Why G and g Are Different: Field Strength, Force and Potential

A 1 kg object near Earth’s surface has a weight of about 9.8 N. The universal gravitational constant, \(G\), is approximately \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\), and it has the same value on Earth, the Moon, and anywhere else in the universe.

Why are these numbers so different?

Because \(G\) and \(g\) describe different physical quantities. \(G\) is the universal gravitational constant that determines the strength of gravitational interactions. \(g\) is the gravitational field strength at a particular location. It also gives the acceleration of an object in free fall under gravity alone.

The distinction becomes much clearer when you connect Newton’s law of gravitation to gravitational fields, and then separate gravitational force from gravitational potential and potential energy.

01One law, two different quantities

Imagine a planet with mass \(M\) and a small object with mass \(m\), positioned outside the planet.

Newton’s law of universal gravitation tells us that the gravitational force between them is:

\[ F=G\frac{Mm}{r^2} \]

Here:

• \(F\) is the magnitude of the gravitational force, measured in newtons (N).
• \(G\) is the universal gravitational constant.
• \(M\) is the mass of the planet, measured in kilograms.
• \(m\) is the mass of the object, measured in kilograms.
• \(r\) is the distance between the centres of the two masses, measured in metres.

Notice that \(G\) is a constant in this equation. It does not change when you change the planet, the object, or their separation.

But \(g\) is a different story.

Gravitational field strength describes the gravitational force experienced per unit mass at a particular location:

\[ g=\frac{F}{m} \]

Substitute Newton’s law:

\[ g=\frac{1}{m}\left(G\frac{Mm}{r^2}\right) \]

The test mass \(m\) cancels:

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

This equation explains the relationship between the two quantities.

• \(G\) is universal and constant.
• \(g\) depends on the mass creating the gravitational field and the distance from its centre.

The test object’s mass does not affect the gravitational field strength at its location.

For example, a 1 kg object and a 10 kg object at the same location experience the same gravitational field strength. However, the 10 kg object experiences ten times the gravitational force.

Comparing G and g

One detail is worth remembering: gravitational field strength and acceleration have equivalent SI units.

\[ 1\ \mathrm{N\,kg^{-1}}=1\ \mathrm{m\,s^{-2}} \]

This follows from Newton’s second law, \(F=ma\). In free fall, when gravity is the only force acting, the acceleration of an object is equal to the gravitational field strength.

That is why \(g\) is sometimes called the acceleration due to gravity.

02How scientists measured G: the Cavendish experiment

There is an interesting experimental reason why \(G\) and \(g\) must be treated separately.

Newton’s law of gravitation described how gravitational force should behave, but scientists initially did not know the numerical value of \(G\).

Measuring the acceleration of falling objects near Earth was not enough to determine it independently.

At Earth’s surface:

\[ g=\frac{GM_E}{R_E^2} \]

Here, \(M_E\) is Earth’s mass and \(R_E\) is Earth’s radius.

Measuring \(g\) gives a relationship involving both \(G\) and Earth’s mass. To determine the universal constant independently, scientists needed another way to measure gravitational attraction.

This led to the famous Cavendish experiment.

The apparatus

In 1798, Henry Cavendish used a torsion balance to investigate the gravitational attraction between known masses.

The apparatus consisted of:

• A horizontal rod with a small lead sphere attached to each end.
• A thin wire suspending the rod from its centre, allowing it to rotate.
• Two much larger lead spheres positioned close to the smaller spheres.
• A measuring arrangement that allowed the tiny rotation of the rod to be observed.

The large spheres were moved close to the smaller ones. Their gravitational attraction pulled the smaller spheres towards them, producing a torque that twisted the suspension wire.

The wire resisted this rotation, eventually balancing the gravitational torque.

What did Cavendish predict?

If Newton’s law was correct, moving the large lead spheres closer to the small ones should produce a measurable rotation.

Moving the large spheres to the opposite positions should reverse the direction of the gravitational torque.

This reversal was particularly useful. A fixed offset in the apparatus would not necessarily reverse when the large masses were moved, but the predicted gravitational effect would.

What did he observe?

The suspended rod twisted by a very small angle and oscillated around a new equilibrium position.

When the large spheres were repositioned, the direction of the deflection changed as predicted.

The twist of the wire provided a way to calculate the tiny gravitational force between the known masses.

The experiment was extremely sensitive. Air currents, temperature differences, vibrations, and other disturbances could overwhelm the gravitational attraction.

Cavendish therefore improved the experimental arrangement by enclosing the apparatus, observing it through a telescope, and using a pulley system to move the heavy spheres from outside the room. He also investigated possible magnetic and other sources of interference.

The important measurement was not simply that the rod moved. It was that its movement changed in the predicted way when the positions of the large masses changed.

A small calculation: extracting G

Suppose an illustrative modern torsion-balance experiment measures an attractive force of \(8.0\times10^{-10}\ \mathrm N\) between two spheres.

The data are:

• Large sphere mass: \(M=1.50\ \mathrm{kg}\).
• Small sphere mass: \(m=0.020\ \mathrm{kg}\).
• Centre-to-centre separation: \(d=0.050\ \mathrm m\).
• Measured gravitational force: \(F=8.0\times10^{-10}\ \mathrm N\).

Rearrange Newton’s law:

\[ F=G\frac{Mm}{d^2} \]

\[ G=\frac{Fd^2}{Mm} \]

Substitute:

\[ G= \frac{(8.0\times10^{-10})(0.050)^2} {(1.50)(0.020)} \]

\[ G=\frac{2.0\times10^{-12}}{0.030} \]

\[ \boxed{G=6.7\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}} \]

These are illustrative values, not Cavendish’s original measurements.

The important idea is that a measured force between known masses allows us to calculate \(G\).

A historical distinction: Cavendish’s original 1798 paper investigated the mean density of Earth. He did not publish a modern measurement of \(G\) in the way physics textbooks present it today. His experiment was subsequently interpreted as a method of determining the universal gravitational constant.

You can read more about the experiment in the American Physical Society’s historical account and the University of Oxford’s history of measurements of G.

03Radius is not altitude: getting the distance right

Now that the distinction between \(G\) and \(g\) is clear, there is another common source of mistakes: confusing a planet’s radius with an object’s altitude.

Consider an object positioned above a spherical planet.

There are three important distances:

• \(R\): the planet’s radius, measured from its centre to its surface.
• \(h\): the object’s altitude, measured from the surface.
• \(r\): the object’s distance from the planet’s centre.

These are related by:

\[ \boxed{r=R+h} \]

This distinction matters because the gravitational field equation uses \(r\), not \(h\):

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

Therefore:

\[ \boxed{g=\frac{GM}{(R+h)^2}} \]

Let’s see what happens at different altitudes.

Let \(g_s\) represent the gravitational field strength at the planet’s surface.

At the surface, \(h=0\), so \(r=R\):

\[ g_s=\frac{GM}{R^2} \]

Now consider the following situations.

Notice something important.

If an object is at an altitude equal to the planet’s radius, its distance from the centre is twice the planet’s radius.

Its gravitational field strength is therefore:

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

\[ \boxed{g=\frac{g_s}{4}} \]

It is one-quarter of the surface field strength, not one-half.

What if the altitude is twice the planet’s radius?

Then:

\[ r=R+2R=3R \]

Therefore:

\[ \boxed{g=\frac{g_s}{9}} \]

The trap is using the altitude as though it were the centre-to-object distance.

Whenever a question gives you an altitude, convert it into \(r=R+h\) before substituting into the gravitational equations.

These equations assume an object outside a spherically symmetric planet. Real Earth also has small variations in gravitational field strength due to its shape, rotation, and local mass distribution, but those effects are normally excluded from this model.

04Gravitational field, force, potential, and energy

So far, we have compared \(G\) and \(g\). However, gravitational questions often introduce two more quantities that are easy to confuse: gravitational potential and gravitational potential energy.

They are related, but they are not interchangeable.

1. Gravitational field strength, g

Gravitational field strength tells us the gravitational force per unit mass at a particular location.

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

Its unit is \(\mathrm{N\,kg^{-1}}\), equivalent to \(\mathrm{m\,s^{-2}}\).

It is a vector quantity, meaning it has both magnitude and direction.

For an isolated spherical planet, the field points towards the planet’s centre.

2. Gravitational force, F

Gravitational force is the actual force experienced by an object of mass \(m\).

\[ \boxed{F=mg=\frac{GMm}{r^2}} \]

Its unit is the newton (N).

Unlike gravitational field strength, gravitational force depends on the mass of the object experiencing the field.

A 100 kg object experiences ten times the gravitational force of a 10 kg object at the same location.

However, both objects experience the same gravitational field strength.

3. Gravitational potential, V

Gravitational potential describes gravitational potential energy per unit mass at a particular location.

Taking gravitational potential to be zero at infinity:

\[ \boxed{V=-\frac{GM}{r}} \]

Its unit is \(\mathrm{J\,kg^{-1}}\).

Notice that gravitational potential depends on the inverse of distance, not the inverse square of distance.

This is a major difference between gravitational field strength and gravitational potential.

• Field strength varies as \(1/r^2\).
• Gravitational potential varies as \(-1/r\).

Why does potential have a different relationship with distance?

Field strength describes the gravitational pull at a particular point. Potential describes the energy per kilogram associated with the object’s position relative to the chosen zero of potential.

A useful analogy is a landscape:

• Gravitational potential is like elevation.
• Gravitational field is related to the slope.

You can stand at the top of a hill where the slope is momentarily zero, even though your elevation is not zero. Likewise, gravitational field strength can be zero at a location where gravitational potential is not zero.

The analogy has limits: gravitational potential is measured in \(\mathrm{J\,kg^{-1}}\), and its zero is a chosen reference, usually infinity.

4. Gravitational potential energy, U

Gravitational potential energy is the total gravitational potential energy of an object at a particular location.

It depends on both the gravitational potential and the object’s mass.

\[ U=mV \]

Substituting:

\[ \boxed{U=-\frac{GMm}{r}} \]

Its unit is the joule (J).

This gives us another important distinction:

• Gravitational potential does not depend on the test object’s mass.
• Gravitational potential energy does depend on the test object’s mass.

A 100 kg object at a particular location has ten times the gravitational potential energy of a 10 kg object at that same location.

Putting all five quantities together

Why field strength and potential do not decrease at the same rate

Return to the situation where an object is at a distance of \(2R\) from a planet’s centre.

We already established that:

\[ g=\frac{g_s}{4} \]

But gravitational potential is:

\[ V=-\frac{GM}{2R} \]

At the surface:

\[ V_s=-\frac{GM}{R} \]

Therefore:

\[ \boxed{V=\frac{V_s}{2}} \]

The magnitude of gravitational potential is now half its surface value, while gravitational field strength is one-quarter of its surface value.

This difference is worth understanding rather than memorising.

There is another important detail about the negative sign in gravitational potential and potential energy.

We choose gravitational potential to be zero at infinity. Since gravity is attractive, an object at a finite distance from the planet has negative gravitational potential.

As the object moves farther away, its potential becomes less negative and approaches zero.

The same applies to gravitational potential energy when the zero is chosen at infinity.

For a slow movement between two positions, the change in potential energy is:

\[ \Delta U=U_f-U_i \]

The work done by gravity is:

\[ W_{\text{gravity}}=-\Delta U \]

If an external force moves the object slowly, with negligible change in kinetic energy, the work done by that external force is:

\[ W_{\text{external}}=\Delta U \]

This relationship becomes useful when calculating the energy needed to move a spacecraft away from a planet.

For a deeper look at the reference point for gravitational potential, see Can Gravitational Potential Be Zero When Gravity Isn’t?.

05Worked examples

06Questions and solutions

The following questions progress from standard HSC applications to more demanding problems involving multiple gravitational quantities.

Question 1

A spherical planet has a mass of \(4.8\times10^{24}\ \mathrm{kg}\) and a radius of \(6.0\times10^6\ \mathrm m\).

A 750 kg payload is positioned at an altitude of twice the planet’s radius above its surface.

Calculate:

a) The distance between the payload and the planet’s centre.

b) The gravitational field strength at the payload.

c) The gravitational force experienced by the payload.

Use \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).

Solution 1

The payload is \(1.8\times10^7\ \mathrm m\) from the planet’s centre. Its gravitational field strength is approximately \(0.99\ \mathrm{N\,kg^{-1}}\), giving a gravitational force of approximately \(741\ \mathrm N\).

Step 1: Calculate the centre-to-payload distance.

The altitude is twice the planet’s radius:

\[ h=2R \]

Therefore:

\[ r=R+h \]

\[ r=R+2R=3R \]

\[ r=3(6.0\times10^6) \]

\[ \boxed{r=1.8\times10^7\ \mathrm m} \]

Step 2: Calculate the gravitational field strength.

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

Substitute:

\[ g= \frac{(6.67\times10^{-11})(4.8\times10^{24})} {(1.8\times10^7)^2} \]

\[ g=\frac{3.2016\times10^{14}} {3.24\times10^{14}} \]

\[ \boxed{g=0.99\ \mathrm{N\,kg^{-1}}} \]

Step 3: Calculate the gravitational force.

\[ F=mg \]

\[ F=750(0.988) \]

\[ \boxed{F\approx741\ \mathrm N} \]

The force acts towards the centre of the planet.

The important step was converting altitude into centre distance before using the inverse-square equation. Using \(2R\) instead of \(3R\) would give an incorrect answer.

Question 2

An ideal spherical planet has mass \(M\) and radius \(R\).

Two probes are positioned outside the planet:

• Probe A has mass \(m\) and is at altitude \(h_A=R\).
• Probe B has mass \(4m\) and is at altitude \(h_B=2R\).

Calculate the following ratios:

a) \(g_A/g_B\)

b) \(F_A/F_B\)

c) \(|V_A|/|V_B|\)

d) \(|U_A|/|U_B|\)

Explain which probe experiences the greater gravitational force and which has the more negative gravitational potential energy.

Take gravitational potential to be zero at infinity.

Solution 2

The ratios are:

\[ \boxed{\frac{g_A}{g_B}=\frac94} \]

\[ \boxed{\frac{F_A}{F_B}=\frac9{16}} \]

\[ \boxed{\frac{|V_A|}{|V_B|}=\frac32} \]

\[ \boxed{\frac{|U_A|}{|U_B|}=\frac38} \]

Probe B experiences the greater gravitational force and has the more negative gravitational potential energy. Probe A is in the stronger gravitational field and has the more negative gravitational potential.

Step 1: Find the centre-to-probe distances.

For probe A:

\[ r_A=R+R=2R \]

For probe B:

\[ r_B=R+2R=3R \]

This distinction is essential because the equations depend on distance from the planet’s centre.

Step 2: Compare gravitational field strengths.

Using:

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

The ratio is:

\[ \frac{g_A}{g_B} = \frac{GM/(2R)^2}{GM/(3R)^2} \]

Cancel the common terms:

\[ \frac{g_A}{g_B} = \frac{9R^2}{4R^2} \]

\[ \boxed{\frac{g_A}{g_B}=\frac94} \]

Probe A experiences a gravitational field strength 2.25 times greater than probe B.

Step 3: Compare gravitational forces.

Use:

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

Remember that probe B has four times the mass of probe A.

\[ \frac{F_A}{F_B} = \frac{GMm/(2R)^2} {GM(4m)/(3R)^2} \]

Simplifying:

\[ \frac{F_A}{F_B} = \frac{9}{16} \]

\[ \boxed{\frac{F_A}{F_B}=\frac9{16}} \]

Probe B experiences the greater gravitational force, even though it is in the weaker gravitational field.

This is because gravitational force depends on both field strength and the object’s mass.

Step 4: Compare gravitational potentials.

Using:

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

Compare the magnitudes:

\[ \frac{|V_A|}{|V_B|} = \frac{GM/(2R)}{GM/(3R)} \]

\[ \boxed{\frac{|V_A|}{|V_B|}=\frac32} \]

Both potentials are negative. Probe A has the more negative gravitational potential because it is closer to the planet.

Step 5: Compare gravitational potential energies.

Use:

\[ U=-\frac{GMm}{r} \]

Therefore:

\[ \frac{|U_A|}{|U_B|} = \frac{GMm/(2R)}{GM(4m)/(3R)} \]

\[ \boxed{\frac{|U_A|}{|U_B|}=\frac38} \]

Probe B has the greater magnitude of gravitational potential energy, meaning its potential energy is more negative.

This is the important distinction: probe A is closer to the planet, but probe B has a much greater mass.

The location determines the potential. The object’s mass determines how much potential energy it has at that location.

Question 3

A 200 kg object is released from rest at a distance of \(3R\) from the centre of a spherical planet.

The planet has:

• Radius \(R=6.0\times10^6\ \mathrm m\).
• Surface gravitational field strength \(g_s=8.9\ \mathrm{N\,kg^{-1}}\).

The object falls radially towards the planet and reaches its surface.

Assume there is no atmosphere, the planet is not rotating, and gravity is the only force doing work.

Calculate:

a) The change in the object’s gravitational potential energy.

b) The object’s speed immediately before reaching the surface.

Explain why using the surface gravitational field strength as a constant throughout the fall would be incorrect.

Solution 3

The object loses \(7.12\times10^9\ \mathrm J\) of gravitational potential energy and reaches the surface with a speed of approximately \(8.4\times10^3\ \mathrm{m\,s^{-1}}\).

Step 1: Express the planet’s mass in terms of its surface gravitational field strength.

At the surface:

\[ g_s=\frac{GM}{R^2} \]

Rearrange:

\[ GM=g_sR^2 \]

This allows us to calculate the energy without needing the planet’s mass.

Step 2: Calculate the initial gravitational potential energy.

The initial distance from the centre is \(3R\).

Using:

\[ U_i=-\frac{GMm}{3R} \]

Substitute \(GM=g_sR^2\):

\[ U_i=-\frac{g_sR^2m}{3R} \]

\[ U_i=-\frac{mg_sR}{3} \]

Step 3: Calculate the final gravitational potential energy.

At the surface, \(r=R\).

\[ U_f=-\frac{GMm}{R} \]

Substitute:

\[ U_f=-mg_sR \]

Step 4: Calculate the change in potential energy.

\[ \Delta U=U_f-U_i \]

\[ \Delta U=-mg_sR+\frac{mg_sR}{3} \]

\[ \Delta U=-\frac23mg_sR \]

Now substitute the values:

\[ \Delta U= -\frac23(200)(8.9)(6.0\times10^6) \]

\[ \boxed{\Delta U=-7.12\times10^9\ \mathrm J} \]

The negative sign indicates that the object’s gravitational potential energy decreases as it moves towards the planet.

Step 5: Apply conservation of mechanical energy.

The object starts from rest, so its initial kinetic energy is zero.

Ignoring energy losses:

\[ K_i+U_i=K_f+U_f \]

Since \(K_i=0\):

\[ K_f=U_i-U_f \]

Therefore:

\[ K_f=7.12\times10^9\ \mathrm J \]

Step 6: Calculate the final speed.

Use kinetic energy:

\[ K_f=\frac12mv^2 \]

Rearrange:

\[ v=\sqrt{\frac{2K_f}{m}} \]

Substitute:

\[ v=\sqrt{\frac{2(7.12\times10^9)}{200}} \]

\[ v=\sqrt{7.12\times10^7} \]

\[ \boxed{v=8.4\times10^3\ \mathrm{m\,s^{-1}}} \]

The object reaches the surface at approximately \(8.4\ \mathrm{km\,s^{-1}}\).

Why can’t we use constant surface gravity?

The gravitational field strength changes as the object falls.

At the initial position:

\[ g_i=\frac{g_s}{(3)^2} \]

\[ g_i=\frac{8.9}{9} \]

\[ g_i\approx0.99\ \mathrm{N\,kg^{-1}} \]

At the surface, it increases to \(8.9\ \mathrm{N\,kg^{-1}}\).

Using the surface value throughout the entire fall would significantly overestimate the gravitational acceleration over most of the journey.

This is why gravitational potential energy and conservation of energy are useful when analysing large changes in altitude.

Question 4

A modern torsion-balance experiment measures the gravitational attraction between two lead spheres.

The experimental data are:

The measured force has been corrected for the effects of the suspension wire and other experimental disturbances.

a) Calculate the universal gravitational constant \(G\).

b) Calculate the gravitational field strength produced by the large sphere at the location of the small sphere.

c) Explain why the field strength in part (b) is not the same as Earth’s gravitational field strength.

Solution 4

The experiment gives \(G=6.7\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\). The field produced by the large sphere at the small sphere is \(4.0\times10^{-8}\ \mathrm{N\,kg^{-1}}\). This is a local gravitational field created by the laboratory sphere, not Earth’s gravitational field.

Step 1: Rearrange Newton’s law to calculate \(G\).

\[ F=G\frac{Mm}{d^2} \]

Rearrange:

\[ G=\frac{Fd^2}{Mm} \]

Step 2: Substitute the experimental data.

\[ G= \frac{(8.0\times10^{-10})(0.050)^2} {(1.50)(0.020)} \]

\[ G= \frac{2.0\times10^{-12}}{0.030} \]

\[ \boxed{G=6.7\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}} \]

The experiment has determined the universal gravitational constant from the attraction between two known masses.

Step 3: Calculate the field strength produced by the large sphere.

Gravitational field strength is force per unit mass.

\[ g_M=\frac{F}{m} \]

Substitute:

\[ g_M=\frac{8.0\times10^{-10}}{0.020} \]

\[ \boxed{g_M=4.0\times10^{-8}\ \mathrm{N\,kg^{-1}}} \]

This is the gravitational field strength produced by the 1.50 kg sphere at the position of the smaller sphere.

Step 4: Explain the difference from Earth’s gravitational field.

The measured field is generated by the nearby laboratory sphere.

Earth also produces a gravitational field at the experimental location, but that field acts predominantly vertically downwards. The horizontal attraction between the lead spheres produces the tiny torque measured by the torsion balance.

The experiment uses the measured attraction between known masses to determine \(G\), rather than simply measuring Earth’s local gravitational field strength.

This is precisely why \(G\) and \(g\) must not be treated as interchangeable quantities.

Question 5

Two identical, spherical planets each have:

• Mass \(M=5.0\times10^{24}\ \mathrm{kg}\).
• Radius \(R=2.0\times10^6\ \mathrm m\).

Their centres are separated by \(1.0\times10^7\ \mathrm m\).

A probe with mass \(1.0\times10^3\ \mathrm{kg}\) is positioned exactly halfway between the two planets.

Assume both planets are spherically symmetric, the probe is outside both planets, and gravitational potential is zero at infinity.

Answer the following:

a) Calculate the net gravitational field strength at the probe.

b) Calculate the total gravitational potential at the probe.

c) Calculate the probe’s gravitational potential energy.

d) A student claims that because the net gravitational field is zero, the gravitational potential and potential energy must also be zero. Evaluate this claim.

e) Is the midpoint a stable equilibrium? Consider what happens if the probe is displaced slightly towards one planet along the line joining the centres, and then if it is displaced slightly perpendicular to that line.

Solution 5

The net gravitational field strength at the midpoint is zero, but the gravitational potential is \(-1.334\times10^8\ \mathrm{J\,kg^{-1}}\), and the probe’s gravitational potential energy is \(-1.334\times10^{11}\ \mathrm J\).

The midpoint is an unstable equilibrium overall: a small displacement towards either planet along the line joining their centres produces an acceleration further towards that planet. A perpendicular displacement produces a restoring acceleration.

This is an important example of why zero gravitational field does not mean zero gravitational potential.

Step 1: Calculate the distance from each planet’s centre to the probe.

The planets are separated by:

\[ d=1.0\times10^7\ \mathrm m \]

The probe is exactly halfway between them.

Therefore:

\[ r=\frac{d}{2} \]

\[ \boxed{r=5.0\times10^6\ \mathrm m} \]

The probe is outside both planets because this distance is greater than their radius of \(2.0\times10^6\ \mathrm m\).

Step 2: Calculate the gravitational field strength produced by each planet.

For either planet:

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

Substitute:

\[ g= \frac{(6.67\times10^{-11})(5.0\times10^{24})} {(5.0\times10^6)^2} \]

\[ g=\frac{3.335\times10^{14}} {2.5\times10^{13}} \]

\[ \boxed{g=13.34\ \mathrm{N\,kg^{-1}}} \]

Each planet produces a field of equal magnitude at the midpoint.

However, their directions are opposite.

The field produced by the left planet points towards the left, while the field produced by the right planet points towards the right.

Step 3: Calculate the net gravitational field.

Gravitational field strength is a vector quantity, so we must account for direction.

At the midpoint, the two fields have equal magnitude and opposite directions:

\[ g_{\text{net}}=13.34-13.34 \]

\[ \boxed{g_{\text{net}}=0\ \mathrm{N\,kg^{-1}}} \]

The gravitational forces also cancel, meaning the probe experiences zero net gravitational force at this exact location.

Step 4: Calculate the gravitational potential.

Here is where the calculation changes.

Gravitational potential is a scalar quantity, not a vector.

The potential produced by one planet is:

\[ V_1=-\frac{GM}{r} \]

Substitute:

\[ V_1= -\frac{(6.67\times10^{-11})(5.0\times10^{24})} {5.0\times10^6} \]

\[ V_1=-6.67\times10^7\ \mathrm{J\,kg^{-1}} \]

The second planet produces exactly the same potential.

Unlike gravitational fields, the two potentials do not cancel because they do not have opposite directions.

Instead, they add:

\[ V_{\text{total}}=V_1+V_2 \]

\[ V_{\text{total}}=-6.67\times10^7-6.67\times10^7 \]

\[ \boxed{V_{\text{total}}=-1.334\times10^8\ \mathrm{J\,kg^{-1}}} \]

Step 5: Calculate the probe’s gravitational potential energy.

Use:

\[ U=mV \]

Substitute:

\[ U=(1.0\times10^3)(-1.334\times10^8) \]

\[ \boxed{U=-1.334\times10^{11}\ \mathrm J} \]

The probe has negative gravitational potential energy relative to infinity, despite experiencing zero net gravitational force at the midpoint.

Step 6: Evaluate the student’s claim.

The claim is incorrect.

The student has confused two different mathematical properties.

• Gravitational field strength is a vector, so equal and opposite contributions can cancel.
• Gravitational potential is a scalar, so the negative contributions from both planets add.

A zero net field does not require zero gravitational potential.

The important distinction is that field strength describes the local gravitational force per unit mass, while potential describes energy per unit mass relative to a chosen reference.

Step 7: Determine whether the midpoint is a stable equilibrium.

This is the more subtle part of the problem.

First, consider a small displacement towards the left planet.

The probe is now slightly closer to the left planet and slightly farther from the right planet.

Since gravitational field strength follows an inverse-square relationship, the closer planet produces the stronger attraction.

The net gravitational force therefore points towards the left planet.

The probe accelerates further away from the midpoint.

The same reasoning applies if it is displaced towards the right planet.

Therefore, the equilibrium is unstable along the line joining the planets.

Now consider a small displacement perpendicular to the line joining the planets.

The probe is now equally farther away from both planets.

The horizontal components of the two gravitational forces still cancel, but their perpendicular components both point back towards the line joining the planetary centres.

The resulting force is therefore a restoring force.

The probe accelerates back towards the midpoint in this perpendicular direction.

The equilibrium is consequently a saddle point: restoring in perpendicular directions but unstable along the line joining the planets.

If the probe were placed exactly at the midpoint and released from perfect rest, it would remain there in the ideal model. However, an arbitrarily small displacement along the line joining the planets would cause it to move further away from that equilibrium.

This final example brings together several ideas:

• \(G\) determines the strength of each gravitational interaction.
• \(g\) is a vector field, so contributions can cancel.
• \(V\) is a scalar potential, so contributions add.
• \(U\) depends on both the potential and the probe’s mass.
• Zero gravitational force does not automatically mean zero potential or stable equilibrium.

07Where to go next

The most useful habit to develop is identifying what a question actually asks you to calculate before choosing an equation.

If it asks for gravitational field strength, use \(g=GM/r^2\). If it asks for force, include the test object’s mass. If it asks for potential, use \(V=-GM/r\). If it asks for energy or work, consider \(U=mV\) and the change in potential energy.

Always convert altitude into distance from the centre first.

The next concept is gravitational superposition: calculating the net field when several masses act on the same object. Remember that gravitational fields add as vectors, while gravitational potentials add as scalars.

For further practice, see How to Calculate the Net Gravitational Field Between Two Masses.