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13.1 Gravitational field

A gravitational field is a region in which a mass experiences gravitational force. At a point, gravitational field strength g is the gravitational force per unit mass on a small test mass placed there.

$g = F/m and F = mg; unit N kg⁻¹ (equivalent to m s⁻²)$

g is a vector directed as the force on the test mass. The field is a property of the source masses and exists without the test mass. For several sources, add their field vectors to obtain the resultant.

At g = 9.8 N kg⁻¹ downward, a 2.0 kg mass experiences F = 19.6 N downward. Between Earth and Moon, a point can have g_resultant = 0 when their field contributions have equal magnitudes in opposite directions; neither individual field is zero there.

Field strength is not the force on every object: at one point g is fixed by the sources, while F=mg changes with test mass. g is not universally 9.8 N kg⁻¹.

Feature of field-line diagram	Meaning
tangent to a line at a point	direction of gravitational force and g on a small test mass
arrowhead	points in that force direction
closer spacing	qualitatively stronger field
non-crossing lines	one unique resultant-field direction at each point

Source/region	Required pattern
isolated point mass or outside a spherical mass	radial lines directed inward toward the centre; symmetrically spaced
approximately uniform field	parallel, equally spaced lines with common arrow direction

To draw a field, identify the direction a small mass would accelerate at several points, sketch smooth lines tangent to those directions, add inward arrowheads and vary spacing only to show relative strength.

Gravitational field lines are a representation, not physical tracks or object trajectories. They refer to a test mass, not a positive test charge, and their density is not an exact numerical scale unless separately calibrated.

13.2 Gravitational force between point masses

For a point outside a uniform sphere, the gravitational force and field are exactly the same as if the sphere's total mass were concentrated at a point at its centre.

$For height h above a sphere of radius R: r = R + h$

Situation	Point-mass use	Distance r
outside uniform spherical planet/star	valid	centre to external point
two non-overlapping uniform spheres	valid for their mutual force	centre-to-centre separation
very distant compact object	approximate when size ≪ separation	centre-to-centre separation
point inside the sphere	this external result is not valid	requires interior mass analysis

A satellite 400 km above a planet of radius 6400 km has orbital radius r = 6800 km = 6.80 × 10⁶ m. Using 400 km as r would overestimate gravity severely.

The source may be represented at its centre; the external test object is not moved there. The theorem's external condition must be checked before using point-mass equations.

The gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation. It acts attractively along the line joining them.

$F = Gm₁m₂/r², G = 6.67 × 10⁻¹¹ N m² kg⁻²$

Convert all quantities to SI, use centre-to-centre separation r, square the complete separation, and report the force magnitude with its attractive direction. The two masses experience equal-magnitude opposite forces.

For m₁ = 3.0 kg, m₂ = 5.0 kg and r = 2.0 m, F = (6.67 × 10⁻¹¹)(3.0)(5.0)/(2.0)² = 2.50 × 10⁻¹⁰ N on each mass, directed toward the other.

Doubling either mass doubles F; doubling r makes F one quarter. The law is inverse-square, not inverse-distance, and Newton's third law prevents different force magnitudes on the two masses.

For a satellite of mass m in a circular orbit of centre radius r around a spherical body of mass M, gravitational attraction is the inward resultant force. Velocity is tangential and gravity is perpendicular to it.

$GMm/r² = mv²/r = mrω²$

$v² = GM/r, so v = √(GM/r)$

$using v = 2πr/T: T² = 4π²r³/(GM) and r³ = GMT²/(4π²)$

Change for the same central mass M	Circular-orbit consequence
larger r	smaller v, smaller acceleration, longer T
smaller r	larger v, larger acceleration, shorter T
different satellite mass m at same r	same v and T because m cancels

Do not add a separate centripetal force to gravity: gravity is the centripetal resultant. If altitude h is given, use r = planet radius + h. The object is not force-free; it is continuously falling around the body.

Required condition	Why it is needed
circular orbit	constant radius and angular speed
directly above the Equator	orbital plane matches Earth's equatorial rotation
west to east	same direction as Earth's rotation
period 24 h	same angular speed as Earth

When all four conditions hold, the satellite remains above the same point/longitude on Earth's surface and appears stationary to an Earth-fixed observer.

A 24-hour polar or tilted orbit crosses different latitudes; a 24-hour westward orbit moves relative to Earth; an elliptical synchronous orbit changes radius and apparent position. Equal period alone is therefore insufficient.

Because its sky position is fixed, a communications dish can point continuously at one geostationary satellite without tracking it across the sky.

Geostationary is Earth-specific wording here: 24 h matches Earth's rotation. A stationary-looking orbit around another planet must match that planet's own rotation period while retaining equatorial, same-direction and circular conditions.

13.3 Gravitational field of a point mass

Place a small test mass m at distance r from a point source mass M. Let F be the gravitational force magnitude and G the universal gravitational constant.

$Newton's law: F = GMm/r²$

$field definition: g = F/m$

$g = (GMm/r²)/m = GM/r²$

The test mass cancels, so g depends only on source mass M and location r. Its vector direction is toward M. The same result applies outside a uniform sphere when r is measured from its centre.

G is universal, but g is not: doubling centre distance quarters g. A heavier test object experiences a larger force F=mg, not a larger field strength.

$g = GM/r², directed toward M; unit N kg⁻¹ = m s⁻²$

Use source mass M and centre distance r in SI units. Calculate the positive magnitude, then state the radial direction or apply a coordinate sign separately. For altitude h above radius R, use r=R+h.

At r = 1.47 × 10¹¹ m from the Sun (M = 1.99 × 10³⁰ kg), g = (6.67 × 10⁻¹¹)(1.99 × 10³⁰)/(1.47 × 10¹¹)² = 6.14 × 10⁻³ N kg⁻¹, directed toward the Sun.

A point Q at r/2 has four times the field magnitude of a point P at r. If Q and P lie on opposite sides of M, their field vectors point in opposite directions because each points toward M.

Do not use altitude alone or scale g as 1/r. Magnitude comparison and vector-direction comparison are separate marks and both may be required.

Near Earth's surface, a vertical height change h is tiny compared with Earth's radius R: h ≪ R. Therefore centre distance changes only from R to R+h by a very small fraction.

$g_h/g_0 = [GM/(R+h)²]/[GM/R²] = [R/(R+h)]² ≈ 1 when h ≪ R$

Globally Earth's field lines are radial. Over a small region and small height range on a sphere with very large R, those radial lines are approximately parallel and their spacing changes negligibly, representing approximately constant field strength and free-fall acceleration.

The constant-g model supports local projectile/free-fall motion and ΔE_p≈mgΔh. For satellites or heights no longer negligible relative to R, use g=GM/(R+h)² instead.

g is approximately—not exactly—constant near the surface. The approximation is local; radial lines are not globally parallel and g decreases with sufficiently large altitude.

13.4 Gravitational potential

Gravitational potential at a point is the work done per unit mass by an external agent bringing a small mass from infinity to that point, with zero potential at infinity.

For an attractive field around a point mass, potential is negative because the field releases energy as the mass moves inward.

Moving a test mass farther from a planet raises its gravitational potential toward zero.

Potential is energy per unit mass, not force; equal potential does not imply equal field strength everywhere.

$φ = −GM/r, unit J kg⁻¹; φ(∞)=0$

Use source mass M and centre distance r in SI units. Keep the negative sign: it records the chosen zero at infinity and attractive binding, not a negative magnitude of force.

For Mars, M = 6.4 × 10²³ kg and surface radius r = 3.4 × 10⁶ m. φ = −(6.67 × 10⁻¹¹)(6.4 × 10²³)/(3.4 × 10⁶) = −1.26 × 10⁷ J kg⁻¹.

Outside a spherical source the φ-r curve stays below zero and rises toward zero as r increases. At 2r, φ is half as negative; at 4r, one quarter as negative. Its slope becomes progressively less steep outward.

Potential follows −1/r, not −1/r²; field-strength magnitude follows 1/r². Use r=R+h when altitude is given, and do not let the potential curve cross zero at finite r for an isolated point mass.

Gravitational potential φ is potential energy per unit test mass. Therefore a mass m at a point of potential φ has gravitational potential energy E_P=mφ for the two-mass system, with E_P=0 at infinite separation.

$E_P = mφ = m(−GM/r) = −GMm/r$

$ΔE_P = m(φ_final − φ_initial); W_external (slow move) = ΔE_P; W_gravity = −ΔE_P$

A 2.63 kg probe at Earth's orbital distance from the Sun, where φ = −9.01 × 10⁸ J kg⁻¹, has E_P = (2.63)(−9.01 × 10⁸) = −2.37 × 10⁹ J.

Slow displacement	φ and E_P	External work	Work by gravity
outward	increase toward 0	positive	negative
inward	become more negative	negative	positive

E_P is the interaction energy of the mass pair, not an intrinsic property of one isolated object. Negative E_P means positive energy must be supplied to separate a bound pair to infinity; it does not mean energy conservation is violated.

13. Gravitational fields

Exam analysis

Most tested topics

13.1 Gravitational field

Gravitational field strength is force per unit mass

Field lines encode gravitational direction and relative strength

13.2 Gravitational force between point masses

Outside a uniform sphere, use its total mass at the centre

Newton's law gives the attraction between two point masses

Gravity supplies the centripetal resultant in a circular orbit

A geostationary orbit satisfies four simultaneous conditions

13.3 Gravitational field of a point mass

Derive the point-mass field from force per unit test mass

Use inverse-square field strength with centre distance and direction

Small height changes barely alter Earth's centre distance or field pattern

13.4 Gravitational potential

Gravitational potential is work done per unit mass in bringing a test mass from infinity

Use negative inverse-distance potential for a point mass

Potential energy of two point masses is mass times potential