Derivation

Gravitational Field, Potential, Circular Orbits and Escape Velocity

D-423 Home PU-101 Threads force · energy · fields Depends on Conservative Forces and Potential Energy, Conservation of Mechanical Energy

Statement

For a spherically symmetric body of mass \(M\) and radius \(R\), the gravitational field at a distance \(r \ge R\) from the centre is \(\vec{g} = -\frac{GM}{r^2}\,\hat{\vec{r}}\), the gravitational potential is \(\phi(r) = -\frac{GM}{r}\) (with \(\phi(\infty)=0\)), the potential energy of a test mass \(m\) is \(U(r) = -\frac{GMm}{r}\), the circular orbital speed is \(v = \sqrt{GM/r}\), and the escape velocity from the surface is \(v_{\text{esc}} = \sqrt{2GM/R}\).

Why it matters

These five results form the backbone of classical orbital mechanics and celestial dynamics. The field and potential give a local and a global description of gravity; the circular orbit formula explains why satellites stay aloft; and the escape velocity sets the threshold between bound and unbound trajectories. Together they underpin satellite design, interplanetary mission planning, and the quantitative understanding of planetary surfaces.

They also illustrate a powerful pattern: a single inverse-square force law generates a potential whose gradient recovers the field, whose energy integral yields orbital speeds, and whose zero-energy boundary defines escape. Mastering this chain of derivations builds the intuition needed for more complex gravitational systems.

Assumptions

Spherical symmetry of the source mass.For a non-spherical body (e.g. an oblate spheroid like Jupiter), the field outside is not exactly \(GM/r^2\,\hat{\vec{r}}\); multipole corrections appear and the potential acquires higher-order terms. The shell theorem, which justifies treating the mass as a point, fails.

The test mass \(m\) is negligible compared to \(M\).If \(m\) is comparable to \(M\), the two-body problem must be solved with the reduced mass and both bodies orbit their common centre of mass. The formula \(v=\sqrt{GM/r}\) then applies to the relative orbit, not to either body individually.

Newtonian gravity — speeds are much less than \(c\), and \(GM/(rc^2)\) is small.When \(GM/(rc^2)\) is not negligible (near a black hole or neutron star), general relativity modifies the potential, the orbital precession becomes significant, and the escape velocity formula requires relativistic corrections.

No atmospheric drag or non-gravitational forces.In a real atmosphere, drag continuously removes mechanical energy, causing orbits to decay. The energy-conservation argument for escape velocity assumes only gravity acts; with drag, the required launch speed is higher and the trajectory is not a conic section.

The reference frame is inertial (or co-moving with the centre of mass).In a rotating frame, centrifugal and Coriolis forces modify the effective potential. The circular-orbit condition \(GMm/r^2 = mv^2/r\) assumes a non-rotating frame centred on \(M\).

Derivation

\[ \vec{F} = -\frac{GMm}{r^2}\,\hat{\vec{r}} \]

Newton's law of universal gravitation: the force on test mass \(m\) due to source mass \(M\) is attractive, directed along \(-\hat{\vec{r}}\), and inversely proportional to \(r^2\). A

\[ \vec{g} \equiv \frac{\vec{F}}{m} = -\frac{GM}{r^2}\,\hat{\vec{r}} \]

Define the gravitational field as force per unit test mass; the test mass cancels, leaving a field that depends only on the source. A

\[ U(r) = -\int_{\infty}^{r} \vec{F}\cdot d\vec{r}' = -\int_{\infty}^{r} \left(-\frac{GMm}{r'^2}\right) dr' \]

Potential energy is defined as the negative of the work done by the conservative force from the reference point (infinity) to \(r\). Along a radial path, \(\vec{F}\cdot d\vec{r}' = F\,dr'\) since force and displacement are collinear. B

\[ U(r) = -\int_{\infty}^{r} \frac{-GMm}{r'^2}\,dr' = -GMm\int_{\infty}^{r} \frac{dr'}{r'^2} = -GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = -\frac{GMm}{r} \]

Evaluate the integral: \(\int r'^{-2}\,dr' = -r'^{-1}\). The lower limit \(\infty\) contributes zero, leaving \(U(r) = -GMm/r\). The negative sign reflects the attractive nature of gravity. B

\[ \phi(r) \equiv \frac{U(r)}{m} = -\frac{GM}{r} \]

Define the gravitational potential as potential energy per unit mass; it depends only on the source and position. A

\[ \vec{g} = -\nabla\phi = -\frac{d\phi}{dr}\,\hat{\vec{r}} = -\frac{d}{dr}\left(-\frac{GM}{r}\right)\hat{\vec{r}} = -\left(\frac{GM}{r^2}\right)\hat{\vec{r}} = -\frac{GM}{r^2}\,\hat{\vec{r}} \]

In spherical symmetry, the gradient reduces to the radial derivative. Differentiating \(\phi = -GM/r\) gives \(d\phi/dr = GM/r^2\), so \(-\nabla\phi = -GM/r^2\,\hat{\vec{r}}\), recovering the field from Step 2. This confirms \(\phi\) is a valid potential for \(\vec{g}\). C

\[ g_{\text{surface}} = \frac{GM}{R^2} \]

Evaluate the field magnitude at the surface \(r=R\); this is the familiar \(g\approx 9.8\ \mathrm{m/s^2}\) on Earth. A

\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]

For a circular orbit, gravity provides the centripetal force. Equate the gravitational force to the required centripetal force \(mv^2/r\). A

\[ v = \sqrt{\frac{GM}{r}} \]

Solve for the orbital speed; it is independent of the orbiting mass \(m\). A

\[ T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}} \]

The period is the circumference divided by the orbital speed. A

\[ T^2 = \frac{4\pi^2 r^3}{GM} \]

Square both sides to obtain Kepler's third law for circular orbits: the period squared is proportional to the cube of the radius. A

\[ E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r} = \frac{1}{2}m\frac{GM}{r} - \frac{GMm}{r} = -\frac{GMm}{2r} \]

Substitute the circular-orbit speed into the kinetic energy and add the potential energy. The total energy is negative (bound orbit) and equals minus the kinetic energy: \(E = -K\). This is the virial theorem for a \(1/r\) potential. C

\[ E_{\text{esc}} = 0 = \frac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{R} \]

Escape means the total mechanical energy is exactly zero at infinity with zero residual speed. Set \(E=0\) at the surface \(r=R\). B

\[ v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = \sqrt{2}\,\sqrt{\frac{GM}{R}} = \sqrt{2}\,v_{\text{circ}} \]

Solve for the escape velocity. It exceeds the circular orbital speed by a factor of \(\sqrt{2}\), since escape requires enough kinetic energy to reach zero total energy rather than just maintaining a bound orbit. B

Result

\[ \vec{g} = -\frac{GM}{r^2}\,\hat{\vec{r}},\quad \phi = -\frac{GM}{r},\quad U = -\frac{GMm}{r},\quad v_{\text{circ}} = \sqrt{\frac{GM}{r}},\quad v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \]

Reading. The gravitational field points inward and falls as \(1/r^2\). The potential is negative and increases toward zero at infinity. A circular orbit requires speed \(\sqrt{GM/r}\), independent of the orbiting mass. Escape from the surface demands \(\sqrt{2}\) times the circular speed, reflecting the extra energy needed to reach zero total energy at infinity.

Units check. \([GM/r^2] = \mathrm{(N\,m^2/kg^2)(kg)/m^2 = N/kg = m/s^2}\) for \(\vec{g}\). \([GM/r] = \mathrm{(N\,m^2/kg^2)(kg)/m = N\,m/kg = J/kg = m^2/s^2}\) for \(\phi\). \([\sqrt{GM/r}] = \mathrm{\sqrt{m^2/s^2} = m/s}\) for speeds. All consistent.

Limiting cases

Far from the source (\(r \gg R\)): The field is indistinguishable from that of a point mass \(M\) at the centre; multipole corrections are of order \((R/r)^2\) or smaller.
At the surface (\(r = R\)): \(g = GM/R^2\), \(v_{\text{circ}} = \sqrt{GM/R}\), and \(v_{\text{esc}} = \sqrt{2GM/R} = \sqrt{2}\,v_{\text{circ}}\). For Earth, \(g\approx 9.8\ \mathrm{m/s^2}\), \(v_{\text{circ}}\approx 7.9\ \mathrm{km/s}\), \(v_{\text{esc}}\approx 11.2\ \mathrm{km/s}\).
Very large \(r\): \(v_{\text{circ}}\to 0\), \(v_{\text{esc}}\to 0\), and the orbit period \(T\to\infty\). The gravitational influence vanishes as \(1/r^2\).
Strong-field limit (\(GM/(Rc^2)\) not small): Newtonian formulas break down; general relativity predicts a different escape condition and orbital precession. For the Sun, \(GM/(Rc^2)\approx 2\times10^{-6}\), so Newtonian results are excellent.

Breaks when

Non-spherical source. An oblate planet (e.g. Saturn, with oblateness \(\approx 0.098\)) produces a field that deviates from \(GM/r^2\,\hat{\vec{r}}\) at the equator versus the poles. The potential acquires a \(J_2\) quadrupole term, and the simple circular-orbit and escape formulas are approximate.
Relativistic regimes. Near a compact object where \(GM/(rc^2)\gtrsim 0.01\), the Newtonian potential \(-GM/r\) is replaced by the Schwarzschild metric. The escape velocity formula \(v_{\text{esc}}=\sqrt{2GM/R}\) exceeds \(c\) for \(R<2GM/c^2\) (the Schwarzschild radius), signalling the breakdown of the Newtonian picture.
Atmospheric drag. In a dense atmosphere, drag force \(\propto v^2\) continuously removes mechanical energy. The escape velocity computed from energy conservation assumes no drag; in practice, a rocket must supply additional thrust to compensate for energy lost to the atmosphere.
Multiple significant masses. In a three-body or many-body system, the potential is not \(-GM/r\) from a single centre. Tidal forces, resonances, and chaotic dynamics invalidate the simple two-body formulas.

Failure modes

Forgetting the negative sign in the potential. Writing \(U = +GMm/r\) instead of \(U = -GMm/r\). The potential must be negative for an attractive force with \(\phi(\infty)=0\); a positive potential would imply a repulsive force.
Confusing escape velocity with circular orbital speed. Using \(v = \sqrt{GM/r}\) for escape. Escape requires \(\sqrt{2}\) times that speed: \(v_{\text{esc}} = \sqrt{2GM/R}\). The factor of \(\sqrt{2}\) comes from the energy condition \(E=0\), not from the force balance.
Assuming the test mass matters. Believing a heavier object needs a higher escape velocity. Both \(v_{\text{circ}}\) and \(v_{\text{esc}}\) are independent of \(m\); the mass cancels in the force and energy equations.
Applying the formulas inside the source. Using \(g = GM/r^2\) for \(r < R\) (inside a uniform sphere, \(g \propto r\)). The shell theorem says only the mass interior to radius \(r\) contributes to the field inside a spherically symmetric body.

Discussion

The derivation chain from Newton's law to escape velocity is a masterclass in how a single force law generates an entire kinematic and energetic structure. The inverse-square form is not arbitrary: it is the unique power law for which the potential is \(1/r\), the field is a gradient of that potential, and the circular-orbit speed scales as \(r^{-1/2}\). This specific scaling is what makes Kepler's third law \(T^2 \propto r^3\) hold, and it is what allows the virial theorem \(E = -K\) to take such a simple form.

The energy perspective unifies the results. A bound orbit has \(E < 0\), a parabolic escape trajectory has \(E = 0\), and a hyperbolic flyby has \(E > 0\). The circular orbit sits at a specific energy \(E = -GMm/(2r)\) where the radial kinetic energy is zero and all kinetic energy is tangential. Escape is the threshold where the kinetic energy exactly cancels the potential energy, leaving zero total energy. This energy bookkeeping is far more robust than force-balance arguments, which only apply to circular orbits.

The shell theorem, though stated without proof here, is the key justification for treating a spherical planet as a point mass. It follows from Gauss's law for gravity, \(\oint \vec{g}\cdot d\vec{A} = -4\pi G M_{\text{enc}}\), applied to a spherical Gaussian surface. Inside a uniform sphere, the enclosed mass scales as \(r^3\), so \(g \propto r\); outside, \(M_{\text{enc}} = M\) is constant, so \(g \propto 1/r^2\). This theorem is what allows the simple formulas to apply to real planets, whose interiors are complex but whose external fields are nearly point-like.

In general relativity the Newtonian potential is replaced by the Schwarzschild metric, yet one Newtonian result survives intact: the speed needed to escape from radius \(r\), as measured by an observer hovering there, is exactly \(\sqrt{2GM/r}\) in Schwarzschild spacetime too. It reaches \(c\) at \(R_s = 2GM/c^2\), the event horizon, which is why the eighteenth-century “dark star” argument of Michell and Laplace landed on the right radius, although its picture of light rising and falling back is wrong. What does not survive is everything else: orbits precess, circular orbits become unstable inside \(3R_s\), and clocks deep in the potential run slow. For \(GM/(Rc^2) \ll 1\) (about \(2\times10^{-6}\) at the Sun's surface, \(7\times10^{-10}\) at Earth's) the Newtonian formulas are excellent.

Common misconceptions. Escape velocity is not a speed that must be maintained; it is the minimum speed needed at the surface (or at any point) to reach infinity with zero residual speed, assuming no further propulsion. A projectile launched at exactly \(v_{\text{esc}}\) will asymptotically approach zero speed at infinity. Also, escape velocity is not the same as the speed needed to “beat gravity” in a local sense; it is a global energy condition. Finally, the circular orbital speed is not the “speed of gravity”; it is the speed at which the gravitational force exactly provides the centripetal acceleration for a circular path.

Worked examples

Example 1. Calculate the escape velocity from Earth's surface. Given: \(G = 6.674\times 10^{-11}\ \mathrm{N\,m^2/kg^2}\), \(M = 5.972\times 10^{24}\ \mathrm{kg}\), \(R = 6.371\times 10^6\ \mathrm{m}\).

\[ v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \]

Use the escape velocity formula derived in Step 14. A

\[ 2GM = 2(6.674\times 10^{-11})(5.972\times 10^{24}) = 7.972\times 10^{14}\ \mathrm{m^3/s^2} \]

Compute the numerator: \(2 \times 6.674 \times 5.972 = 79.72\), and \(10^{-11}\times 10^{24} = 10^{13}\), so \(79.72\times 10^{13} = 7.972\times 10^{14}\). A

\[ \frac{2GM}{R} = \frac{7.972\times 10^{14}}{6.371\times 10^6} = 1.251\times 10^8\ \mathrm{m^2/s^2} \]

Divide: \(7.972/6.371 = 1.251\), and \(10^{14}/10^6 = 10^8\). A

\[ v_{\text{esc}} = \sqrt{1.251\times 10^8} = 1.118\times 10^4\ \mathrm{m/s} = 11.2\ \mathrm{km/s} \]

Take the square root: \(\sqrt{1.251} = 1.118\), \(\sqrt{10^8} = 10^4\). Round to three significant figures. A

\[ v_{\text{esc}} = 11.2\ \mathrm{km/s} \]

Reading. An object launched from Earth's surface at 11.2 km/s (ignoring air resistance) will escape Earth's gravitational field and coast to infinity with zero residual speed.

Units check. \([\sqrt{GM/R}] = \mathrm{\sqrt{(N\,m^2/kg^2)(kg)/m} = \sqrt{N\,m/kg} = \sqrt{J/kg} = \sqrt{m^2/s^2} = m/s}\). Correct.

Example 2. A satellite orbits Earth in a circular orbit at an altitude of 35,786 km (geostationary orbit). Find its orbital speed and period. Given: \(G = 6.674\times 10^{-11}\ \mathrm{N\,m^2/kg^2}\), \(M = 5.972\times 10^{24}\ \mathrm{kg}\), \(R = 6.371\times 10^6\ \mathrm{m}\), altitude \(h = 3.5786\times 10^7\ \mathrm{m}\).

\[ r = R + h = 6.371\times 10^6 + 3.5786\times 10^7 = 4.2157\times 10^7\ \mathrm{m} \]

The orbital radius is the distance from Earth's centre, not the altitude. A

\[ v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{4.2157\times 10^7}} \]

Use the circular orbital speed formula with the full orbital radius. A

\[ GM = 3.986\times 10^{14}\ \mathrm{m^3/s^2},\quad \frac{GM}{r} = \frac{3.986\times 10^{14}}{4.2157\times 10^7} = 9.455\times 10^6\ \mathrm{m^2/s^2} \]

Compute \(GM\) and divide by \(r\): \(3.986/4.2157 = 0.9455\), \(10^{14}/10^7 = 10^7\), so \(9.455\times 10^6\). A

\[ v = \sqrt{9.455\times 10^6} = 3.075\times 10^3\ \mathrm{m/s} = 3.08\ \mathrm{km/s} \]

Square root: \(\sqrt{9.455} = 3.075\), \(\sqrt{10^6} = 10^3\). A

\[ T = \frac{2\pi r}{v} = \frac{2\pi(4.2157\times 10^7)}{3.075\times 10^3} = \frac{2.649\times 10^8}{3.075\times 10^3} = 8.615\times 10^4\ \mathrm{s} \]

Circumference divided by speed. \(2\pi \times 4.2157\times 10^7 = 2.649\times 10^8\). Divide by \(3.075\times 10^3\): \(2.649/3.075 = 0.8615\), \(10^8/10^3 = 10^5\), so \(8.615\times 10^4\ \mathrm{s}\). A

\[ T = 8.615\times 10^4\ \mathrm{s} \approx 23.9\ \mathrm{h} \approx 24\ \mathrm{h} \]

Convert to hours: \(8.615\times 10^4 / 3600 = 23.93\ \mathrm{h}\). This matches the sidereal day, confirming the geostationary orbit. A

\[ v = 3.08\ \mathrm{km/s},\qquad T \approx 24.0\ \mathrm{h} \]

Reading. A geostationary satellite orbits at 3.08 km/s with a period of one sidereal day, appearing fixed above a point on Earth's equator.

Units check. \([\sqrt{GM/r}] = \mathrm{m/s}\). \([2\pi r/v] = \mathrm{m/(m/s) = s}\). Both consistent.

Problems

Calculate the gravitational field strength at the surface of the Moon. Given: \(G = 6.674\times 10^{-11}\ \mathrm{N\,m^2/kg^2}\), \(M_{\text{Moon}} = 7.348\times 10^{22}\ \mathrm{kg}\), \(R_{\text{Moon}} = 1.737\times 10^6\ \mathrm{m}\).
Solution
\(g = GM/R^2 = (6.674\times 10^{-11})(7.348\times 10^{22})/(1.737\times 10^6)^2\). Numerator: \(6.674\times 7.348 = 49.04\), so \(4.904\times 10^{12}\). Denominator: \((1.737)^2 = 3.017\), so \(3.017\times 10^{12}\). \(g = 4.904/3.017 = 1.625\ \mathrm{m/s^2}\). Rounded to three significant figures: \(1.63\ \mathrm{m/s^2}\).
An asteroid of mass \(1.20\times 10^{16}\ \mathrm{kg}\) and radius \(4.00\times 10^3\ \mathrm{m}\) is spherical. Find the escape velocity from its surface. (\(G = 6.674\times 10^{-11}\ \mathrm{N\,m^2/kg^2}\).)
Solution
\(v_{\text{esc}} = \sqrt{2GM/R} = \sqrt{2(6.674\times 10^{-11})(1.20\times 10^{16})/(4.00\times 10^3)}\). Numerator: \(2\times 6.674\times 1.20 = 16.02\), so \(1.602\times 10^6\). Divide by \(4.00\times 10^3\): \(1.602\times 10^6 / 4.00\times 10^3 = 400.5\ \mathrm{m^2/s^2}\). \(v_{\text{esc}} = \sqrt{400.5} = 20.0\ \mathrm{m/s}\).
A satellite orbits at a radius of \(2.00\times 10^7\ \mathrm{m}\) from Earth's centre. Find its orbital speed and period. (\(G = 6.674\times 10^{-11}\ \mathrm{N\,m^2/kg^2}\), \(M = 5.972\times 10^{24}\ \mathrm{kg}\).)
Solution
\(v = \sqrt{GM/r} = \sqrt{(6.674\times 10^{-11})(5.972\times 10^{24})/(2.00\times 10^7)}\). \(GM = 3.986\times 10^{14}\). \(GM/r = 3.986\times 10^{14}/2.00\times 10^7 = 1.993\times 10^7\ \mathrm{m^2/s^2}\). \(v = \sqrt{1.993\times 10^7} = 4.464\times 10^3\ \mathrm{m/s} = 4.46\ \mathrm{km/s}\). \(T = 2\pi r/v = 2\pi(2.00\times 10^7)/(4.464\times 10^3) = 1.257\times 10^8/4.464\times 10^3 = 2.816\times 10^4\ \mathrm{s} = 7.82\ \mathrm{h}\).
At what altitude above Earth's surface is the escape velocity half the surface escape velocity? (Surface \(v_{\text{esc}} = 11.2\ \mathrm{km/s}\), \(R = 6.371\times 10^6\ \mathrm{m}\).)
Solution
\(v_{\text{esc}}(r) = \sqrt{2GM/r}\). We want \(v_{\text{esc}}(r) = v_{\text{esc}}(R)/2\). So \(\sqrt{2GM/r} = \frac{1}{2}\sqrt{2GM/R}\). Square both sides: \(2GM/r = \frac{1}{4}(2GM/R)\). Cancel \(2GM\): \(1/r = 1/(4R)\), so \(r = 4R = 4(6.371\times 10^6) = 2.548\times 10^7\ \mathrm{m}\). Altitude \(h = r - R = 2.548\times 10^7 - 6.371\times 10^6 = 1.911\times 10^7\ \mathrm{m} = 19{,}100\ \mathrm{km}\).
A planet has twice the mass and half the radius of Earth. Compare its surface gravitational field, circular orbital speed at the surface, and escape velocity to Earth's values. Express each as a multiple of the corresponding Earth value.
Solution
Let \(M' = 2M\), \(R' = R/2\). Surface field: \(g' = GM'/R'^2 = G(2M)/(R/2)^2 = 2GM/(R^2/4) = 8GM/R^2 = 8g\). So \(g'\) is 8 times Earth's. Circular orbital speed: \(v' = \sqrt{GM'/R'} = \sqrt{G(2M)/(R/2)} = \sqrt{4GM/R} = 2\sqrt{GM/R} = 2v\). So \(v'\) is 2 times Earth's. Escape velocity: \(v'_{\text{esc}} = \sqrt{2GM'/R'} = \sqrt{2G(2M)/(R/2)} = \sqrt{8GM/R} = \sqrt{8}\sqrt{GM/R} = \sqrt{8}\cdot v_{\text{circ}}\). Earth's \(v_{\text{esc}} = \sqrt{2GM/R} = \sqrt{2}\,v_{\text{circ}}\). Ratio: \(v'_{\text{esc}}/v_{\text{esc}} = \sqrt{8}/\sqrt{2} = \sqrt{4} = 2\). So \(v'_{\text{esc}}\) is 2 times Earth's. Summary: \(g' = 8g\), \(v' = 2v\), \(v'_{\text{esc}} = 2v_{\text{esc}}\).