Gravitational Fields

01

Mass, force and field strength

Gravitational force depends on both interacting masses. Field strength describes the force per unit test mass at a position, so it depends on the source and the position rather than the small test mass you choose.

Newtonian attraction

Two point masses M and m attract each other with force magnitude:

G is the universal gravitational constant, in N m² kg^-2. Here M names the source mass and m the small test mass, both in kg. The distance r joins the masses, in metres. The force acts along that joining line, towards the other mass.

The forces on the two bodies are equal and opposite, but act on different bodies. They do not cancel in the equation for either body. Treating a much more massive source as fixed neglects its very small motion, not its interaction with the test mass.

For a point outside a spherically symmetric body, use the body's mass as concentrated at its centre. At altitude h above a surface of radius R, the required separation is r = R + h. This rule applies outside the body; it does not give the field at points inside it.

At fixed masses, doubling r quarters the force. At fixed r, doubling either mass doubles the force on each partner.

Derive the point-mass field

Gravitational field strength is gravitational force per unit test mass at a point. Dividing the force law by m gives:

Here g is the magnitude, in N/kg. The field points towards the source centre. The small test mass cancels: a heavier test body has more gravitational force, but the same force per unit mass at that position.

If outward radial direction is positive, the signed component is g_r = -GM/r². Keep the positive magnitude g separate from the negative outward-axis component g_r. A minus sign in that component means inward, not repulsion.

Estimate the field scale

Using supplied rough scales GM about 4 × 10¹⁴ m³/s² and surface radius R about 6 × 10⁶ m, the ratio GM/R² is of order 10 N/kg. These rounded scales give a plausibility check before the more precise model calculation.

Why a near-surface field can be approximately uniform

At height h, the exact ratio in the spherical model is:

When the height change is much smaller than R, the radius barely changes. Within a small surface region, the directions towards the distant centre are also nearly parallel. These are the conditions behind a locally uniform downward field and the approximation ΔE_p = mgΔh with nearly constant g.

The approximation does not extend to all altitudes. In the worked example, r = 1.25R, so g = g(R)/1.25² = 0.64g(R). The field has changed substantially, and constant surface g would be unsuitable.

Field strength and free-fall acceleration

If gravity is the only force on a body, Newton's second law gives mg = ma, so its acceleration has magnitude g and points inward. The units are equivalent:

Field strength is defined through force per unit mass; acceleration is defined through velocity change per unit time. They are equal for free fall under gravity alone. Support or drag can change the resultant acceleration without removing the gravitational field.

An astronaut and spacecraft can fall together with little or no usual support force between them. That apparent weightlessness is compatible with the substantial 6.25 N/kg field at the quoted orbital radius; it is not evidence of zero gravity.

02

Potential and potential energy

Gravitational potential is an energy-per-mass quantity at a point. Gravitational potential energy belongs to a specified interacting system and also depends on the test body's mass.

Choose the zero at infinity

Gravitational potential φ at a point is the work done per unit mass by an external force in bringing a small test mass from infinity to that point, without a change in kinetic energy. Its unit is J/kg.

Set the potential to zero at infinite separation. This is an energy reference. Bringing the test mass inward slowly lets gravity do positive work. To prevent a kinetic-energy increase, the external agent does negative work, so the potential at the finite point is negative.

This is the point-source potential and the exterior potential of a spherical source with zero at infinity. The radius is centre distance. Negative potential does not mean negative mass or an outward force: it describes the work relative to the chosen zero.

Multiply by mass to obtain system energy

For source mass M and test mass m separated by r, the gravitational potential energy is:

U_G names the gravitational potential energy of the two-mass interaction, in joules. It is the gravitational case of the potential energy E_p used earlier, so E_p = U_G in this account. It is not an additional energy store to add again.

Changing the small test mass at a fixed position leaves φ unchanged in this source-field model, but changes U_G in proportion to that mass. Keep potential in J/kg distinct from potential energy in J.

Estimate the potential and energy scales

Using the rough inputs GM about 4 × 10¹⁴ m³/s² and R about 6 × 10⁶ m gives |φ| about 7 × 10⁷ J/kg. For a roughly 400 kg test body, |U_G| is of order 3 × 10¹⁰ J. Both signed values are negative with the infinity reference.

Give work its correct sign and recipient

For movement from r₁ to r₂, subtract the two potential energies:

During the outward transfer above, gravity does -5.00 × 10⁹ J of work because its force opposes the outward displacement. If an external agent moves the body with no kinetic-energy change and no other transfer, its work is +5.00 × 10⁹ J.

If kinetic energy changes, the external work must account for that too. With no dissipative transfer, W_external = ΔE_k + ΔU_G. Include any relevant losses or other stores in a more complete account. The field-work relationship does not say every applied work equals ΔU_G under all conditions.

In particular, raising the body to a satellite's radius is not enough to put it into a circular orbit. That final state also needs the orbital kinetic energy.

03

Field from a potential gradient

The field at a point comes from how quickly potential changes with position, not from the potential's value there. Use a local tangent and include the negative sign.

Take radial distance r as positive outwards from the source centre, and set φ = 0 at infinity. For the exterior field of an isolated spherical source, φ = -GM/r. Moving outwards makes φ less negative, so its graph rises towards zero.

The notation dφ/dr means the local gradient of potential against outward radius. The signed radial field component is its negative. A positive potential gradient therefore gives a negative g_r, meaning an inward field.

This agrees with work: a small outward displacement raises gravitational potential energy, so gravity does negative work. Gravity acts towards decreasing potential. The vector direction is not obtained simply by asking whether the potential itself is positive or negative.

Read graph height and gradient as different quantities

Use GM = 4.0 × 10¹⁴ m³/s² and source radius R = 6.4 × 10⁶ m. The table values are calculated from the exterior model; the displayed decimals are rounded where necessary.

Both curves stay below zero under this convention, approaching zero as r increases. Their zero lines represent the infinity limit; neither quantity becomes exactly zero at a finite radius in this isolated-source model. The curved functions should not be replaced by straight segments joining the table entries.

Reflecting the negative curves across their horizontal zero lines gives the corresponding positive magnitudes: |φ| = GM/r has a 1/r shape, while |g_r| = GM/r² has a 1/r² shape. At twice the centre distance, the potential magnitude halves and the field magnitude quarters. This comparison uses the same exterior domain r ≥ R; it says nothing about the source's interior.

A long chord between two curve points gives an average potential gradient over that interval, not necessarily the field at a particular point. Use the tangent at the required radius, not the potential's graph height or an arbitrary chord.

Zero field need not mean zero potential

For a separate comparison, hold two equal source masses at equal distances on opposite sides of a midpoint. At that point their equal field vectors point in opposite directions, so the total field is zero.

Potential contributions add as scalars. With zero at infinity, each source contributes a negative potential at the midpoint, so the total potential is still negative. Vector cancellation of force per mass does not require the energy per mass to vanish.

04

Escape using an energy account

Minimum escape means reaching indefinitely large separation with speed tending to zero. Set that final condition in an energy account rather than extending constant-g mgh to an infinite height.

Use a spherical source of mass M, treated as fixed, and a much smaller body of mass m starting at centre distance r. Neglect atmosphere, other massive bodies, dissipative effects and propulsion after release. Choose a path that avoids collision with the source; begin with an outward launch.

With zero gravitational potential energy at infinity, U_G = -GMm/r. Mechanical energy E_k + U_G is conserved during this unpowered motion under the stated assumptions.

Derive the minimum speed from the final condition

For the minimum escape case, both U_G and E_k tend to zero at indefinitely large separation. The initial mechanical energy must therefore be zero:

The test mass cancels from the speed threshold. Its required kinetic energy still depends on mass. This equation gives a speed magnitude, not a unique velocity direction; the proposed direction must also avoid the source and satisfy the model assumptions.

Estimate the scale

With rough GM about 4 × 10¹⁴ m³/s² and R about 6 × 10⁶ m, the square root of 2GM/R is of order 10⁴ m/s. For a roughly 400 kg body, the required kinetic energy is of order 3 × 10¹⁰ J, matching the magnitude of its initial negative potential energy.

Compare launch speeds using total energy

• At the minimum speed: total mechanical energy is zero, and speed tends to zero at infinity.
• Above the minimum: total mechanical energy is positive, leaving nonzero kinetic energy at large separation in this model.
• Below the minimum: total mechanical energy is negative, so the body cannot reach infinity without further energy input under the stated assumptions.

For a greater launch speed u and residual far-away speed v_∞, the account becomes ½mu² - GMm/r = ½mv_∞². Minimum escape is the special case v_∞ = 0.

Real launch energy can involve source rotation, atmospheric losses and continuing propulsion. The ideal threshold does not include those effects. Surface g also decreases with altitude; a constant-g mgh expression cannot describe lifting to infinity.

05

Circular orbits and geostationary satellites

A circular orbit needs the correct tangential speed at its radius. Gravity supplies the inward acceleration; an energy account then distinguishes reaching that radius from reaching the complete orbital state.

Model a small satellite m outside a spherical central mass M that is effectively fixed. Neglect drag, thrust and other gravitational sources. The radius r is measured from the source centre, not its surface.

On this page: orbital energy; radius-period data; geostationary conditions.

Use gravity as the inward resultant

Uniform circular motion requires inward acceleration v²/r = rω². Gravity supplies the actual force:

The small satellite mass cancels. At a fixed radius around the same source, its required speed is independent of its mass, although the gravitational force is proportional to that mass.

As in the radial-force method, centripetal describes the resultant's role, not an extra interaction to add to gravity. There is no outward balancing force in this inertial-frame account. Constant speed still involves changing velocity and inward acceleration.

The period T is time for one complete revolution. Substitute v = 2πr/T into the radial equation:

Larger circular orbits around the same source have lower speed and longer period. The relation v = rω does not imply greater speed at larger r here, because the different orbits do not have the same angular speed.

Estimate the orbit scale

For supplied rough scales GM about 4 × 10¹⁴ m³/s² and centre distance r about 8 × 10⁶ m, the speed is about 7 × 10³ m/s. The circumference divided by that speed gives a period of about 7 × 10³ s, of order 2 hours. These checks precede the more precise calculation.

At this same radius, circular speed is 7.07 km/s, while minimum escape speed is 10.0 km/s. A maintained circular orbit and an escape to indefinitely large separation have different final conditions.

Test the radius-period relationship

For a common central mass, T² is proportional to r³. The table gives periods generated from the supplied GM model and rounded to the nearest second. They are model values, not satellite observations.

Plot T² vertically against r³ horizontally. Before rounding, the exact model is a straight line through the origin with gradient:

The gradient represents the inverse of GM multiplied by 4π². It is not an orbital speed. Rearranging gives GM = 4π²/gradient.

For genuine observations, differences from the circular prediction could reflect uncertainty in radius or period, a noncircular orbit, or other neglected influences. Check those conditions before attributing every mismatch to a measurement zero error.

A geostationary orbit needs more than a matching period

In the ideal model, a geostationary satellite remains above the same equatorial longitude. It has a circular orbit in Earth's equatorial plane, moves west to east in Earth's rotational sense, and has the same period as Earth's rotation.

These conditions give equal angular rates and a fixed apparent ground position. A matching period alone is insufficient: an inclined orbit changes the apparent position, while an eccentric orbit does not maintain the required constant angular rate. Reversing the orbital sense also prevents the fixed alignment.

The satellite is still moving and accelerating in an Earth-centred inertial description. Gravity provides its inward resultant. Its angular speed matches Earth's rotation; its tangential speed is not the same as that of a ground marker at a smaller radius.

Communications: the fixed apparent direction allows a ground antenna to keep pointing the same way. Weather observation: a continuing view lets changes in the same broad visible region be followed over time.

The satellite is not directly overhead every observer: its fixed ground location is equatorial. The viewing geometry is also less useful for far polar regions. The application depends on the region that must be seen, not just on the period.