}{r}$, so the answer is the same $1.12 \times 10^4\ \text{m s}^{-1}$ for the $2000\ \text{kg}$ probe.

Marks: one for the formula $v_{\text{esc}} = \sqrt{2GM/r}$, one for $v_{\text{esc}} = 1.12 \times 10^4\ \text{m s}^{-1}$ with the unit, one for correctly stating the mass independence.

(a) Reading the data point on the curve at $r = 2.0 \times 10^7\ \text{m}$ gives $U \approx -1.99 \times 10^{10}\ \text{J}$.

(b) $U = -GMm/r$ is negative for every finite $r$ and only approaches zero as $r \to \infty$ (the chosen reference point). The curve rises toward, but never reaches, $U = 0$, so it never crosses the axis.

Marks: one for reading $U \approx -1.99 \times 10^{10}\ \text{J}$ (accept $-1.9$ to $-2.0 \times 10^{10}\ \text{J}$) directly off the labelled point, one for stating $U < 0$ for all finite $r$, one for linking this to the zero-at-infinity reference point.

$\Delta U = U(r_2) - U(r_1) = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$.

$\dfrac{1}{r_1} - \dfrac{1}{r_2} = \dfrac{1}{7.0 \times 10^6} - \dfrac{1}{9.0 \times 10^6} = 1.429 \times 10^{-7} - 1.111 \times 10^{-7} = 3.17 \times 10^{-8}\ \text{m}^{-1}$.

$\Delta U = (6.67 \times 10^{-11})(5.97 \times 10^{24})(600)(3.17 \times 10^{-8}) = 7.58 \times 10^{9}\ \text{J}$.

Marks: one for the correct $\Delta U$ formula (from subtracting two negative terms), one for the reciprocal-bracket working, one for $\Delta U = 7.58 \times 10^9\ \text{J}$ with the correct positive sign.

$v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} = \sqrt{\dfrac{2(6.67 \times 10^{-11})(7.35 \times 10^{22})}{1.74 \times 10^6}} = \sqrt{5.63 \times 10^{6}} = 2.37 \times 10^3\ \text{m s}^{-1}$ (about $2.37\ \text{km s}^{-1}$).

The Moon's mass is much smaller than Earth's, which alone would reduce $v_{\text{esc}}$; although its radius is also smaller (which alone would increase $v_{\text{esc}}$), the mass effect dominates, so the ratio $M/r$ is far smaller for the Moon and its escape velocity is roughly one-fifth of Earth's.

Marks: one for the correct substitution, one for $v_{\text{esc}} = 2.37 \times 10^3\ \text{m s}^{-1}$ with the unit, one for explaining the size of the difference in terms of $M/r$ rather than mass alone.

(a) Total mechanical energy is kinetic plus gravitational potential: $E = \tfrac{1}{2}mv^2 - \dfrac{GMm}{r}$.

$\tfrac{1}{2}mv^2 = \tfrac{1}{2}(2000)(1.0 \times 10^4)^2 = 1.00 \times 10^{11}\ \text{J}$.

$\dfrac{GMm}{r} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(2000)}{7.5 \times 10^6} = 1.06 \times 10^{11}\ \text{J}$.

$E = 1.00 \times 10^{11} - 1.06 \times 10^{11} = -6.2 \times 10^{9}\ \text{J}$.

(b) Since $E < 0$, the spacecraft does not have enough kinetic energy to reach infinity with $v \geq 0$: it is gravitationally bound and will remain in orbit (an ellipse) rather than escaping.

Marks: one for the kinetic energy term, one for the potential energy term with correct sign, one for $E = -6.2 \times 10^{9}\ \text{J}$, one for correctly linking $E < 0$ to being bound.

Band-6 plan. (1) State $U = -GMm/r$ and total energy $E = K + U$. (2) Explain the three regimes by the sign of $E$: $E < 0$ bound, $E = 0$ marginal (just escapes), $E > 0$ escapes with speed to spare. (3) Connect each regime to a physical outcome (falls back / orbits / escapes). (4) Use escape velocity as the boundary case and justify why it depends only on $M$ and $r$, not the object's own mass.

Model answer. A spacecraft's total mechanical energy is $E = \tfrac{1}{2}mv^2 - \dfrac{GMm}{r}$, the sum of its kinetic energy and its gravitational potential energy in the radial field. Because $U$ is negative and $K$ is always positive, the sign of $E$ determines the spacecraft's fate as it moves outward, since $E$ is conserved along its trajectory.

If $E < 0$, the spacecraft can never reach $r \to \infty$, because $K = E - U$ would have to become negative once $U$ (which rises toward zero) exceeds $E$ in magnitude - impossible for a real kinetic energy. The spacecraft is gravitationally bound: at low launch speed it falls back to Earth, and at a higher but still sub-escape speed with the right direction it settles into a closed elliptical or circular orbit, oscillating between a minimum and maximum radius (or a fixed radius for a circular orbit) while $E$ stays constant and negative.

If $E = 0$ exactly, the spacecraft has just enough energy to reach $r \to \infty$ with $v \to 0$: this is the escape velocity condition, $\tfrac{1}{2}mv_{\text{esc}}^2 = \dfrac{GMm}{r}$, giving $v_{\text{esc}} = \sqrt{2GM/r}$. The spacecraft's own mass $m$ cancels from this condition, so escape velocity depends only on the source mass $M$ and the launch radius $r$, not on what is being launched.

If $E > 0$, the spacecraft still has positive kinetic energy left over even as $r \to \infty$ ($U \to 0$), so it escapes Earth's gravity permanently and continues to recede, following a hyperbolic trajectory relative to Earth rather than a closed orbit.

Marker's note: the top band explicitly ties each of the three outcomes (falls back, orbits, escapes) to the sign of total energy, derives the escape-velocity boundary case with the mass cancelling, and treats $U$ as always negative throughout rather than switching to $+GMm/r$. A response that lists the outcomes without the energy-sign reasoning caps in the middle band.

Band-6 plan. (1) Show the two formulas are not directly comparable (one is absolute, one is a change, with different zero references). (2) Derive $mgh$ as the small-$h$ limit of $\Delta U$ from the radial formula. (3) State the condition under which the approximation is valid (small $h$ compared with the planet's radius) and give a rough error estimate. (4) Reach a judgement: true but conditionally, not universally.

Model answer. The claim is only partly accurate. $U = -GMm/r$ gives the absolute potential energy relative to a zero at infinity, while $U = mgh$ gives a change in potential energy relative to an arbitrary local zero (usually the ground) - they are not measuring the same quantity, so $mgh$ cannot simply replace $-GMm/r$ in every calculation.

However, $mgh$ can be derived as a limiting case of the change $\Delta U = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ for a small rise $h = r_2 - r_1$ near a planet's surface ($r_1 \approx r_2 \approx R$). Combining the fractions gives $\Delta U = GMm \dfrac{h}{r_1 r_2} \approx \dfrac{GMm}{R^2}h$, and since $g = GM/R^2$ at the surface, this is $\Delta U \approx mgh$. So $mgh$ is a good approximation precisely because $g$ barely changes over the small height $h$.

The approximation breaks down once $h$ is a significant fraction of the planet's radius, because $g$ falls off as $1/r^2$ and is no longer constant: using $mgh$ for a satellite raised from the surface to an altitude of, say, $1000\ \text{km}$ (about 16% of Earth's radius) already introduces a few per cent error, and the error grows without bound as $r \to \infty$, where $mgh$ predicts unbounded energy while the true $\Delta U$ approaches a finite value $GMm/R$. For orbital and escape-velocity problems the radial formula must be used.

Marker's note: the top band recognises the reference-point mismatch, performs the small-$h$ derivation linking $g = GM/R^2$ to the local field, and gives a concrete condition (or a quantified case) for when the approximation fails rather than a vague "it's less accurate far away." A response that only asserts "use $mgh$ near the surface, use $-GMm/r$ far away" without the derivation or a failure condition caps in the middle band.