How to Calculate Gravity at Twice a Planet’s Radius

A planet's gravitational field is much weaker at twice its radius than it is at the surface. But there's a catch: **twice the planet's radius is not necessarily the same as an altitude of twice the planet's radius.**

If an object is a distance of \(2R\) from the planet's centre, its gravitational field strength is one-quarter of the surface value:

\[
\boxed{g_{2R}=\frac{g_{\text{surface}}}{4}}
\]

If an object is instead at an altitude of \(2R\) above the surface, its distance from the centre is \(3R\), so the field strength is one-ninth of the surface value.

This distinction is easy to miss in HSC Physics questions. Let's work through why it happens, how to calculate the answer, and how to recognise which distance a question actually means.

## 1. Start with the distance from the centre

Imagine a planet with radius \(R\). Its radius is the distance from its centre to its surface.

An object sitting on the surface is therefore already a distance \(R\) from the centre. It isn't at a distance of zero.

Now move the object away from the planet. The gravitational field becomes weaker as the distance from the centre increases.

[IMAGE: Scientific illustration of a spherical planet with its centre marked O, radius R extending to the surface, and an object positioned along a radial line at distance 2R from the centre. Show the surface-to-object altitude as R, with clear dimension arrows distinguishing radius, radial distance and altitude.]

For the object at \(2R\):

- Distance from the planet's centre: \(2R\).
- Distance from the surface: \(R\).
- Gravitational field strength: one-quarter of the surface value.

The important point is that gravity depends on the object's distance from the planet's centre, not its altitude above the surface.

## 2. Why does gravity become one-quarter as strong?

Think about the light spreading out from a lamp. As you move further away, the same light spreads across a larger area, so the intensity decreases.

Gravity follows an inverse-square relationship. The analogy helps picture the spreading effect, although gravitational fields are not light and do not depend on a physical beam travelling outwards.

For a spherical planet, the gravitational field strength at a distance \(r\) from its centre is:

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

where:

- \(g\) is gravitational field strength, measured in newtons per kilogram (\(\mathrm{N\,kg^{-1}}\)).
- \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
- \(M\) is the mass of the planet, measured in kilograms.
- \(r\) is the distance from the planet's centre, measured in metres.

Notice the square on the distance. If the distance doubles, gravity doesn't halve. It becomes:

\[
g_{\text{new}}=\frac{g_{\text{old}}}{2^2}
=\frac{g_{\text{old}}}{4}
\]

That's the entire reason the answer is one-quarter.

03How was the gravitational constant measured?

There’s an interesting problem hiding in Newton’s equation:

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

The universal gravitational constant \(G\) is extremely small. The gravitational attraction between everyday objects is so weak that it’s difficult to measure directly.

So how did physicists establish the value needed to make gravitational calculations?

One important experiment was Henry Cavendish’s torsion balance experiment in 1798.

The original measurement problem

An earlier approach involved measuring how much a mountain pulled a nearby plumb bob sideways.

A plumb bob is simply a small weight hanging from a string. Normally, it hangs vertically because of Earth’s gravitational field. But a nearby mountain also attracts the bob, producing a tiny sideways deflection.

The prediction was straightforward: if the mountain attracts the bob, the string should tilt slightly towards the mountain.

The difficulty was estimating the mountain’s mass accurately. Its shape and internal density were not known well enough, so the calculated gravitational attraction depended on uncertain assumptions.

Cavendish wanted a more controlled measurement.

Cavendish’s torsion balance

Cavendish used an apparatus designed originally by John Michell.

Picture a horizontal rod suspended from its centre by a thin wire. A small lead sphere hangs from each end of the rod.

Now place two much larger lead spheres nearby, one beside each small sphere. The larger spheres attract the smaller ones gravitationally.

The attractions create a turning effect on the suspended rod. As the rod turns, it twists the suspension wire. The wire resists this twisting, eventually balancing the gravitational turning effect.

The apparatus worked in a sequence:

1. The large lead spheres were positioned close to the smaller spheres.
2. Their gravitational attraction caused the suspended rod to twist.
3. A telescope and scale allowed the tiny angular displacement to be observed without disturbing the apparatus.
4. The large spheres were moved into reversed positions. The equilibrium position of the rod shifted in the opposite direction.
5. The measured twisting and oscillations allowed the gravitational force to be calculated.

The experiment was extremely delicate. Air currents and temperature differences could disturb the rod, so the apparatus was enclosed and observed remotely.

What did Cavendish actually discover?

Cavendish published his results in 1798 in a paper titled Experiments to Determine the Density of the Earth.

He reported Earth’s mean density as approximately 5.48 times the density of water.

Historically, his stated objective was to determine Earth’s mean density. In modern physics, the experiment is understood as an experimental determination of the universal gravitational constant \(G\).

You can read the historical account from the American Physical Society or examine the original paper preserved by the Royal Society.

It’s important to separate two things here:

• Newton’s inverse-square law describes how gravitational attraction changes with separation.
• Cavendish’s experiment provided a way to measure the strength of that attraction between known masses.

Cavendish did not physically measure gravity at twice Earth’s radius. The calculation at \(2R\) follows from applying Newton’s gravitational model.

Worked scale estimate: How weak is gravitational attraction?

Suppose two laboratory objects have masses of \(1.0\ \mathrm{kg}\) and \(100\ \mathrm{kg}\), separated by a centre-to-centre distance of \(0.20\ \mathrm{m}\).

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

Using Newton’s law:

\[ F=\frac{Gm_1m_2}{r^2} \]

Substitute:

\[ F= \frac{(6.67\times10^{-11})(1.0)(100)} {(0.20)^2} \]

\[ \boxed{F=1.67\times10^{-7}\ \mathrm{N}} \]

That’s an incredibly small force.

For comparison, a \(1.0\ \mathrm{kg}\) object near Earth’s surface experiences a gravitational force of approximately \(9.8\ \mathrm{N}\).

The laboratory attraction is about 59 million times smaller.

This explains why the torsion balance needed such a sensitive design.

04Calculate gravity at twice the planet’s radius

Now we can answer the original question mathematically.

Let:

• \(M\) be the planet’s mass in kilograms.
• \(R\) be the planet’s radius in metres.
• \(G\) be the universal gravitational constant.
• \(g_s\) be the gravitational field strength at the planet’s surface.

At the surface, the distance from the centre is \(R\).

Therefore:

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

At twice the planet’s radius, the distance from the centre is \(2R\).

Substitute this into the gravitational field equation:

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

Expand the square:

\[ g_{2R}=\frac{GM}{4R^2} \]

But we already know that \(GM/R^2=g_s\).

Therefore:

\[ \boxed{g_{2R}=\frac{g_s}{4}} \]

The gravitational field strength is one-quarter of its surface value.

But what if the question means an altitude of twice the radius?

This is where students often lose marks.

Suppose a question states:

An object is at an altitude of twice the planet’s radius above its surface.

The altitude is \(h=2R\), not the distance from the centre.

We must add the planet’s radius:

\[ r=R+h \]

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

Now calculate:

\[ g=\frac{GM}{(3R)^2} \]

\[ g=\frac{GM}{9R^2} \]

Therefore:

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

The difference is substantial.

A useful habit is to write \(r=R+h\) before substituting anything into the gravitational equation.

05Gravitational field, force, potential and energy are different

The inverse-square calculation gives us gravitational field strength. But it is easy to confuse this with other gravitational quantities.

They are related, but they do not all decrease at the same rate.

Gravitational field strength

Gravitational field strength describes the gravitational force per unit mass at a location.

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

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

The field points towards the planet’s centre.

Gravitational force

The gravitational force on an object depends on both the field strength and the object’s mass:

\[ F_g=mg \]

Combining this with the field equation:

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

Its units are newtons.

If you double the object’s mass, the gravitational force doubles. The gravitational field strength does not change.

Gravitational potential

Gravitational potential describes gravitational potential energy per unit mass.

Taking zero potential at infinity:

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

Its units are \(\mathrm{J\\,kg^{-1}}\).

Notice that potential depends on \(1/r\), not \(1/r^2\).

At twice the planet’s radius:

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

This is half the surface potential:

\[ \boxed{V_{2R}=\frac{V_R}{2}} \]

Because gravitational potential is negative, this means it becomes less negative as you move further away.

Gravitational potential energy

Gravitational potential energy is the energy associated with an object in a gravitational field.

It depends on the object’s mass:

\[ U=mV \]

Therefore:

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

At twice the planet’s radius:

\[ \boxed{U_{2R}=\frac{U_R}{2}} \]

This comparison assumes the same object mass and the same zero-energy reference at infinity.

Put the four quantities side by side

For the same test mass, compare the values at the surface and at \(2R\).

The important pattern is:

• Field strength and force follow the inverse-square law.
• Potential and potential energy follow an inverse-first-power relationship.

Don’t apply the inverse-square rule to every gravitational quantity. That’s a common conceptual error.

Why is gravitational potential negative?

We conventionally choose gravitational potential to be zero at infinity.

As an object moves towards a planet, gravity attracts it. Work must be done against gravity to move the object back to infinity.

Consequently, the gravitational potential at any finite distance is negative under this convention.

At \(2R\), the potential is less negative than at the surface, but it is not zero.

You can explore the distinction between a zero potential and a non-zero gravitational field in Can Gravitational Potential Be Zero When Gravity Isn’t?.

What happens to gravitational potential energy when an object moves to \(2R\)?

This is another useful distinction.

At the surface:

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

At twice the radius:

\[ U_{2R}=-\frac{GMm}{2R} \]

The change in gravitational potential energy is:

\[ \Delta U=U_{2R}-U_R \]

Substituting:

\[ \Delta U= -\frac{GMm}{2R} +\frac{GMm}{R} \]

Therefore:

\[ \boxed{\Delta U=\frac{GMm}{2R}} \]

The change is positive.

An object moved slowly from the surface to \(2R\) gains gravitational potential energy. An external agent must supply energy, while gravity does negative work.

Notice that the potential energy at \(2R\) is still negative. It is the change in potential energy that is positive.

06Worked examples

07A reliable method for HSC questions

When you see a question involving gravity at twice a planet’s radius, work through these decisions in order.

1. Identify the distance. Is the question giving a distance from the centre or an altitude above the surface?
2. Convert the distance if necessary. Use \(r=R+h\).
3. Identify the requested quantity. Is it field strength, force, potential, or potential energy?
4. Choose the correct equation. Check whether the relationship is inverse-square or inverse-first-power.
5. Check the units and direction. Field strength is measured in \(\mathrm{N\\,kg^{-1}}\), force in N, potential in \(\mathrm{J\\,kg^{-1}}\), and potential energy in J.
6. Check the result. Moving further from the planet should make the field weaker and the potential less negative.

For the standard calculation, assume the planet is spherical, the object is outside the planet, and other gravitational influences are negligible. The planet is treated as a fixed central mass. Real planets are not perfectly spherical, and rotation and nearby masses can produce small variations, but those effects are normally excluded unless a question supplies additional information.

## 8. Questions and solutions

### Question 1

A planet has a radius of \(5.0\times10^6\ \mathrm{m}\) and a surface gravitational field strength of \(9.6\ \mathrm{N\,kg^{-1}}\).

A \(5.0\ \mathrm{kg}\) object is positioned at a distance of twice the planet's radius from its centre.

Calculate:

(a) the gravitational field strength at the object's position;

(b) the gravitational force acting on the object.

[IMAGE: Spherical planet with radius R and a 5 kg test object positioned at radial distance 2R from its centre. Label the planet radius and centre-to-object distance. Do not include force arrows or calculated values.]

### Solution 1

The gravitational field strength is \(2.4\ \mathrm{N\,kg^{-1}}\), and the gravitational force is \(12\ \mathrm{N}\) towards the planet's centre.

**Part (a)**

Use the inverse-square relationship:

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

Substitute:

\[
g_{2R}=\frac{9.6}{4}
\]

\[
\boxed{g_{2R}=2.4\ \mathrm{N\,kg^{-1}}}
\]

**Part (b)**

Use:

\[
F_g=mg
\]

Substitute:

\[
F_g=5.0(2.4)
\]

\[
\boxed{F_g=12\ \mathrm{N}}
\]

The force acts towards the planet's centre because gravity is attractive.

Notice that the field strength was calculated without using the object's mass. The mass was only needed when calculating the gravitational force.

### Question 2

A planet has a surface gravitational field strength of \(10.8\ \mathrm{N\,kg^{-1}}\).

A \(250\ \mathrm{kg}\) probe is positioned at an altitude of twice the planet's radius above its surface.

Calculate:

(a) the probe's distance from the planet's centre in terms of \(R\);

(b) the gravitational field strength at the probe;

(c) the gravitational force acting on the probe.

Explain how your answer would differ if the probe were instead positioned at a radial distance of \(2R\).

[IMAGE: Spherical planet showing its surface radius R and a probe positioned at altitude h=2R above the surface. Include a dimension line from the centre to the probe, but leave its final numerical expression unlabelled.]

### Solution 2

The probe is \(3R\) from the planet's centre, experiences a field strength of \(1.20\ \mathrm{N\,kg^{-1}}\), and experiences a gravitational force of \(300\ \mathrm{N}\) towards the planet.

**Part (a)**

The distance from the centre is the planet's radius plus the altitude:

\[
r=R+h
\]

Since \(h=2R\):

\[
r=R+2R
\]

\[
\boxed{r=3R}
\]

**Part (b)**

Use the inverse-square relationship:

\[
g=\frac{g_s}{(r/R)^2}
\]

Substituting \(r=3R\):

\[
g=\frac{10.8}{3^2}
\]

\[
g=\frac{10.8}{9}
\]

\[
\boxed{g=1.20\ \mathrm{N\,kg^{-1}}}
\]

**Part (c)**

The gravitational force is:

\[
F_g=mg
\]

\[
F_g=250(1.20)
\]

\[
\boxed{F_g=300\ \mathrm{N}}
\]

The force acts towards the planet's centre.

**Comparison with a radial distance of \(2R\)**

If the probe were at \(2R\) from the centre instead:

\[
g=\frac{10.8}{4}
=2.70\ \mathrm{N\,kg^{-1}}
\]

The force would be:

\[
F_g=250(2.70)
\]

\[
F_g=675\ \mathrm{N}
\]

The two positions produce very different answers.

The tempting mistake is to substitute the altitude \(2R\) directly into the equation. That incorrectly treats the planet's surface as the centre of the gravitational field.

Always measure \(r\) from the centre of the planet.

### Question 3

A spherical planet has a mass of \(4.00\times10^{24}\ \mathrm{kg}\) and a radius of \(5.00\times10^6\ \mathrm{m}\).

A \(2.00\ \mathrm{kg}\) probe is moved slowly from the planet's surface to a distance of \(2R\) from its centre.

Assume gravitational potential energy is zero at infinity.

Calculate:

(a) the gravitational potential at the surface and at \(2R\);

(b) the gravitational potential energy of the probe at both positions;

(c) the change in gravitational potential energy;

(d) the work done by gravity during the transfer.

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

[IMAGE: Spherical planet with a 2 kg probe shown at the surface and at radial distance 2R. Include a radial displacement path between the two positions, with no energy arrows or calculated values.]

### Solution 3

The gravitational potential at \(2R\) is \(-2.67\times10^7\ \mathrm{J\,kg^{-1}}\), and the probe gains \(5.34\times10^7\ \mathrm{J}\) of gravitational potential energy during the transfer. Gravity does \(-5.34\times10^7\ \mathrm{J}\) of work.

**Part (a)**

Use the gravitational potential equation:

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

At the surface, \(r=R=5.00\times10^6\ \mathrm{m}\).

\[
V_R=
-\frac{
(6.67\times10^{-11})(4.00\times10^{24})
}{
5.00\times10^6
}
\]

\[
\boxed{V_R=-5.34\times10^7\ \mathrm{J\,kg^{-1}}}
\]

At twice the radius:

\[
r=2R=1.00\times10^7\ \mathrm{m}
\]

Therefore:

\[
V_{2R}=
-\frac{
(6.67\times10^{-11})(4.00\times10^{24})
}{
1.00\times10^7
}
\]

\[
\boxed{V_{2R}=-2.67\times10^7\ \mathrm{J\,kg^{-1}}}
\]

The potential is half as negative at twice the radius.

**Part (b)**

Use:

\[
U=mV
\]

At the surface:

\[
U_R=2.00(-5.34\times10^7)
\]

\[
\boxed{U_R=-1.07\times10^8\ \mathrm{J}}
\]

At \(2R\):

\[
U_{2R}=2.00(-2.67\times10^7)
\]

\[
\boxed{U_{2R}=-5.34\times10^7\ \mathrm{J}}
\]

**Part (c)**

The change in gravitational potential energy is:

\[
\Delta U=U_f-U_i
\]

Substitute:

\[
\Delta U=
(-5.34\times10^7)
-
(-1.07\times10^8)
\]

\[
\boxed{\Delta U=+5.34\times10^7\ \mathrm{J}}
\]

The probe gains gravitational potential energy as it moves away from the planet.

**Part (d)**

Gravity does the opposite work to the change in gravitational potential energy:

\[
W_g=-\Delta U
\]

Therefore:

\[
\boxed{W_g=-5.34\times10^7\ \mathrm{J}}
\]

The negative sign makes physical sense. Gravity acts towards the planet, while the probe is being moved away from it.

Because the probe is moved slowly with no change in kinetic energy, the external work is positive and equal in magnitude to the energy gained.

A common mistake is to report the potential energy at \(2R\) as positive. Remember that moving away from the planet increases gravitational potential energy, but its value remains negative when zero is defined at infinity.

### Question 4

A spherical planet has a gravitational field strength of \(2.4\ \mathrm{N\,kg^{-1}}\) at a radial distance of \(2R\).

The gravitational potential at the same location is \(-18\ \mathrm{MJ\,kg^{-1}}\), with zero potential defined at infinity.

Calculate:

(a) the gravitational field strength at the planet's surface;

(b) the gravitational potential at the surface;

(c) the gravitational field strength at \(4R\);

(d) the gravitational potential at \(4R\).

A student argues that both gravitational field strength and gravitational potential must decrease by a factor of four when the distance doubles. Evaluate this statement.

[IMAGE: Spherical planet with three marked radial positions R, 2R and 4R. Show the surface, the intermediate measurement position and the outer measurement position along one radial line. Include no field arrows or potential values.]

### Solution 4

The surface field strength is \(9.6\ \mathrm{N\,kg^{-1}}\), the surface potential is \(-36\ \mathrm{MJ\,kg^{-1}}\), the field at \(4R\) is \(0.60\ \mathrm{N\,kg^{-1}}\), and the potential at \(4R\) is \(-9.0\ \mathrm{MJ\,kg^{-1}}\).

The student's statement is incorrect because gravitational field strength and potential follow different distance relationships.

**Part (a)**

The field strength at \(2R\) is one-quarter of the surface value:

\[
g_{2R}=\frac{g_s}{4}
\]

Rearrange:

\[
g_s=4g_{2R}
\]

Substitute:

\[
g_s=4(2.4)
\]

\[
\boxed{g_s=9.6\ \mathrm{N\,kg^{-1}}}
\]

**Part (b)**

Gravitational potential follows an inverse-first-power relationship:

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

At \(2R\), its value is half the surface potential.

Therefore:

\[
V_{2R}=\frac{V_R}{2}
\]

Rearranging:

\[
V_R=2V_{2R}
\]

\[
V_R=2(-18)
\]

\[
\boxed{V_R=-36\ \mathrm{MJ\,kg^{-1}}}
\]

**Part (c)**

Compare \(2R\) and \(4R\):

\[
\frac{g_{4R}}{g_{2R}}
=
\left(\frac{2R}{4R}\right)^2
\]

\[
\frac{g_{4R}}{2.4}
=
\frac14
\]

Therefore:

\[
\boxed{g_{4R}=0.60\ \mathrm{N\,kg^{-1}}}
\]

**Part (d)**

For gravitational potential:

\[
\frac{V_{4R}}{V_{2R}}
=
\frac{2R}{4R}
\]

\[
V_{4R}=\frac{-18}{2}
\]

\[
\boxed{V_{4R}=-9.0\ \mathrm{MJ\,kg^{-1}}}
\]

**Evaluate the student's statement.**

The student has applied the inverse-square relationship to both quantities.

However:

\[
g\propto\frac{1}{r^2}
\]

while:

\[
V\propto-\frac{1}{r}
\]

Doubling the distance quarters the field strength but only halves the magnitude of the gravitational potential.

The negative sign also matters. The potential changes from \(-18\) to \(-9\ \mathrm{MJ\,kg^{-1}}\), meaning it becomes less negative.

### Question 5

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

Two small probes, A and B, have masses \(m\) and \(3m\), respectively. Both are initially stationary at a radial distance of \(2R\) from the planet's centre.

The probes are released simultaneously and fall radially towards the planet's surface.

Assume:

- The masses of the probes are negligible compared with the planet's mass.
- The planet does not move appreciably.
- There is no atmospheric resistance or other external force.
- Both probes begin with zero kinetic energy.
- Gravitational potential is defined to be zero at \(2R\), rather than at infinity.

Answer the following:

(a) Determine the gravitational field strength at the release position and the gravitational force acting on each probe.

(b) Determine the gravitational potential at the planet's surface using the chosen reference level.

(c) Calculate the change in gravitational potential energy of each probe as it reaches the surface.

(d) Determine the final speed of each probe.

(e) A student makes two claims:

- Probe B must fall faster because it experiences three times the gravitational force.
- Gravity must be zero at the release position because gravitational potential is defined as zero there.

Explain why both claims are incorrect.

[IMAGE: Scientific diagram of a spherical planet with two probes, A of mass m and B of mass 3m, initially positioned at the same radial distance 2R from the centre. Show the planet radius R, the common starting distance 2R and the surface. Include the radial paths towards the surface but no force arrows, acceleration arrows, velocity arrows or calculated results.]

### Solution 5

Both probes reach the surface with the same final speed:

\[
\boxed{v_A=v_B=\sqrt{\frac{GM}{R}}}
\]

However, probe B experiences three times the gravitational force and undergoes three times the change in gravitational potential energy.

The reason is that gravitational field strength is independent of the probe's mass, while gravitational force and potential energy depend on it. Also, choosing a zero potential at \(2R\) does not remove the gravitational field.

**Part (a): Field strength and force**

Both probes begin at:

\[
r=2R
\]

The gravitational field strength is:

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

Therefore:

\[
\boxed{g=\frac{GM}{4R^2}}
\]

The field points towards the planet's centre.

For probe A:

\[
F_A=mg
\]

\[
\boxed{F_A=\frac{GMm}{4R^2}}
\]

For probe B:

\[
F_B=3mg
\]

\[
\boxed{F_B=\frac{3GMm}{4R^2}}
\]

Therefore:

\[
F_B=3F_A
\]

Probe B experiences three times the gravitational force.

However, it also has three times the mass.

Using Newton's second law:

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

Both probes have the same initial acceleration:

\[
\boxed{a_A=a_B=\frac{GM}{4R^2}}
\]

This is the first important result.

**Part (b): Potential at the surface**

We have changed the reference level for gravitational potential.

Normally, with zero potential at infinity:

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

But we now want the potential to be zero at \(r=2R\).

To achieve this, subtract the potential at \(2R\) from the original potential:

\[
V'(r)=V_\infty(r)-V_\infty(2R)
\]

Therefore:

\[
V'(r)=-\frac{GM}{r}+\frac{GM}{2R}
\]

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

\[
V'(R)=-\frac{GM}{R}+\frac{GM}{2R}
\]

Hence:

\[
\boxed{V'(R)=-\frac{GM}{2R}}
\]

The potential at the surface is negative relative to our newly chosen zero.

**Part (c): Change in gravitational potential energy**

For probe A:

\[
\Delta U_A=m\Delta V
\]

The initial potential is zero, and the final potential is \(-GM/(2R)\).

Therefore:

\[
\boxed{\Delta U_A=-\frac{GMm}{2R}}
\]

For probe B:

\[
\Delta U_B=3m\Delta V
\]

Therefore:

\[
\boxed{\Delta U_B=-\frac{3GMm}{2R}}
\]

Both probes lose gravitational potential energy as they fall.

Probe B loses three times as much because its mass is three times larger.

**Part (d): Final speeds**

Assume mechanical energy is conserved.

For probe A, the loss in potential energy becomes kinetic energy:

\[
\frac12 mv_A^2=\frac{GMm}{2R}
\]

Cancel \(m\):

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

Therefore:

\[
\boxed{v_A=\sqrt{\frac{GM}{R}}}
\]

For probe B:

\[
\frac12(3m)v_B^2
=
\frac{3GMm}{2R}
\]

Cancel the common factor \(3m/2\):

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

Therefore:

\[
\boxed{v_B=\sqrt{\frac{GM}{R}}}
\]

The two probes reach the surface with the same speed.

This is consistent with their equal accelerations at every position along their shared radial path.

**Part (e): Evaluate the two claims**

The first claim is incorrect because a larger gravitational force does not automatically mean a larger acceleration.

Newton's second law gives:

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

Probe B has three times the force but also three times the mass. These effects cancel.

The second claim is incorrect because gravitational potential and gravitational field strength are different quantities.

Setting the potential to zero at \(2R\) simply changes the reference level used to measure potential energy.

It does not change the gravitational attraction between the planet and either probe.

At the release position:

\[
g=\frac{GM}{4R^2}
\]

which is clearly non-zero.

The potential is zero there only because we selected that location as our reference.

[IMAGE: Two gravitational potential versus radial distance curves for the same spherical planet. Show the conventional curve V=-GM/r with zero at infinity and a vertically shifted curve V'=GM/(2R)-GM/r with zero at 2R. Mark r=R and r=2R. Show that both curves have the same shape and rate of change despite having different zero levels.]

This final problem brings together three ideas that are easy to confuse:

- Changing the reference level changes the numerical value of gravitational potential and potential energy, but not their differences.
- Increasing an object's mass increases its gravitational force and potential-energy change.
- Increasing an object's mass does not change its gravitational acceleration or its final speed when released from the same position under the same conditions.

The distinction between field strength, force, potential and energy is much more important than simply remembering that gravity decreases with distance.

Once you're comfortable with these relationships, the next useful step is applying gravitational force to circular motion. In particular, you can investigate how gravity supplies the centripetal force that keeps a satellite in orbit in [Gravity as Centripetal Force in Circular Orbits for HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-gravity-as-centripetal-force-in-an-orbit/).