Can Gravitational Potential Be Zero When Gravity Isn’t?

Imagine marking a point in mid-air as having zero gravitational potential. You hold a ball at that point, then let go. Should it stay there because its gravitational potential is zero?

No. The ball still accelerates downwards.

Gravitational potential can be zero where gravitational field strength is not zero. Potential tells you about energy per kilogram relative to a chosen reference. Gravitational field strength tells you about the force per kilogram at that location. Choosing a zero for potential does not switch off the force.

There is one important qualification. If we choose gravitational potential to be zero at infinity, as we normally do for an isolated planet, potential is negative at every finite distance from that planet. The answer therefore depends on the reference we are using.

01Zero potential is not the same as zero gravitational field

Think about a hillside.

You can decide that the school oval is at an elevation of zero metres. That does not make the oval’s surroundings perfectly flat. A slope can still exist at the chosen zero height.

Gravitational potential works in a similar way:

• The numerical value of potential depends on where you put the zero.
• The gravitational field depends on how potential changes as you move between positions.

The analogy has a limit. Gravitational potential is energy per unit mass, not physical height, and gravitational fields exist in three dimensions. But the distinction between a level and its slope is useful.

Four quantities that students often mix up

Gravitational potential is defined by

\[ \phi=\frac{U}{m} \]

where \(m\) is the mass of the object in kilograms.

Rearranging gives \(U=m\phi\). This is a useful way of extending the gravitational potential energy calculations you meet in HSC Physics.

The distinction is that \(\phi\) describes the gravitational potential at a location, while \(U\) also depends on the mass of the object placed there.

If the gravitational potential is zero, then the gravitational potential energy is also zero under that reference convention. Neither statement tells us that the gravitational force is zero.

02How could an experiment distinguish them?

Suppose you wanted to investigate the claim that zero gravitational potential means zero gravitational force.

You cannot simply place a potential meter at a point and read an absolute value. Gravitational potential differences are physically meaningful, but the overall zero must be chosen.

So what can we actually measure?

First design: measure the energy required to lift a mass

Imagine an apparatus consisting of a vertical measuring rail, a suspended metal mass, and a mechanism for raising the mass slowly.

There are two marked positions:

• A is the lower position.
• B is exactly \(2.00\ \mathrm{m}\) vertically above A.

We choose A as our zero-potential reference.

If we measure the work required to move the mass from A to B, we can determine the increase in gravitational potential energy.

But there is a limitation. Work only gives us the difference in potential energy between A and B. It cannot establish an absolute potential at either point.

Revised design: measure the force at the zero-potential point

Add a calibrated spring balance between the lifting mechanism and the suspended mass.

The balance measures the upward tension needed to support the mass. When the mass is stationary, this tension equals its weight, provided we neglect buoyancy and other forces.

Now the apparatus gives us two different measurements:

1. The local gravitational force, from the spring balance.
2. The change in gravitational potential energy, from the work done while lifting.

What happens if we change the reference?

Suppose we now define B, rather than A, as the zero-potential point.

The physical apparatus does not change. Neither does the spring balance reading.

Both conventions give the same increase of \(19.6\ \mathrm{J\,kg^{-1}}\) when moving from A to B.

The field strength is unchanged.

The graph illustrates the relationship more precisely. Near Earth’s surface, gravitational potential increases by approximately \(g\) joules per kilogram for every metre moved upwards.

For a small height increase \(\Delta h\),

\[ \Delta\phi\approx g\Delta h \]

Here, \(g\) is the positive magnitude of the gravitational field strength, while \(\Delta h\) is positive upwards.

A different zero shifts the whole potential graph up or down. It does not change its slope or the downward direction of the gravitational field.

An actual experiment would also have measurement uncertainty. The uniform-field assumption is excellent over two metres but would not remain accurate over planetary-scale height changes.

03Why the usual radial-field equation seems to disagree

You may have seen gravitational potential written as

\[ \phi=-\frac{GM}{r} \]

That equation never equals zero at a finite positive value of \(r\).

So why have we just shown that gravitational potential can be zero in a non-zero field?

Because the equation includes a reference convention: gravitational potential is defined as zero at infinite separation.

For a point mass or a location outside a spherically symmetric planet, the corresponding gravitational field-strength magnitude is

\[ g=\frac{GM}{r^2} \]

The symbols are:

• \(G\): universal gravitational constant, approximately \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
• \(M\): mass of the planet or central body, in kilograms.
• \(r\): distance from the body’s centre, in metres.
• \(g\): magnitude of gravitational field strength, in \(\mathrm{N\,kg^{-1}}\).
• \(\phi\): gravitational potential, in \(\mathrm{J\,kg^{-1}}\).

The field points towards the planet’s centre. We use the positive value \(g\) for its magnitude.

Radius is not altitude

For a planet of radius \(R\), an object at altitude \(h\) has radial distance

\[ r=R+h \]

Altitude is measured above the surface. Radius, in these gravitational equations, is measured from the planet’s centre.

With zero at infinity

For an isolated spherical planet of positive mass,

\[ \phi_\infty(r)=-\frac{GM}{r} \]

At any finite exterior radius, \(\phi_\infty\) is negative and \(g\) is non-zero.

As \(r\) increases towards infinity, both the potential and the field-strength magnitude approach zero.

This is why a student using the infinity-zero convention correctly concludes that gravitational potential cannot be zero at a finite radius in the single-planet model.

With zero at the surface

Now choose the planet’s surface, \(r=R\), to have zero potential.

We add the same constant, \(GM/R\), to the potential everywhere:

\[ \phi_R(r)=GM\left(\frac1R-\frac1r\right) \]

At the surface,

\[ \phi_R(R)=0 \]

Yet

\[ g(R)=\frac{GM}{R^2}>0 \]

We have found exactly what the question asks about: a location with zero gravitational potential and a non-zero gravitational field.

This does not contradict the infinity-zero equation. We changed the reference, not the gravitational interaction.

For a fuller treatment of gravitational potential energy and its reference convention, see Gravitational Potential Energy in a Radial Field for HSC Physics.

04A radial-field example where the surface potential is zero

The laboratory model used an almost constant gravitational field. A satellite provides a stronger test because the field changes significantly with altitude.

05The opposite case: zero gravitational field but non-zero potential

Now reverse the problem.

Imagine two identical massive spheres fixed in space, separated by some distance. Consider the exact midpoint between their centres.

Each sphere produces a gravitational field pointing towards itself.

What is the net gravitational field at the midpoint?

Zero. The two fields have equal magnitudes and opposite directions, so they cancel.

But gravitational potential behaves differently.

Potential is a scalar quantity. It has no direction, so contributions from the two masses add algebraically rather than cancelling as vectors.

If each sphere has mass \(M\) and the midpoint is a distance \(a\) from each centre, then with zero potential at infinity,

\[ \phi(O)=-\frac{GM}{a}-\frac{GM}{a} \]

So

\[ \phi(O)=-\frac{2GM}{a} \]

The net gravitational field is zero, but the potential is negative.

That is the converse of our original situation.

There is another subtle point. A test mass placed exactly at rest at this midpoint would initially experience zero net gravitational force. But a slight displacement towards one sphere would bring it closer to that sphere, causing the forces to become unequal. The midpoint is not a stable equilibrium along the line joining the two spheres.

A quick diagnostic for exam questions

The last two rows lead to an especially useful distinction: zero work and zero force are not the same thing.

A force can be non-zero while doing no work over a displacement perpendicular to its direction. It can also do positive work over one part of a journey and negative work over another, giving zero net work.

06Questions and solutions

Questions 1-4 develop the HSC gravitational field and energy ideas. Question 5 is a harder extension combining two gravitational sources, a changed potential reference, and equilibrium.

Unless stated otherwise, assume Newtonian gravity, negligible test-mass effects, and no energy losses. All numerical bodies and measurements are illustrative.

Question 1

A \(2.00\ \mathrm{kg}\) ball is held at point A in a uniform gravitational field of strength \(9.80\ \mathrm{N\,kg^{-1}}\), directed vertically downwards.

The gravitational potential is defined as zero at A.

Point B is \(3.00\ \mathrm{m}\) vertically above A.

(a) Calculate the gravitational force on the ball at A.

(b) Calculate the gravitational potential at B and the ball’s gravitational potential energy at B relative to A.

(c) State whether the ball would accelerate if released from rest at A, with no support or resistance. Explain.

Solution 1

The gravitational force is \(19.6\ \mathrm{N}\) downwards, the potential at B is \(+29.4\ \mathrm{J\,kg^{-1}}\), and the ball’s potential energy at B is \(+58.8\ \mathrm{J}\) relative to A. The ball would accelerate downwards if released.

(a) Gravitational force is

\[ \begin{aligned} F_g&=mg\\ &=(2.00)(9.80)\\ &=19.6\ \mathrm{N} \end{aligned} \]

The force is directed downwards, in the direction of the gravitational field.

(b) With constant \(g\), the potential difference is

\[ \begin{aligned} \Delta\phi&=g\Delta h\\ &=(9.80)(3.00)\\ &=29.4\ \mathrm{J\,kg^{-1}} \end{aligned} \]

Because \(\phi_A=0\), we have \(\phi_B=+29.4\ \mathrm{J\,kg^{-1}}\).

The corresponding potential energy is

\[ \begin{aligned} U_B&=m\phi_B\\ &=(2.00)(29.4)\\ &=58.8\ \mathrm{J} \end{aligned} \]

(c) The ball accelerates downwards at \(9.80\ \mathrm{m\,s^{-2}}\).

The tempting mistake is to treat \(\phi_A=0\) as meaning \(F_g=0\). But the force depends on the local gravitational field and the ball’s mass, not on the arbitrary numerical value assigned to potential.

Question 2

A fictional spherical planet has radius \(R=3.00\times10^6\ \mathrm{m}\) and surface gravitational field strength \(12.0\ \mathrm{N\,kg^{-1}}\).

Its surface is defined as zero gravitational potential.

A \(2.00\ \mathrm{kg}\) probe is moved slowly from the surface to an altitude equal to the planet’s radius.

(a) Determine the gravitational field strength at the final position.

(b) Calculate the change in gravitational potential and the work needed to move the probe, assuming it starts and finishes at rest.

(c) A student estimates the work using the surface value of \(g\) in \(W=mgh\). Determine the factor by which this estimate differs from the correct result, and explain the source of the error.

Solution 2

The field strength falls to \(3.00\ \mathrm{N\,kg^{-1}}\), the potential increases by \(1.80\times10^7\ \mathrm{J\,kg^{-1}}\), and the work required is \(3.60\times10^7\ \mathrm{J}\). The constant-\(g\) estimate is twice the correct value.

(a) At the final position, the radial distance is

\[ r=R+h=2R \]

The gravitational field follows an inverse-square relationship:

\[ \begin{aligned} g(2R)&=\frac{GM}{(2R)^2}\\ &=\frac{g(R)}4\\ &=\frac{12.0}{4}\\ &=3.00\ \mathrm{N\,kg^{-1}} \end{aligned} \]

The important constraint is that the altitude has increased by \(R\), but the radial distance has doubled from \(R\) to \(2R\).

(b) First calculate the planet’s gravitational parameter:

\[ \begin{aligned} GM&=g(R)R^2\\ &=(12.0)(3.00\times10^6)^2\\ &=1.08\times10^{14}\ \mathrm{m^3\,s^{-2}} \end{aligned} \]

Using the surface-zero reference,

\[ \begin{aligned} \Delta\phi &=GM\left(\frac1R-\frac1{2R}\right)\\ &=\frac{GM}{2R}\\ &=\frac{1.08\times10^{14}}{6.00\times10^6}\\ &=1.80\times10^7\ \mathrm{J\,kg^{-1}} \end{aligned} \]

The work supplied to the probe is

\[ \begin{aligned} W_{\mathrm{ext}}&=m\Delta\phi\\ &=(2.00)(1.80\times10^7)\\ &=3.60\times10^7\ \mathrm{J} \end{aligned} \]

(c) The student’s calculation gives

\[ \begin{aligned} W_{\mathrm{approx}}&=mg(R)h\\ &=(2.00)(12.0)(3.00\times10^6)\\ &=7.20\times10^7\ \mathrm{J} \end{aligned} \]

The ratio is

\[ \frac{W_{\mathrm{approx}}}{W_{\mathrm{ext}}} =\frac{7.20}{3.60}=2.00 \]

The estimate is twice as large as the correct work.

The student has treated the gravitational field as constant across a height change comparable to the planet’s radius. In reality, the field decreases continuously throughout the journey.

The zero-potential surface does not cause the work to vanish. Work depends on the difference in potential between the two positions.

Question 3

Two identical spherical bodies are held fixed in space. Their centres are separated by \(6.00\times10^5\ \mathrm{m}\), and each has gravitational parameter \(GM=9.00\times10^{10}\ \mathrm{m^3\,s^{-2}}\).

Point O is exactly halfway between their centres. Each body’s radius is much smaller than its distance from O.

Gravitational potential is defined as zero at infinity.

(a) Calculate the net gravitational field strength at O.

(b) Calculate the gravitational potential at O.

(c) A \(2.00\ \mathrm{kg}\) probe is moved slowly from O to infinity, starting and finishing at rest. Calculate the work supplied by an external agent.

(d) Explain why zero field strength at O does not mean that moving the probe to infinity requires zero energy.

Solution 3

The net field at O is zero, but its potential is \(-6.00\times10^5\ \mathrm{J\,kg^{-1}}\). Moving the \(2.00\ \mathrm{kg}\) probe from O to infinity requires \(1.20\times10^6\ \mathrm{J}\) of external work.

(a) The distance from O to either body is

\[ a=3.00\times10^5\ \mathrm{m} \]

The field-strength magnitude from either body is

\[ \begin{aligned} g_{\mathrm{each}}&=\frac{GM}{a^2}\\ &=\frac{9.00\times10^{10}} {(3.00\times10^5)^2}\\ &=1.00\ \mathrm{N\,kg^{-1}} \end{aligned} \]

The two gravitational field vectors point in opposite directions.

They cancel, so

\[ \vec g_{\mathrm{net}}=\vec 0 \]

(b) Gravitational potential is a scalar. The contributions from both bodies add.

\[ \begin{aligned} \phi(O)&=-\frac{GM}{a}-\frac{GM}{a}\\ &=-\frac{2GM}{a}\\ &=-6.00\times10^5\ \mathrm{J\,kg^{-1}} \end{aligned} \]

Both contributions are negative because potential is defined as zero at infinity.

(c) The probe’s initial gravitational potential energy is

\[ \begin{aligned} U_i&=m\phi(O)\\ &=(2.00)(-6.00\times10^5)\\ &=-1.20\times10^6\ \mathrm{J} \end{aligned} \]

At infinity, the potential energy approaches zero.

Therefore,

\[ \begin{aligned} W_{\mathrm{ext}}&=\Delta U\\ &=0-(-1.20\times10^6)\\ &=1.20\times10^6\ \mathrm{J} \end{aligned} \]

(d) Zero net field at O means the two gravitational forces cancel at that point. It does not mean the gravitational interaction has no potential energy.

The tempting reasoning is that if gravity exerts no net force initially, escaping should require no work.

But as the probe moves away from O, the field generally becomes non-zero. The external agent must supply energy to increase the gravitational potential energy from its negative initial value to zero at infinity.

Question 4

A spherical planet has radius \(R=4.00\times10^6\ \mathrm{m}\) and gravitational parameter \(GM=1.20\times10^{14}\ \mathrm{m^3\,s^{-2}}\).

Three points are located outside the planet:

• A is due east of the planet’s centre, at radial distance \(2R\).
• B is due north of the centre, also at radial distance \(2R\).
• C is due north of the centre, at radial distance \(3R\).

Gravitational potential is defined as zero at A.

A \(2.00\ \mathrm{kg}\) probe is moved slowly from A to B along a quarter-circle path centred on the planet. It is then moved radially outwards from B to C. The probe starts and finishes each stage at rest, and all non-gravitational forces are assumed to do only the work needed to change its potential energy.

(a) Determine the gravitational potential at B and C.

(b) Calculate the gravitational field-strength magnitude at A and B, and state the direction of the field at each point.

(c) Calculate the work done by gravity during A to B and during B to C.

(d) Explain why the work done by gravity from A to B is zero even though the gravitational field is non-zero throughout that journey.

Solution 4

Points A and B have the same potential, zero, even though their gravitational fields point in different directions. Point C has potential \(+5.00\times10^6\ \mathrm{J\,kg^{-1}}\). Gravity does zero work from A to B and \(-1.00\times10^7\ \mathrm{J}\) from B to C.

(a) Gravitational potential around a single spherical planet depends only on radial distance.

Since A and B are both at \(2R\),

\[ \phi_A=\phi_B=0 \]

For C, the increase in potential relative to B is

\[ \begin{aligned} \phi_C-\phi_B &=GM\left(\frac1{2R}-\frac1{3R}\right)\\ &=\frac{GM}{6R} \end{aligned} \]

Substituting,

\[ \begin{aligned} \phi_C&=\frac{1.20\times10^{14}} {6(4.00\times10^6)}\\ &=5.00\times10^6\ \mathrm{J\,kg^{-1}} \end{aligned} \]

(b) At both A and B,

\[ \begin{aligned} g&=\frac{GM}{(2R)^2}\\ &=\frac{1.20\times10^{14}} {(8.00\times10^6)^2}\\ &=1.875\ \mathrm{N\,kg^{-1}} \end{aligned} \]

At A, the field points west towards the planet’s centre.

At B, the field points south towards the centre.

The magnitudes are identical, but the vectors are not identical because they have different directions.

(c) From A to B,

\[ \begin{aligned} W_{g,AB}&=-m(\phi_B-\phi_A)\\ &=-2.00(0-0)\\ &=0\ \mathrm{J} \end{aligned} \]

Here, \(W_g\) denotes work done by gravity.

From B to C,

\[ \begin{aligned} W_{g,BC}&=-m(\phi_C-\phi_B)\\ &=-2.00(5.00\times10^6)\\ &=-1.00\times10^7\ \mathrm{J} \end{aligned} \]

Gravity does negative work during the outward radial movement because the displacement is opposite to the gravitational pull.

The external agent must supply \(+1.00\times10^7\ \mathrm{J}\) for the B-to-C stage.

(d) Along the quarter-circle from A to B, the probe’s instantaneous displacement is tangent to the circular path.

Gravitational force always points radially towards the planet’s centre.

The force is therefore perpendicular to the displacement throughout the circular part of the journey.

Work depends on the component of force along the displacement. That component is zero, so gravity does no work along the arc.

The tempting mistake is to conclude that zero work requires zero force. In fact, a non-zero force perpendicular to motion does zero work.

This also explains why two different points can have equal gravitational potential and non-zero gravitational fields.

Question 5

Two identical small spherical bodies are held fixed in space, with centres at coordinates \(x=-a\) and \(x=+a\).

Each body has gravitational parameter \(GM=1.00\times10^{12}\ \mathrm{m^3\,s^{-2}}\), and \(a=1.00\times10^6\ \mathrm{m}\). Each body’s radius is less than \(0.1a\).

Point O is the midpoint, \(x=0\).

A scientist chooses the gravitational potential reference so that

\[ \phi(O)=0 \]

A second point P lies on the positive \(x\)-axis beyond the right-hand body, so \(x_P>a\).

(a) Calculate the net gravitational field and the gravitational potential at O using the infinity-zero convention. Explain what changes when the scientist defines \(\phi(O)=0\).

(b) Determine whether a point P can exist where \(\phi(P)=0\) under the same reference convention as O. If so, find its coordinate \(x_P\).

(c) Calculate the magnitude and direction of the gravitational field at P.

(d) A \(2.00\ \mathrm{kg}\) probe is guided slowly from O to P along a path avoiding both bodies, starting and finishing at rest. Calculate the net external work done against gravity.

(e) Suppose instead that the probe is placed exactly at rest at O, with no guide or external disturbance. Will it begin moving towards P? What happens if it is displaced a tiny distance to the right and then released from rest?

Solution 5

Yes, P exists. It lies at \(x_P\approx1.618a\), where the gravitational potential is zero under the same reference as O, but the gravitational field is approximately \(2.76\ \mathrm{N\,kg^{-1}}\) to the left. The net external work from O to P is zero. A probe placed exactly at rest at O, however, will not spontaneously begin moving.

This result depends on separating three ideas: the scalar addition of gravitational potential, the vector addition of gravitational fields, and the difference between gravitational work and gravitational force.

(a) At O, the net gravitational field is zero by symmetry.

Each body produces an equal gravitational field in the opposite direction, so

\[ \vec g(O)=\vec0 \]

Under the infinity-zero convention,

\[ \begin{aligned} \phi_\infty(O)&=-\frac{2GM}{a}\\ &=-\frac{2(1.00\times10^{12})} {1.00\times10^6}\\ &=-2.00\times10^6\ \mathrm{J\,kg^{-1}} \end{aligned} \]

The scientist’s new reference adds \(2.00\times10^6\ \mathrm{J\,kg^{-1}}\) to the gravitational potential at every point.

That changes \(\phi(O)\) to zero but leaves the gravitational field unchanged everywhere.

(b) We need a point P with the same gravitational potential as O.

Since the reference shift is the same everywhere, this means

\[ \phi_\infty(P)=\phi_\infty(O) \]

At P, where \(x>a\), the distances from the right-hand and left-hand bodies are \(x-a\) and \(x+a\), respectively.

Therefore,

\[ -\frac{GM}{x-a}-\frac{GM}{x+a} =-\frac{2GM}{a} \]

Cancelling the common factor \(-GM\),

\[ \frac1{x-a}+\frac1{x+a}=\frac2a \]

Combining the fractions gives

\[ \frac{2x}{x^2-a^2}=\frac2a \]

Rearranging,

\[ x^2-ax-a^2=0 \]

Dividing by \(a^2\), and letting \(y=x/a\),

\[ y^2-y-1=0 \]

Using the quadratic formula,

\[ y=\frac{1\pm\sqrt5}{2} \]

We require \(x>a\), so \(y>1\). Only the positive root satisfies this condition.

Thus,

\[ \begin{aligned} x_P&=\frac{1+\sqrt5}{2}a\\ &\approx1.618a\\ &=1.618\times10^6\ \mathrm{m} \end{aligned} \]

Point P genuinely exists outside the right-hand body.

Its gravitational potential is zero under the scientist’s chosen convention, just like O.

(c) At P, both gravitational field contributions point left, towards their respective source bodies.

The distances to the bodies are

\[ \begin{aligned} d_{\mathrm{right}}&=x_P-a\\ &=0.618\times10^6\ \mathrm{m} \end{aligned} \]

and

\[ \begin{aligned} d_{\mathrm{left}}&=x_P+a\\ &=2.618\times10^6\ \mathrm{m} \end{aligned} \]

The field-strength magnitude due to the nearer, right-hand body is

\[ \begin{aligned} g_{\mathrm{right}}&=\frac{GM}{d_{\mathrm{right}}^2}\\ &=\frac{1.00\times10^{12}} {(0.618\times10^6)^2}\\ &\approx2.62\ \mathrm{N\,kg^{-1}} \end{aligned} \]

The farther, left-hand body contributes approximately \(0.146\ \mathrm{N\,kg^{-1}}\), using the same inverse-square equation with \(d_{\mathrm{left}}\).

The contributions point in the same direction, so their magnitudes add:

\[ \begin{aligned} g(P)&\approx2.62+0.146\\ &=2.76\ \mathrm{N\,kg^{-1}} \end{aligned} \]

The net field points left, in the negative \(x\)-direction.

This is the central surprise. O and P both have zero gravitational potential, but O has zero net field while P has a substantial non-zero field.

(d) The work done against gravity depends on the change in potential energy, not the local field strength at either endpoint.

Since

\[ \phi(O)=\phi(P)=0 \]

we have

\[ \begin{aligned} W_{\mathrm{ext}}&=\Delta U\\ &=m[\phi(P)-\phi(O)]\\ &=(2.00)(0-0)\\ &=0\ \mathrm{J} \end{aligned} \]

The net external work is zero.

This does not mean the guiding force is zero throughout the journey. Nor does it mean gravity does zero work during every small displacement.

Gravity can do positive work during one part of the journey and negative work during another. Those contributions cancel over the complete path.

Because gravitational force is conservative, the total gravitational work depends only on the endpoints, provided the bodies remain fixed and the path stays outside them.

(e) A probe placed exactly at rest at O remains at rest in this idealised arrangement.

At O, the two gravitational field vectors cancel exactly. With no initial velocity and no net force, there is nothing to start its motion.

However, this equilibrium is unstable along the line joining the two bodies.

If the probe is displaced slightly to the right, it becomes closer to the right-hand body than to the left-hand body. The right-hand gravitational attraction becomes stronger.

The net field then points right, and the probe accelerates towards the right-hand body.

It does not automatically reach P; without a guide, its subsequent motion is governed by the gravitational forces and may bring it into contact with that body.

The trap in this problem is assuming that the same potential value at two positions implies the same gravitational force, or that an object must move simply because another position has equal potential.

Neither is true.

Equal potential determines the net gravitational work between positions. The gravitational field determines the local acceleration. Initial conditions determine whether motion begins at all.

06What to remember when choosing an equation

The quickest way to handle these questions is to identify exactly which quantity the question is asking about.

If the question asks whether a mass accelerates, calculate the gravitational field or net gravitational force. A zero value of potential tells you nothing directly about acceleration.

If the question asks how much energy is gained or lost during movement, calculate the potential difference and use \(\Delta U=m\Delta\phi\).

If the question asks where gravitational potential is zero, first identify the reference convention. With zero at infinity, the potential of a single isolated spherical planet is negative at every finite radius. With a surface-zero convention, potential can be zero at the surface while gravity remains non-zero.

And if multiple gravitational sources are involved, remember that fields add as vectors, while potentials add as scalars.

The next useful distinction is how changing radial distance affects field strength and gravitational potential energy differently. That relationship is developed in Gravitational Field Strength vs Potential Energy: How Distance Changes Each.

The main idea is simple but powerful: a zero on an energy scale is not a point where gravity has switched off. What matters for motion is the local gravitational field, and what matters for gravitational work is the difference in potential.