How to Calculate the Net Gravitational Field Between Two Masses

Two planets can pull on the same satellite in opposite directions. If one planet produces a gravitational field strength of \(8.0\ \mathrm{N\,kg^{-1}}\) and the other produces \(2.0\ \mathrm{N\,kg^{-1}}\) at the satellite's position, what is the net gravitational field strength?

You might be tempted to add them and get \(10.0\ \mathrm{N\,kg^{-1}}\). That would be correct if the fields pointed in the same direction. Between two masses, however, their gravitational fields point towards opposite masses, so you must subtract their magnitudes.

**The key method is to calculate each gravitational field separately, assign its direction, and then add the fields as vectors.** The direction of the net field tells you which mass has the stronger gravitational influence at that position.

## 1. The method: calculate both fields, then combine them

For a point mass, gravitational field strength is calculated using

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

where:

- \(g\) is gravitational field strength, measured in \(\mathrm{N\,kg^{-1}}\).
- \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
- \(M\) is the mass producing the gravitational field, measured in kilograms.
- \(r\) is the distance from the centre of that mass to the point where the field is being calculated, measured in metres.

For two masses, calculate the field produced by each mass at the same point:

\[
g_1=\frac{GM_1}{r_1^2}
\qquad
g_2=\frac{GM_2}{r_2^2}
\]

Then combine the results using vector addition.

For a point directly between two masses, the fields point in opposite directions. Therefore, the net field strength is the difference between their magnitudes:

\[
g_{\mathrm{net}}=|g_1-g_2|
\]

The direction is towards the mass producing the stronger field.

This subtraction rule only applies when the point is between the masses and the fields are directly opposite. In other arrangements, you must use vector addition with the appropriate directions.

[IMAGE: Two spherical masses of different sizes separated horizontally, with a small test mass positioned between them. Draw gravitational field arrows at the test mass pointing left towards the left-hand mass and right towards the right-hand mass. Label the two individual field contributions g1 and g2, and show a separate net-field arrow pointing towards the stronger contribution. Include distances measured from each sphere's centre to the test mass.]

## 2. Get the distances right: radius is not altitude

This is one of the easiest ways to lose marks in gravitational field calculations.

In the equation

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

\(r\) is the distance from the centre of the mass producing the field to the point being investigated.

It is not necessarily the radius of the planet, and it is not necessarily the altitude above its surface.

Use these symbols consistently:

- \(R\): physical radius of a spherical planet, measured from its centre to its surface.
- \(h\): altitude above the planet's surface.
- \(r\): radial distance from the planet's centre to the point where the field is calculated.

If a satellite is at altitude \(h\) above a planet of radius \(R\), then

\[
r=R+h
\]

[IMAGE: A spherical planet shown in cross-section with its centre, surface, and a satellite above the surface. Draw and label the planet radius R from centre to surface, altitude h from surface to satellite, and radial distance r from centre to satellite. Clearly show that r = R + h.]

For example, suppose a planet has a radius of \(6.0\times10^6\ \mathrm{m}\), and a satellite is \(4.0\times10^5\ \mathrm{m}\) above its surface.

The distance used in the gravitational field equation is

\[
r=6.0\times10^6+4.0\times10^5
\]

\[
r=6.4\times10^6\ \mathrm{m}
\]

Using the altitude alone would give you the wrong gravitational field by a very large factor.

### What if there are two spherical masses?

Measure the distance from each centre separately.

Suppose two spherical planets have centre-to-centre separation \(D\), and a point P lies directly between them.

Then

\[
r_A+r_B=D
\]

where \(r_A\) and \(r_B\) are the distances from the respective centres to P.

If a question gives the gap between the planets' surfaces instead, you must also account for both planetary radii when finding their centre-to-centre separation.

For the calculations in this article, assume both source masses are spherical and the field point is outside each mass. By spherical symmetry, each planet's external gravitational field can be calculated as though its entire mass were concentrated at its centre.

## 3. Choose a direction before calculating

Let's make the subtraction method reliable.

Imagine two masses, A on the left and B on the right, with point P between them.

Choose right as the positive direction.

- Mass A attracts a test mass towards the left, so its field contribution is negative.
- Mass B attracts a test mass towards the right, so its field contribution is positive.

The net gravitational field is therefore

\[
g_{\mathrm{net}}=g_B-g_A
\]

Here, \(g_A\) and \(g_B\) represent the positive magnitudes of the individual field contributions.

The sign of the answer tells you the direction:

| Result | Meaning |
|---|---|
| \(g_{\mathrm{net}}>0\) | Net field points towards B |
| \(g_{\mathrm{net}}<0\) | Net field points towards A |
| \(g_{\mathrm{net}}=0\) | The two fields cancel |

If you only need the magnitude of the field, you can write

\[
g_{\mathrm{net}}=|g_B-g_A|
\]

But always state the direction separately.

A useful mental picture is a tug-of-war. Two people pull a rope in opposite directions, and the stronger pull determines the direction of the resultant force.

The analogy has a limit: gravitational fields exist even when there is no rope or object present. The field describes the gravitational influence at a location, not a force that requires a test object to exist.

### Three arrangements you need to recognise

The subtraction rule is not universal. The direction of each field determines how the contributions combine.

| Position of P | Direction of individual fields | What to do |
|---|---|---|
| Between two masses, on their joining line | Opposite | Subtract |
| Outside both masses, on their joining line | Same | Add |
| Off the joining line | At an angle | Resolve into components and add vectors |

[IMAGE: Three side-by-side physical arrangements of two spherical masses and a test point. First, P lies between the masses on their centre line, with the two gravitational field contributions pointing oppositely. Second, P lies outside both masses, with both contributions pointing in the same direction. Third, P lies above the midpoint, with two angled field contributions directed towards the separate masses. Label only the positions and individual field directions, not the resultant answers.]

The third arrangement is particularly important.

If the two field contributions are at an angle, you cannot simply add or subtract their magnitudes. You must resolve them into horizontal and vertical components, then combine the components separately.

We will practise this in the questions.

## 4. Gravitational field strength is not gravitational force or potential

These quantities are related, but they describe different things.

| Quantity | Symbol | Meaning | Unit |
|---|---|---|---|
| Gravitational field strength | \(\vec g\) | Gravitational force per unit mass at a location | \(\mathrm{N\,kg^{-1}}\) |
| Gravitational force | \(\vec F_g\) | Actual force acting on a particular mass | \(\mathrm{N}\) |
| Gravitational potential | \(\phi\) | Gravitational potential energy per unit mass | \(\mathrm{J\,kg^{-1}}\) |
| Gravitational potential energy | \(U\) | Energy associated with the gravitational interaction | \(\mathrm{J}\) |

### Field strength and force

Gravitational field strength describes the gravitational influence at a location.

The gravitational force depends on the mass placed in that field.

Their relationship is

\[
\vec F_g=m_t\vec g_{\mathrm{net}}
\]

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

For example, a net field strength of \(0.20\ \mathrm{N\,kg^{-1}}\) produces a gravitational force of \(2.0\ \mathrm{N}\) on a \(10\ \mathrm{kg}\) object.

A \(20\ \mathrm{kg}\) object at exactly the same location experiences \(4.0\ \mathrm{N}\).

The field strength has not changed. Only the force has changed.

### Field strength and gravitational potential

This distinction is even more important when two gravitational fields cancel.

Gravitational field strength is a vector. It has both magnitude and direction.

Gravitational potential is a scalar. It has magnitude and sign, but no direction.

For a point mass, gravitational potential, with zero defined at infinity, is

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

For two masses, the total potential is

\[
\phi_{\mathrm{net}}
=
-\frac{GM_A}{r_A}
-\frac{GM_B}{r_B}
\]

Notice that the two contributions are added, not subtracted.

Both are negative because both masses attract the test object.

This creates an interesting result: **the net gravitational field can be zero while the gravitational potential is still negative.**

At a point where the two field vectors cancel, the two potential contributions do not cancel. They simply add.

If you want to explore the energy implications in more detail, see [Gravitational Potential Energy in a Radial Field for HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-gravitational-potential-energy-in-a-radial-field/).

Gravitational potential energy is related to potential by

\[
U=m_t\phi
\]

It depends on the test mass, whereas gravitational potential does not.

This distinction will become important when we reach Question 4.

## 5. Worked calculations

### Worked example: Calculate the net field between two planets

Two spherical planets, A and B, have masses of \(4.0\times10^{24}\ \mathrm{kg}\) and \(1.0\times10^{24}\ \mathrm{kg}\), respectively.

Their centres are separated by \(1.00\times10^8\ \mathrm{m}\).

A point P lies directly between them, \(4.00\times10^7\ \mathrm{m}\) from the centre of A and \(6.00\times10^7\ \mathrm{m}\) from the centre of B.

Calculate:

a. The net gravitational field strength at P, including its direction.

b. The gravitational force on a \(500\ \mathrm{kg}\) probe placed at P.

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

**Step 1**: Calculate the field produced by each planet separately.

Start with

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

For planet A,

\[
\begin{aligned}
g_A&=\frac{GM_A}{r_A^2}\\
&=\frac{(6.67\times10^{-11})(4.0\times10^{24})}
{(4.00\times10^7)^2}\\
&=0.16675\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

For planet B,

\[
\begin{aligned}
g_B&=\frac{GM_B}{r_B^2}\\
&=\frac{(6.67\times10^{-11})(1.0\times10^{24})}
{(6.00\times10^7)^2}\\
&=0.01853\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

**Step 2**: Determine the directions.

Planet A is to the left of P, so its field points left.

Planet B is to the right of P, so its field points right.

The two contributions oppose each other.

**Step 3**: Calculate the net field.

Choose right as positive:

\[
\begin{aligned}
g_{\mathrm{net}}
&=g_B-g_A\\
&=0.01853-0.16675\\
&=-0.14822\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

Therefore,

\[
\boxed{g_{\mathrm{net}}=0.148\ \mathrm{N\,kg^{-1}}}
\]

towards planet A.

The negative sign indicates leftward direction. It does not mean that gravitational field strength is physically negative.

**Step 4**: Calculate the force on the probe.

Now use

\[
F_g=m_tg_{\mathrm{net}}
\]

Taking the magnitude,

\[
\begin{aligned}
F_g&=500(0.14822)\\
&=74.11\ \mathrm{N}
\end{aligned}
\]

The probe experiences a gravitational force of

\[
\boxed{74.1\ \mathrm{N}\text{ towards A}}
\]

Notice the order of the calculation. We found the field first, then used the probe's mass to calculate its force.

Changing the probe's mass would change the force, but it would not change the net gravitational field.

### Worked example: Find the point where the net field is zero

Two spherical planets have masses:

\[
M_A=9.0\times10^{24}\ \mathrm{kg}
\]

\[
M_B=1.0\times10^{24}\ \mathrm{kg}
\]

Their centres are separated by \(1.20\times10^8\ \mathrm{m}\).

Determine the position between the planets where the net gravitational field strength is zero.

**Step 1**: Recognise the condition for zero net field.

At the required point, the two field contributions must have equal magnitudes and opposite directions.

Therefore,

\[
g_A=g_B
\]

Using the gravitational field equation,

\[
\frac{GM_A}{r_A^2}
=
\frac{GM_B}{r_B^2}
\]

The universal gravitational constant cancels:

\[
\frac{M_A}{r_A^2}
=
\frac{M_B}{r_B^2}
\]

**Step 2**: Rearrange to find the distance ratio.

Taking the positive square root,

\[
\frac{r_A}{r_B}
=
\sqrt{\frac{M_A}{M_B}}
\]

Substitute the mass ratio:

\[
\frac{r_A}{r_B}
=
\sqrt{\frac{9.0\times10^{24}}{1.0\times10^{24}}}
\]

\[
\frac{r_A}{r_B}=3
\]

Therefore,

\[
r_A=3r_B
\]

**Step 3**: Use the total separation.

Because the point lies between the planets,

\[
r_A+r_B=1.20\times10^8
\]

Substitute \(r_A=3r_B\):

\[
\begin{aligned}
3r_B+r_B&=1.20\times10^8\\
4r_B&=1.20\times10^8\\
r_B&=3.00\times10^7\ \mathrm{m}
\end{aligned}
\]

Therefore,

\[
r_A=9.00\times10^7\ \mathrm{m}
\]

[IMAGE: Two spherical planets A and B separated by 1.20 x 10^8 m between their centres. Planet A is nine times as massive as planet B. Show an unknown point P between them, with a distance x from A and the remaining distance from P to B labelled 1.20 x 10^8 - x. Do not show the calculated position or field arrows.]

**Step 4**: Interpret the result.

The net field is zero at a point:

- \(9.00\times10^7\ \mathrm{m}\) from A's centre.
- \(3.00\times10^7\ \mathrm{m}\) from B's centre.

The point is closer to the smaller planet.

This makes physical sense. Planet A is nine times as massive, so its gravitational influence is stronger at equal distances. The point must be farther from A to reduce its field contribution enough to balance B's.

We can check the result numerically:

\[
\begin{aligned}
g_A&=\frac{(6.67\times10^{-11})(9.0\times10^{24})}
{(9.00\times10^7)^2}\\
&=0.0741\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

and

\[
\begin{aligned}
g_B&=\frac{(6.67\times10^{-11})(1.0\times10^{24})}
{(3.00\times10^7)^2}\\
&=0.0741\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

The magnitudes are equal, and the directions are opposite.

Therefore,

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

The important shortcut is that you can find the zero-field position without calculating \(G\) or the individual field strengths. The ratio method is much quicker.

## Questions and solutions

Use

\[
G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}
\]

unless stated otherwise.

### Question 1

Two spherical bodies, A and B, have masses of \(2.0\times10^{24}\ \mathrm{kg}\) and \(5.0\times10^{24}\ \mathrm{kg}\), respectively.

Their centres are separated by \(1.00\times10^8\ \mathrm{m}\).

A \(250\ \mathrm{kg}\) probe is positioned between them, \(4.00\times10^7\ \mathrm{m}\) from A's centre and \(6.00\times10^7\ \mathrm{m}\) from B's centre.

Calculate:

a. The net gravitational field strength at the probe.

b. The magnitude and direction of the gravitational force on the probe.

[IMAGE: Two spherical bodies A and B arranged horizontally, with centre separation 1.00 x 10^8 m. Label their masses as 2.0 x 10^24 kg and 5.0 x 10^24 kg. Place a small probe P between them, 4.00 x 10^7 m from A and 6.00 x 10^7 m from B. Show the centre positions and supplied distances, but no gravitational field or force arrows.]

### Solution 1

The net gravitational field is \(9.26\times10^{-3}\ \mathrm{N\,kg^{-1}}\) towards B, and the probe experiences a gravitational force of \(2.32\ \mathrm{N}\) towards B.

First, calculate the individual field contributions.

For A,

\[
\begin{aligned}
g_A&=\frac{GM_A}{r_A^2}\\
&=\frac{(6.67\times10^{-11})(2.0\times10^{24})}
{(4.00\times10^7)^2}\\
&=0.083375\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

For B,

\[
\begin{aligned}
g_B&=\frac{GM_B}{r_B^2}\\
&=\frac{(6.67\times10^{-11})(5.0\times10^{24})}
{(6.00\times10^7)^2}\\
&=0.092639\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

The fields point in opposite directions because the probe is between the two bodies.

Choose right as positive:

\[
\begin{aligned}
g_{\mathrm{net}}
&=g_B-g_A\\
&=0.092639-0.083375\\
&=0.009264\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

Therefore,

\[
g_{\mathrm{net}}
=9.26\times10^{-3}\ \mathrm{N\,kg^{-1}}
\]

towards B.

Now calculate the force:

\[
F_g=m_tg_{\mathrm{net}}
\]

\[
\begin{aligned}
F_g&=250(9.264\times10^{-3})\\
&=2.316\ \mathrm{N}
\end{aligned}
\]

Therefore,

\[
F_g=2.32\ \mathrm{N}
\]

towards B.

The important detail is that B is more massive, but the probe is also farther from B. You must compare the actual field contributions, not just the masses.

### Question 2

Two identical spherical bodies each have a mass of \(6.0\times10^{24}\ \mathrm{kg}\).

Their centres are separated by \(8.0\times10^7\ \mathrm{m}\).

Point P is located \(3.0\times10^7\ \mathrm{m}\) directly above the midpoint between their centres.

Calculate the magnitude and direction of the net gravitational field strength at P.

Explain why the two horizontal field components cancel while the vertical components reinforce each other.

[IMAGE: Two identical spherical masses positioned symmetrically on a horizontal axis, with centre separation 8.0 x 10^7 m. Place P directly above their midpoint by 3.0 x 10^7 m. Show the right triangle from either mass centre to P, with horizontal side 4.0 x 10^7 m and vertical side 3.0 x 10^7 m. Label the diagonal distance as r, without calculating it. Do not draw gravitational field vectors.]

### Solution 2

The net gravitational field strength is \(0.192\ \mathrm{N\,kg^{-1}}\), vertically downwards towards the midpoint between the two masses.

This is not a simple subtraction problem. The individual field vectors are at an angle.

First, use the geometry to calculate the distance from either mass to P.

Half the centre separation is

\[
\frac{8.0\times10^7}{2}
=4.0\times10^7\ \mathrm{m}
\]

Using Pythagoras,

\[
\begin{aligned}
r^2
&=(4.0\times10^7)^2+(3.0\times10^7)^2\\
&=16\times10^{14}+9\times10^{14}\\
&=25\times10^{14}
\end{aligned}
\]

Therefore,

\[
r=5.0\times10^7\ \mathrm{m}
\]

**Step 1**: Calculate the magnitude of each field contribution.

Because the masses and distances are identical,

\[
g_1=g_2=\frac{GM}{r^2}
\]

\[
\begin{aligned}
g_1=g_2
&=\frac{(6.67\times10^{-11})(6.0\times10^{24})}
{(5.0\times10^7)^2}\\
&=0.16008\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

**Step 2**: Resolve each field into components.

The triangle has side lengths in the ratio \(3:4:5\).

The vertical component of each field is therefore

\[
g_y=\frac35g
\]

The horizontal component is

\[
g_x=\frac45g
\]

**Step 3**: Combine the horizontal components.

The left mass produces a field with a horizontal component towards the left.

The right mass produces an equal horizontal component towards the right.

Therefore,

\[
g_{x,\mathrm{net}}=0
\]

**Step 4**: Combine the vertical components.

Both vertical components point downwards, towards the line joining the masses.

Therefore,

\[
\begin{aligned}
g_{y,\mathrm{net}}
&=2\left(\frac35g\right)\\
&=2\left(\frac35\times0.16008\right)\\
&=0.192096\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

Hence,

\[
\boxed{g_{\mathrm{net}}=0.192\ \mathrm{N\,kg^{-1}}}
\]

vertically downwards.

The tempting mistake is to say that the fields cancel because the masses are identical. They only cancel horizontally. Both masses attract towards themselves, so their vertical contributions point in the same direction.

### Question 3

Two spherical masses A and B have masses of \(2.0\times10^{24}\ \mathrm{kg}\) and \(8.0\times10^{24}\ \mathrm{kg}\), respectively.

Their centres are separated by \(6.0\times10^7\ \mathrm{m}\).

A point P lies \(3.0\times10^7\ \mathrm{m}\) to the left of A.

Calculate the net gravitational field strength at P, including its direction.

Explain why adding the two field magnitudes is appropriate in this arrangement.

[IMAGE: Two spherical masses A and B on a horizontal line, with A to the left of B and centre separation 6.0 x 10^7 m. Place P to the left of A, 3.0 x 10^7 m from A. Show the centre-to-point distances from P to A and P to B as spatial distances, without drawing field arrows.]

### Solution 3

The net gravitational field strength is \(0.214\ \mathrm{N\,kg^{-1}}\) towards the right.

The key difference from Question 1 is the position of P.

Both masses are to the right of P. Since gravitational fields point towards their source masses, both field contributions point right.

Therefore, the magnitudes add.

First, calculate the distance from P to each mass.

For A,

\[
r_A=3.0\times10^7\ \mathrm{m}
\]

For B,

\[
\begin{aligned}
r_B&=3.0\times10^7+6.0\times10^7\\
&=9.0\times10^7\ \mathrm{m}
\end{aligned}
\]

Calculate the individual fields.

For A,

\[
\begin{aligned}
g_A&=\frac{GM_A}{r_A^2}\\
&=\frac{(6.67\times10^{-11})(2.0\times10^{24})}
{(3.0\times10^7)^2}\\
&=0.14822\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

For B,

\[
\begin{aligned}
g_B&=\frac{GM_B}{r_B^2}\\
&=\frac{(6.67\times10^{-11})(8.0\times10^{24})}
{(9.0\times10^7)^2}\\
&=0.06588\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

Both contributions point right, so

\[
\begin{aligned}
g_{\mathrm{net}}
&=g_A+g_B\\
&=0.14822+0.06588\\
&=0.21410\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

Therefore,

\[
\boxed{g_{\mathrm{net}}=0.214\ \mathrm{N\,kg^{-1}}}
\]

to the right.

A common misconception is that two gravitational fields should always be subtracted when calculating a net field. They should not. Vector addition depends on direction, not simply on the number of masses involved.

### Question 4

Two spherical planets A and B have masses of

\[
M_A=4.0\times10^{24}\ \mathrm{kg}
\]

and

\[
M_B=1.0\times10^{24}\ \mathrm{kg}.
\]

Their centres are separated by \(1.00\times10^8\ \mathrm{m}\).

A \(200\ \mathrm{kg}\) probe is placed at the point between them where the net gravitational field strength is zero.

Assume gravitational potential is defined to be zero at infinity.

Calculate:

a. The distance of the probe from each planet's centre.

b. The net gravitational field strength at the probe.

c. The net gravitational potential at the probe.

d. The probe's gravitational potential energy.

Explain why the gravitational potential energy is not zero even though the net gravitational field strength is zero.

[IMAGE: Two spherical planets A and B separated by 1.00 x 10^8 m between their centres. Show a 200 kg probe at an unspecified point P between them. Label the planetary masses and total separation. Do not show the calculated position, field vectors, or potential values.]

### Solution 4

The probe is \(6.67\times10^7\ \mathrm{m}\) from A and \(3.33\times10^7\ \mathrm{m}\) from B. Its net gravitational field strength and gravitational force are zero, but its gravitational potential is approximately \(-6.00\times10^6\ \mathrm{J\,kg^{-1}}\), giving a gravitational potential energy of approximately \(-1.20\times10^9\ \mathrm{J}\).

**Step 1**: Find the zero-field position.

At the zero-field point, the two field magnitudes are equal:

\[
\frac{GM_A}{r_A^2}
=
\frac{GM_B}{r_B^2}
\]

Cancel \(G\) and take the square root:

\[
\frac{r_A}{r_B}
=
\sqrt{\frac{M_A}{M_B}}
\]

Substitute:

\[
\frac{r_A}{r_B}
=
\sqrt{\frac{4.0}{1.0}}
=2
\]

Therefore,

\[
r_A=2r_B
\]

The total centre separation is

\[
r_A+r_B=1.00\times10^8
\]

Substituting,

\[
\begin{aligned}
2r_B+r_B&=1.00\times10^8\\
3r_B&=1.00\times10^8\\
r_B&=3.33\times10^7\ \mathrm{m}
\end{aligned}
\]

Hence,

\[
r_A=6.67\times10^7\ \mathrm{m}
\]

**Step 2**: Confirm that the net field is zero.

Calculate the field produced by A:

\[
\begin{aligned}
g_A&=\frac{(6.67\times10^{-11})(4.0\times10^{24})}
{(6.67\times10^7)^2}\\
&\approx0.0600\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

For B,

\[
\begin{aligned}
g_B&=\frac{(6.67\times10^{-11})(1.0\times10^{24})}
{(3.33\times10^7)^2}\\
&\approx0.0600\ \mathrm{N\,kg^{-1}}
\end{aligned}
\]

The fields have equal magnitudes and opposite directions.

Therefore,

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

The gravitational force on the probe is also zero:

\[
F_g=m_tg_{\mathrm{net}}=0\ \mathrm{N}
\]

**Step 3**: Calculate the gravitational potential.

This is where the calculation changes.

Field strength is a vector, but gravitational potential is a scalar.

The total potential is

\[
\phi_{\mathrm{net}}
=
-\frac{GM_A}{r_A}
-\frac{GM_B}{r_B}
\]

Substitute the values:

\[
\begin{aligned}
\phi_{\mathrm{net}}
=&-\frac{(6.67\times10^{-11})(4.0\times10^{24})}
{6.67\times10^7}\\
&-\frac{(6.67\times10^{-11})(1.0\times10^{24})}
{3.33\times10^7}
\end{aligned}
\]

The individual contributions are approximately

\[
\phi_A=-4.00\times10^6\ \mathrm{J\,kg^{-1}}
\]

and

\[
\phi_B=-2.00\times10^6\ \mathrm{J\,kg^{-1}}
\]

Therefore,

\[
\boxed{
\phi_{\mathrm{net}}
=-6.00\times10^6\ \mathrm{J\,kg^{-1}}
}
\]

**Step 4**: Calculate gravitational potential energy.

Use

\[
U=m_t\phi_{\mathrm{net}}
\]

\[
\begin{aligned}
U&=200(-6.00\times10^6)\\
&=-1.20\times10^9\ \mathrm{J}
\end{aligned}
\]

Therefore,

\[
\boxed{U=-1.20\times10^9\ \mathrm{J}}
\]

The important conceptual distinction is that the field contributions cancel because they point in opposite directions.

The potential contributions do not have directions. Both planets produce negative gravitational potential, so the contributions add.

Zero net field does not mean zero potential or zero potential energy.

### Question 5

Two identical spherical masses, each of mass \(M\), are fixed a distance \(2a\) apart.

A small test object is placed at their midpoint O and released from rest.

At O, the gravitational field contributions from the two masses are equal and opposite.

A student concludes:

"The net gravitational field is zero, so the object is in stable equilibrium. If it moves slightly, gravity will pull it back to the midpoint."

Investigate this claim.

Consider two separate small displacements:

a. The object moves a distance \(x\) along the line joining the two masses, towards the right-hand mass.

b. The object moves a distance \(y\) perpendicular to the line joining the masses.

For each displacement, determine the direction of the net gravitational field.

Use your results to decide whether the midpoint is a stable equilibrium, an unstable equilibrium, or a saddle equilibrium.

Assume the test object is sufficiently small, and both source masses remain fixed.

[IMAGE: Two identical spherical masses placed symmetrically on a horizontal axis, separated by 2a. Mark their midpoint O. Show a small test object at O, a possible displacement x along the horizontal axis towards the right-hand mass, and a separate possible displacement y vertically above O. Include the equal distances a from O to each mass. Do not draw gravitational field arrows or indicate whether either displacement is restoring.]

### Solution 5

The midpoint is a saddle equilibrium, not a stable equilibrium.

The net field is zero at the exact midpoint, but a small displacement produces different behaviour depending on its direction.

Along the line joining the masses, the object is accelerated further away from the midpoint.

Perpendicular to that line, the object is accelerated back towards the line joining the masses.

This means the equilibrium is stable in one direction but unstable in another.

**Step 1**: Consider a displacement along the line joining the masses.

Place the left mass at \(x=-a\) and the right mass at \(x=+a\).

Initially, the object is at \(x=0\).

Now move the object a small distance \(x\) towards the right-hand mass, where \(0<x<a\).

Its distance from the right-hand mass becomes

\[
r_R=a-x
\]

Its distance from the left-hand mass becomes

\[
r_L=a+x
\]

The field from the right-hand mass points right, towards that mass:

\[
g_R=\frac{GM}{(a-x)^2}
\]

The field from the left-hand mass points left:

\[
g_L=\frac{GM}{(a+x)^2}
\]

The net field towards the right is

\[
g_{\mathrm{net},x}
=
\frac{GM}{(a-x)^2}
-
\frac{GM}{(a+x)^2}
\]

Since

\[
a-x<a+x,
\]

the first field contribution is stronger.

Therefore,

\[
g_{\mathrm{net},x}>0
\]

The net field points towards the right-hand mass, in the same direction as the displacement.

The object accelerates further away from the midpoint.

So the midpoint is unstable along the line joining the masses.

**Step 2**: Consider a displacement perpendicular to the line joining the masses.

Now return to the midpoint and move the object a small distance \(y\) upwards.

Both masses are the same distance from the object:

\[
r=\sqrt{a^2+y^2}
\]

The two gravitational field magnitudes are therefore equal.

Their horizontal components point in opposite directions and cancel.

Their vertical components both point downwards, towards the line joining the masses.

The vertical component from either mass is

\[
g_y^{(1)}
=
\frac{GM}{a^2+y^2}
\left(
\frac{y}{\sqrt{a^2+y^2}}
\right)
\]

Combining the two vertical contributions gives

\[
g_{\mathrm{net},y}
=
-\frac{2GMy}{(a^2+y^2)^{3/2}}
\]

The negative sign indicates a downward field when \(y\) is positive.

Therefore, the net gravitational field points back towards the line joining the masses.

The object experiences a restoring acceleration in this direction.

So the equilibrium is stable against small perpendicular displacements.

**Step 3**: Combine the two results.

| Direction of displacement | Direction of net field | Behaviour |
|---|---|---|
| Along the line joining the masses | Further away from midpoint | Unstable |
| Perpendicular to the line joining the masses | Back towards the line | Restoring |

At the exact midpoint, an object released from rest remains there in the ideal model because the net gravitational force is zero.

However, a small displacement along the line joining the masses causes the object to accelerate further away.

The midpoint is therefore a saddle equilibrium.

The student's mistake was assuming that zero net gravitational field automatically means stable equilibrium.

Zero field tells you the instantaneous net force is zero. It does not tell you what happens after a small displacement.

That is a much more demanding question, and it is an important distinction in understanding gravitational fields.

## 6. What to do next

When you encounter two gravitational masses in an HSC Physics question, keep three decisions separate:

1. Calculate the individual field contributions using centre-to-point distances.
2. Combine the contributions as vectors, using their actual directions.
3. If a force or energy is requested, use the appropriate relationship rather than treating field strength, potential, and potential energy as interchangeable.

Once you can calculate the net field at any position, you can use

\[
\vec F_g=m_t\vec g_{\mathrm{net}}
\]

to predict the acceleration of an object, provided gravity is the only significant force.

The next useful step is connecting gravitational fields to gravitational potential and potential energy. That allows you to move beyond finding the force at one position and start analysing how objects move between different positions in a radial gravitational field.