Gravitational Field and Escape Speed

01 / UNDERSTAND & APPLY

1. Field is force per mass; potential is energy per mass

Newton’s gravitational interaction between two point masses has magnitude GMm/r² and is attractive along their joining line. For a source mass M at the origin, with outward unit vector r̂, the gravitational field is g = −GM r̂/r². A small test mass m experiences force mg. Field has units N kg⁻¹, equivalent to m s⁻², and is a vector; using the scalar magnitude g does not specify its direction.

Gravitational potential Φ is potential energy per unit test mass. With zero chosen at infinity for an isolated finite source, Φ = −GM/r outside a spherical source, and U = mΦ = −GMm/r. The negative sign means the configuration has lower potential energy than the infinite-separation reference. Whether a trajectory is bound depends on total energy, including kinetic energy. It does not mean the gravitational field points in a negative direction in every coordinate system.

In one-dimensional radial notation, g_r = −dΦ/dr. Since Φ rises from a more negative value toward zero as r increases, its derivative is positive and g_r is inward. Force depends on potential slope rather than potential value. At a symmetry point the field can vanish while potential remains negative and finite.

For several sources, add fields as vectors but potentials as scalars. Between two positive masses, their fields can cancel; their potentials, with the common zero at infinity, are both negative and do not cancel. At the source point of an ideal point mass the formula diverges; a real extended source requires its actual mass distribution.

02 / UNDERSTAND & APPLY

2. Spherical symmetry determines what mass contributes

The shell theorem states that a uniform thin spherical shell produces the same external gravitational field as a point mass at its centre and zero field everywhere inside the shell. Inside, its potential is constant at −GM/R, not zero under the infinity reference. A zero field means no spatial potential gradient, not necessarily zero potential.

For a spherically symmetric extended body, shells outside radius r exert zero net field at an interior point, while the enclosed mass acts as if concentrated at the centre. Thus g_r = −GM_enclosed(r)/r². The density need not be uniform for this enclosed-mass relation, but it must be spherically symmetric. An arbitrary nonspherical body cannot be replaced by its total mass at its centre at every exterior point.

For a uniform solid sphere of radius R and total mass M, M_enclosed = M(r/R)³. Inside, g_r = −GMr/R³ and field magnitude increases linearly from zero at the centre to GM/R² at the surface. Outside it decreases as 1/r². This linear interior law is not an exact model of the layered Earth.

The corresponding uniform-sphere interior potential is Φ(r) = −GM(3R² − r²)/(2R³). It joins continuously to −GM/R at the surface and equals −3GM/(2R) at the centre. Differentiating reproduces the linear field. Both potential and radial field are continuous at the boundary for this finite volume-density model; the derivative of the field changes there.

03 / UNDERSTAND & APPLY

3. Height, depth and apparent weight use different models

At height h above a spherical nonrotating planet of radius R, r = R + h and g_h = g_surface[R/(R + h)]². For h much smaller than R, binomial expansion gives g_h ≈ g_surface(1 − 2h/R). Use the exact ratio for heights comparable with R; putting h itself into the denominator treats distance from the surface as distance from the centre.

At depth d inside a uniform-density spherical planet, r = R − d and g_d = g_surface(1 − d/R). The uniform-density assumption is essential. The real Earth’s density increases inward, so its internal gravity profile cannot be inferred from this simple linear formula alone. A question giving no uniform-density model should not silently be treated as an exact Earth calculation.

Apparent weight is the support force measured by a scale, not necessarily the gravitational force. In free fall, the object and scale accelerate together, so the reading can vanish while gravity remains substantial. Orbiting astronauts are in continuous free fall; they do not require a region where gravitational attraction is zero.

For a simplified spherical planet rotating at angular speed Ω, a stationary object at the equator moves in a circle of radius R. The inward equation mg − N = mΩ²R gives scale reading N = m(g − Ω²R), ignoring oblateness and other complications. Away from the equator, centrifugal geometry and the direction of the local effective vertical need care; the equatorial expression is not universal.

04 / UNDERSTAND & APPLY

4. Escape speed is an energy threshold, not an orbit speed

For a point source, or a launch at or outside the surface of an isolated spherical body, the minimum escape speed from radius r follows by setting the final speed at infinity to zero. Conservation gives mv_escape²/2 − GMm/r = 0, so v_escape = √(2GM/r). The test mass cancels. Inside an extended body, use its actual interior potential in v_escape² = −2Φ(r) with zero potential at infinity rather than substituting total mass into the exterior formula. The result assumes no further propulsion, no air resistance, no other attracting bodies and a viable outgoing path that does not intersect the planet.

Escape is about reaching unbounded distance with nonnegative total orbital energy. It does not require the object to be launched continuously at constant escape speed: it slows as it climbs the gravitational well. A speed below this threshold can still reach a substantial finite height. A rocket can escape with a smaller initial speed if it supplies additional energy later, so the formula is not a general minimum speed for every propulsion history.

For radial launch from a surface of radius R with initial speed u less than escape speed, the turning radius satisfies u²/2 − GM/R = −GM/r_max. Thus 1/r_max = 1/R − u²/(2GM). The maximum height is r_max − R. Using constant-g height u²/(2g) is accurate only when the height is small compared with R.

If the launch speed exceeds escape speed and the other assumptions hold, asymptotic speed obeys v_infinity² = u² − 2GM/r_launch. The kinetic energy remaining at infinity is not all the launch kinetic energy, because part of the energy has raised gravitational potential from negative to zero. An inward-directed high-speed object may hit the source before any escape trajectory can be realised.

05 / UNDERSTAND & APPLY

5. Circular orbit speed comes from radial force balance

For a small satellite in a circular orbit of radius r around a much more massive spherical body, gravity supplies centripetal acceleration: GMm/r² = mv²/r. Therefore v_circular = √(GM/r), angular speed is √(GM/r³), and orbital period T = 2π√(r³/GM). Radius is measured from the central body’s centre, so an altitude h means r = R + h.

A higher circular orbit has lower orbital speed but a longer path and substantially longer period. Increasing radius by a factor k decreases speed by √k and increases period by k^(3/2). The orbital period is independent of the satellite mass in the test-mass approximation. For two comparable bodies, use the total mass in the relative two-body orbit and distinguish each body’s path about their shared centre of mass.

In a circular orbit, kinetic energy is GMm/(2r), potential energy is −GMm/r and total energy is −GMm/(2r). Thus U = −2K and E = −K. A more distant circular orbit has less kinetic energy but greater total energy, because its potential becomes less negative by more than kinetic energy decreases. Raising an orbit therefore needs added total energy despite the lower final circular speed.

At the same radius, escape speed is √2 times circular speed. If a satellite already moves in a circular orbit, an ideal instantaneous tangential speed increase in the direction of motion of (√2 − 1)v_circular brings it to the escape-energy threshold. This is a particular impulsive maneuver under an ideal isolated model; it does not describe every transfer or propulsion plan.

06 / UNDERSTAND & APPLY

6. Kepler’s laws connect energy, area and the changing speed

Bound two-body inverse-square orbits are ellipses with the attracting centre at one focus in the fixed-central-body approximation. A circle is the special case of zero eccentricity. The semimajor axis a determines total orbital energy E = −GMm/(2a) and period T² = 4π²a³/(GM). Using the instantaneous radius in the circular energy formula for an elliptical orbit is wrong.

The gravitational force is central, so torque about the source vanishes and angular momentum is conserved. Areal velocity is constant: dA/dt = L/(2m). Equal time intervals sweep equal areas, which makes the satellite move faster near closest approach and more slowly near farthest approach. Equal areas do not mean equal arc lengths.

At periapsis and apoapsis, velocity is perpendicular to the radius. Hence r_pv_p = r_av_a. Elsewhere, angular momentum is mr v_tangential, not generally mrv using the total speed, because radial motion can be present. The turning radii satisfy r_p = a(1 − e) and r_a = a(1 + e), so a = (r_p + r_a)/2.

Combining energy with the kinetic-energy expression gives the vis-viva relation v² = GM(2/r − 1/a) for an ellipse. It reduces to GM/r for a circle when a = r. In the parabolic escape limit, 1/a tends to zero and it gives 2GM/r. These checks connect circular motion and escape as different energy cases of the same inverse-square model.

07 / UNDERSTAND & APPLY

7. Matching a planet’s rotation requires more than a period

A synchronous satellite has orbital period equal to the planet’s rotation period. To remain fixed above one surface longitude in the ideal model, its orbit must also be circular, equatorial and prograde. A synchronous inclined or eccentric orbit does not remain at a fixed point in the sky. These extra geometry conditions distinguish a stationary appearance from merely a matching period.

For a circular synchronous orbit, r_sync = [GMT_rotation²/(4π²)]^(1/3). The altitude is r_sync − R. The relevant rotation period is the inertial or sidereal rotation period, rather than automatically the solar day. In numerical exercises use the supplied period and state that simplification; do not silently replace one definition with another.

A polar orbit passes over or near the poles and is useful for viewing changing strips of a rotating planet, but it cannot be geostationary. Orbital inclination changes viewing geometry, not just altitude. An energy-only calculation cannot determine whether an orbit remains above a particular latitude or longitude.

In a real environment, drag, nonspherical gravity, third-body perturbations and propulsion affect long-term orbital motion. The elementary formulas deliberately omit them. They are excellent for understanding ideal ratios and first estimates, but should not be mistaken for a complete operational spacecraft-navigation model. Apparent weightlessness remains a free-fall property even in an orbit that is only approximately circular.

08 / UNDERSTAND & APPLY

8. Original practice with centre-distance and sign checks

Use an isolated spherical source and a negligible-mass test body unless the question says otherwise. Depth formulas additionally require uniform density. Distinguish scalar potential from vector field at every symmetry point.

09 / UNDERSTAND & APPLY

9. Separate geometry, energy and the meaning of weight

Begin with the correct radius from the source centre and identify whether the point is outside a sphere, inside a uniform sphere or inside a shell. Use vector addition for field and scalar addition for potential. Then decide whether the motion is circular, elliptical, radial or at the escape threshold; each case has different valid shortcuts.

Finish by testing the limiting behavior. External field must weaken as 1/r², potential should approach zero from below at infinity, circular total energy must be negative, and a claimed geostationary orbit must meet the geometry as well as the period condition. Weightlessness describes support-force absence, not a failure of gravity.

• Field is a vector; potential is a scalar with an explicit zero reference.
• Surface altitude and centre distance are not interchangeable.
• Linear depth dependence requires uniform density.
• Circular speed and escape speed differ by √2 at the same radius.
• Elliptical energy depends on semimajor axis, not instantaneous radius.
• A synchronous period alone does not make an orbit geostationary.